Physical Pharmaceutics 1 Lecture Notes PDF

COURSE TITTLE: PHYSICAL PHARMACEUTICS 1 COURSE CODE: PCT 301 OUTLINE  Thermodynamics, Thermochemistry & chemical equilibrium, Chemical Kinetics. THERMODYNAMICS. Thermodynamics is the study of the laws that govern the conversion of energy from one form to anot...

COURSE TITTLE: PHYSICAL PHARMACEUTICS 1 COURSE CODE: PCT 301 OUTLINE  Thermodynamics, Thermochemistry & chemical equilibrium, Chemical Kinetics. THERMODYNAMICS. Thermodynamics is the study of the laws that govern the conversion of energy from one form to another, the direction in which heat will flow, and the availability of energy to do work. It is based on the concept that in an isolated system anywhere in the universe there is a measurable quantity of energy called the internal energy (E) of the system. Thus, thermodynamics is concerned with the quantitative relationships between heat and other forms of energy, comprising mechanical, chemical, electrical and radiant energy. The kinetic energy of a body is due to its molecules, atoms and electrons and possess potential energy by virtue of the configuration of its part or its position. The value of the total energy of a system may not be known but the changes in energy when a system undergoes some transformation can be noted. Thermodynamics is based on three ‘laws’ or facts of experience that have never been proven in a direct way, in part due to the ideal conditions for which they were derived. Various conclusions, usually expressed in the form of mathematical equations, however, may be deduced from these 3 principles, and the results consistently agree with observations. Consequently, the laws of thermodynamics, from which these equations are obtained are accepted as valid for systems involving large numbers of molecules. BASIC DEFINITIONS A System: A system in thermodynamics is a well-defined part of the universe that one is interested in studying. The system is separated from surroundings, the rest of the universe and from which the observations are made, by physical barriers defined as boundaries. Work (W) and heat (Q) also have precise thermodynamic meanings, Work is a transfer of energy that can be used to change the height of a weight somewhere in the surroundings and Heat is a transfer of energy resulting from a temperature difference between the system and the surroundings.  It is important to consider that both Work and Heat appear only at the system’s boundaries where the energy is being transferred.  THE FIRST LAW OF THERMODYNAMICS. The first law of thermodynamics is all about the conservation of energy. It states that energy can neither be created nor destroyed but can only be transformed from one kind into another. Thus, the total energy of a system and its surroundings is a constant irrespective of environmental changes. In other words, the various forms of energy are equivalent and when one kind of energy is formed, an equal amount of another kind must be displaced. The value of E can only be changed if the system ceases to be isolated. In this case E can change by the transfer of mass to or from the system, the transfer of heat (Q) to or from the system or by the work (W) being done on or by the system. According to the first law for work done by the system ∆E = Q – W ---------------------------------------- 1 Where ∆E = increase in internal energy Q = heat absorbed W = work done by the system. For work to be done by the system i. Input of heat is necessary ii. Work done on the system is accompanied by the evolution of heat. iii. The internal energy results from the motion of the molecules, electrons and nuclei in a system and it depends on the measurable properties: pressure (P), volume (V) and temperature (T). iv. Any two of these variables must be specified in order to define the internal energy. For an infinitesimal increase in the energy dE, equation 1 becomes dE = q – w ---------------------------------------------------- 2 where q = the heat absorbed w = the work done during the very small change of the system. Thermodynamic is concerned with the changes of internal energy and not the determination of absolute value. The finite change of internal energy is given by ∆E = E2 – E1 ----------------------------------------------------- 3 Where E2 = energy of the system in its final state E1 = energy of the system in its initial state. For instance, 750 g of water at 1 atm and 100C at E2 and 750 g at 5 atm and 1500C as E1.  OPEN AND CLOSED SYSTEMS A system is defined as a set of units/parts or entities which interact with one another to perpetuate the functional existence of the whole. An open system, involves a transfer of matter in addition to the exchange of heat and work. A closed system I one which may exchange heat and work but not matter with its surroundings. Consider two immiscible solvents, water and kerosene confined in closed container and potassium permanganate is distributed between the two phases. Each phase is an open system, yet the total system made up of the two phases is closed as it does not exchange matter with its surrounding. ISOTHERMAL AND ADIABATIC PROCESSES. Any process that takes place at constant temperature, the reaction is said to be conducted isothermally. An isothermal reaction may be conducted by placing the system in a large constant temperature bath so that heat is removed from or supplied at just the right rate to maintain constant temperature. When heat is neither leaving nor entering a system during a process, the reaction is said to occur adiabatically. A reaction carried inside a sealed vacuum bottle is adiabatic since the system is thermally insulated from its surroundings. In thermodynamic terms, an adiabatic process is one in which q = 0, and the first law under adiabatic conditions reduces to W = -dE ---------------------------------------------------------- 4 When work is done by the system the internal energy decreases, according to equation 4 since heat cannot be absorbed in an adiabatic process, the temperature must fall. Thus, the work done becomes a thermodynamic property dependent on the initial and final states of the system. REVERSIBLE AND IRREVERSIBLE PROCESSES Reversible process is any process in which the variable that define the state of the system can be made to change in such a way that they pass through the same values in the reverse order when the process is reversed. It is also a condition of a reversible process that any exchanges of energy, work or matter with the surroundings should be reversed in direction and order when the process is reversed. Any process that does not follow these conditions when it is reversed is said to be irreversible process. All natural processes are irreversible, although some processes can be made to approach close to a reversible process using an external agent. Gases expand naturally from higher to lower pressures and solute molecules diffuse from a region of higher to lower concentration. These spontaneous processes will not proceed in reverse without the intervention of some external agency. Although spontaneous processes are not thermodynamically reversible, they can be carried out in nearly reversible manner by an outside agency. Maximum work is obtained by conducting a spontaneous process reversibly, however the frictional losses and the necessity of carrying out the process at an infinitely slow rate preclude the possibility of complete reversibility in real processes.  REVERSIBLE PROCESS AND MAXIMUM WORK The work done by a system in an isothermal process is at a maximum when it is done reversibly. This is so, as no work is accomplished if an ideal gas expands freely into a vacuum, where P = 0, since any work accomplished depends on the external pressure. As the external pressure becomes greater, more work is done by the system, and it rise to a maximum when the external pressure is infinitesimally less than the pressure of the gas which occurs when the process is reversible. Thus, if the external pressure is continually increased, the gas is compressed and work is done on the system rather than by the system in an isothermal reversible process. IDEAL GAS EQUATION. Four quantities are important in all experimental work measurements or calculations involving gases. They are: i. Volume ii. Pressure iii. Temperature and iv. Number of moles or mass. Three of these parameters appear in the general gas equation which is obtained from Boyle’s, Gay-Lussac and Charles laws. P1V1/T1 = P2V2/T2 ---------------------------------------------------- 5 Where p1 = initial pressure V1 = initial volume T1 = Kelvin Temperature V2 = new volume T2 = new Kelvin Temperature P2 = new pressure This equation states that for an ideal gas PV/T is a constant. Experimentally, it has been found that equal volume of all gases at the same temperature and pressure contain the same number of molecules, and that one mole of any gas at standard temperature and pressure (S.T.P) occupies a volume of 22.4 dm 3. This fact, together with Boyle’s and Charles’ laws lead to the equation PV = RT ------------------------------------------------------------------ 6 Where R = a constant termed as the molar gas constant. The equation is called the ideal gas equation or the equation of state of an ideal gas and holds true for all gases. For n moles of gas, we get PV = nRT ----------------------------------------------------------------- 7 Where P I in atm, V in dm3, T in K. At S.T.P for 1 mole of gas R = PV/nR = (1atm) (22.4 dm3)/(1 mole)(273K) = 0.83 atm dm3K-1mole or 0.08 Litre/atm mole deg. The molar gas constant may also be given in energy units by expressing the pressure in dyne/cm2 (1 atm = 1.0133 x 106 dyne/cm2) and the volume in the corresponding units of cm3 (22.414 Litres = 22,414 cm3) then R = PV/T = 1.0133 x 106 x 22,414/273.16 = 8.314 x 107 erg/mole deg or Since 1 joule = 107 erg R = 8.314 joule/mole deg. The constant can also be expressed in cal/mole deg employing the equivalent 1 cal = 4.184 joules R = 8.314/mole deg/4.184 joule/cal = 1.987 cal/mole deg. The value of R depends on the unit under consideration in each problem. In gas law problem R is usually expressed in litre atm/mole deg. Also at S.T.P, the standard temperature is 00C which is the same as 273K. the standard pressure is one atmosphere expressed as 760 mmHg or 1013278.85 dyne/cm 2 (1.0133 x 106 dyne/cm2) i.e. 76 x 13.595 x 980 7 = 1013278.85 = 1.0133 x 106. Where P = 760 mmHg = 76 cm g = 980.7 cm/sec2 = 9.1 m/sec2 and density of mercury at 00C = 1.595 cm/cm3. WORK OF EXPANSION AGAINST A CONSTANT PRESSURE. Consider work done resulting from an expansion or compression of a gas against a constant opposing pressure, P of a vapor confined in a cylinder fitted with a weightless, frictionless piston of area A. if a constant external pressure P is exerted on the piston, the total force is P x A, since P = force/area. The vapor in the cylinder is now made to expand by increasing the temperature and the piston moves a distance h. The work done against the opposing pressure is W = P x A x h --------------------------------------------------------------- 8 Now A x h is the increase in volume, W = P ∆V -------------------------------------------------------------------- 9 W = P (V2 – V1) ------------------------------------------------------------ 10 CALCULATION INVOLVING THE APPLICATION OF FIRST LAW. Example 1: A gas expands to 1 litre when a constant pressure of 1 atm is applied at 25 0C. compute the work done in joules by the system. Answer: The formula to use is; W = P ∆V 1 atm = 1.013 x 106 dyne/cm2 W = (1.013 x 106 dyne/cm2) x 1000 cm3 = 1.013 x 109 ergs. But 1 joule = 107 erg i.e. 1.013 x 109 erg = 101.3 joules. Example 2: One mole of water is converted by boiling into steam at 100 0C at 1 atm. The heat of vaporization of water at 1000C is about 7720 cal/mole. In this case compute the value of Q, W, and ∆E. Answer: From the question the heat of vaporization is the amount of heat absorbed and given as 7720 cal/mole. Therefore, Q = 7720 cal/mole. The work W performed against the constant atmospheric pressure is obtained by using equation 10. W = P (V2 – V1). Now V1, is the volume of 1 mole of liquid water at 1000C or about 0.018 litre. The volume V2 of 1 mole of steam at 1000C and 1 atm is given by the gas law, assuming the vapour behaves ideally: V2 = RT/P = 8.314 x 373/1 = 3101.133 litres. The work done is obtained thus; W = P (V2 – V1) = 1 x (3101.12 – 0.018) = 3101.10 litres. 1 cal = 4.184 joule So, 3101.10/4.184 = 741.18 cal/mole. The internal energy change ∆E = Q – W = 7720 – 741 = 6979 cal/mole. From the data above, the heat absorbed by 1 mole of water, during the work done was 7720 cal/mole, 741 cal was the work done by the system against an external pressure of 1 atm. So 6979 cal is the internal energy of the system. This quantity of heat supplies potential energy to the vapor molecules for the work done against the intermolecular forces of attraction. Internal energy is the total of potential and kinetic which include rotational, vibrational, translational energy of atoms and the electrons which constitutes the molecules. IDEAL GAS AND THE FIRST LAW OF THERMODYNAMICS. An ideal gas consists of molecules that occupy negligible space and have negligible forces between them. An ideal gas has no internal pressure, and hence no work need be done to separate the molecules against their cohesive forces when the gas expands. Therefore, for an ideal gas, the first law (∆E = Q – w) is reduced (dE/dV)T ----------------------------------------------------------- 11 The internal energy of an ideal gas is a function of the temperature only, from equation 11, it follows that for an ideal gas involved in an isothermal process (dT = 0) and (dE = 0) and the first law reduces to q = w ----------------------------------------------------------------- 12 Hence, the work done in the isothermal expansion of an ideal gas is equal to the heat absorbed or evolved by the gas. THE SECOND LAW OF THERMODYNAMICS. The first law of thermodynamics only deliberates on the conservation of energy. It however, does not elaborate on the rate processes or the usefulness of heat in doing work. The second law holds that there is an inherent direction in which any system not at equilibrium moves. This brings about the concept of a. Free energy of a reaction b. Spontaneity of a reaction and c. Entropy change. Free energy is the energy at the course of a chemical reaction that is available for work. Spontaneity of a reaction is the natural course that is taken by a reaction before it occurs. Entropy on the other hand is the state of disorderliness of a system. The second law of thermodynamics states that a spontaneous process occurs only if there is an increase in the entropy of a system and its surrounding. Mathematically put ∆S = ∆H/T Where ∆S = entropy change ∆H = heat change and T = absolute temperature All natural processes conform to the second law of thermodynamics, but not all processes conforming to it can occur in nature. Most natural processes are irreversible i.e. they only proceed in one direction. The direction that a natural process can take is the subject of the second law of thermodynamics. The second law can be stated in 2 ways: 1. “Heat cannot be transferred from one body to a second body at a higher temperature without producing some other effect”. In other words, heat flows spontaneously only from hotter to colder bodies. 2. “The entropy of a closed system increases with time”. These statements introduce the thermodynamics concepts of temperature (T) and entropy (S) both of which are parameters determining the directions in which an irreversible process can go. The temperature of a body or system determines whether heat will flow into it or out of it. Its entropy is a measurement of the unavailability of its energy to do work. Thus. T and S determine the relationship between Q and W in the statement of the first law of thermodynamics. Thus, the second law can be presented in the form ∆E = T ∆S – W ------------------------------------------------------------ 1 Which signified that the second law is concerned with the changes in entropy (∆S). The second law considers the probability of a process occurring. No useful work can be obtained from heat at constant temperature. The first law simply makes the observation that energy must be conserved when it is converted from one form to another. It has nothing to say about the probability of the occurrence of a process, based on the observed tendency of a system to approach a state of energy equilibrium. The energy which may be freed for useful work in a gas, liquid or solid or any reaction mixture is known as the free energy of the system. The free energy decreases as a physical or chemical reaction proceeds. For example, solute molecule diffuses from a region of higher concentration to lower concentration. Usually, spontaneous processes at  constant temperature and pressure are accompanied by a loss in free energy, and this decrease signified the natural tendency for the transformation to occur. Thus, the statement of second law of thermodynamics constitutes the spontaneous character of natural processes and the limitations on the conversion of heat into work. Also when a substance melts, it passes from a condition of low heat content and a highly ordered arrangement to a condition of higher heat content and more randomness. This change from orderliness to randomness to the physical chemist represents an increase in the entropy of the system.  The statement of the second law of thermodynamics means that the probability of the occurrence of a process is based on the observed tendency of the system to move toward the phase of increase entropy. FACTORS AFFECTING SPONTANEITY OF CHEMICAL REACTIONS. The factors are Enthalpy, Entropy and free energy. i. Enthalpy (H): This is the heat content of a system and is denoted as H. Enthalpy change is denoted by ∆H. ii. Entropy (S): It is a measure of the degree of disorder or randomness of a system. The greater the degree of disorder of a system the higher will be the entropy of the system. When ∆S = 0, it means energy is conserved. iii. Gibb’s free energy (G): Free energy is a measure of the energy available to do work. It is the driving force that brings about a chemical change. Thus, the relationship is mathematically given as ∆G = ∆H - T∆S Where ∆G = free energy change ∆H = enthalpy change T = absolute temperature and ∆S = entropy change. THE EFFICIENCY OF A HEAT ENGINE. Here, we are concerned with the possibility of converting heat to work. Heat is isothermally unavailable for conversion completely to work, as mentioned in the second part of the law of thermodynamics emphasizing the limitation on the conversion of heat to work. A heat engine (steam engine) can do useful work in using two heat reservoirs a “source” and a “sink”, at two different temperatures. However, only part of the heat at the source is converted into work, the remainder being returned to the sink (the surroundings) at the lower temperature. The fraction of the heat Q at the source converted into work W is known as the efficiency of the engine: Efficiency = W/Q ---------------------------------------------------- 2 According to the second law of thermodynamics, the efficiency of the heat engine operating without friction cannot be unity, for W is always less than Q in a continuous conversion of heat into work. A steam engine operating reversibly between an upper temperature T 2 and lower temperature T1, absorbs heat Q2 from the hot boiler or source, and by means of,  steam, converts the quantity W into work, and returns heat Q 1 to the cold reservoir or sink. For this type of engine, the efficiency of operating reversibly at every stage and returning to the initial state (cyclical process), can be given by the expression: W/Q2 = Q2 – Q1/Q2 --------------------------------------------------- 3 The heat flow in the operation of the engine follows the temperature gradient and the heat absorbed and rejected is directly related to the temperatures. Q2/Q1 = T2/T1 ---------------------------------------------------------- 4 When equation 2, 3, and 4 are combined therefore, Efficiency = Q2 – Q1/Q2 = T2 – T1/T2 ----------------------------- 5 Thus, equation 5 shows that the higher T2 becomes the lower T1 becomes, the greater is the efficiency of the engine. When T1 reaches absolute zero on the Kelvin scale, the reversible heat engine converts heat completely to work and its theoretical efficiency becomes unity. This can be seen by setting T1 = 0 in equation 5. However, since absolute zero is considered to be unattainable, an efficiency of unity is impossible, and heat can never be completely converted to work. This statement can be written using the notation of limits as follows: T1 = 0 W/Q = 1 ------------------------------------------- 6 If T2 = T1 in equation 5, the cycle is isothermal and efficiency is zero, confirming the second law of thermodynamics statement that there is limitation on the conversion of heat into work. Example A steam engine operates between the temperature of 3780K and 3030K. compute the theoretical efficiency of the engine. If the engine is supplied with 800 cal of heat Q 2. What is the theoretical work in joules? Answer The formula to use is T2 – T1/T2 a) Efficiency = W/Q2 = T2 – T1/T2 = 0.198 or 19.8%. b) W = 800 x 0.198 = 158.4 158.4 X 4.184j = 662.5j. ENTROPY Entropy (S) is a measure of the unavailability of a system’s energy to do work; in a closed system, an increase in entropy is accompanied by a decrease in energy availability. When a system undergoes a reversible change, the entropy (S) changes by an amount equal to the energy (Q) transferred to the system by heat divided by the thermodynamic temperature (T) at which this occurs i.e. ∆S = ∆Q/T. Entropy may be defined as the molar energy per degree of absolute temperature which is unavailable for work. However, all real processes are to a certain extent irreversible changes and in any closed system an irreversible change is always accompanied by an increase in entropy. In a wider sense, entropy can be interpreted as a measure of disorder, the higher the entropy the greater the disorder. As any real change to a closed system tends towards higher entropy, and therefore higher disorder, it follows that the entropy of the universe when considered as closed system is increasing and its available energy is decreasing. This increase in entropy of the universe is one way of stating the second law of thermodynamics. CALCULATIONS INVOLVING THE APPLICATION OF ENTROPY. Example Compute the entropy change following the vaporization of 1 mole of water in equilibrium with its vapor at 250C. In this process the heat of vaporization ∆Hv is 8500 cal/mole. Answer The process is carried out at constant pressure so that Q = ∆HP and since it is reversible process the equation to use for entropy change is ∆S = ∆Hv/T = 8500/298 = 28.52 cal/mole deg. THE THIRD LAW OF THERMODYNAMICS. st nd The 1 and 2 laws of thermodynamics have introduced 2 new concepts of energy content and entropy. The 3rd law however, does not lead to a new concept but only places a limitation on the value of entropy of a crystalline substance. The law states that the entropy of a pure crystalline substance is zero at absolute zero because the crystal arrangement must show the greatest orderliness at this temperature. Since it is not possible to attain absolute zero, a better statement of the law is that – the entropy of a perfectly crystalline substance approaches zero as the temperature approaches absolute zero.  The 3rd law cannot be applied to super cooled liquids, because their entropy at 0 0K is probably not zero.  The 3rd law helps in calculating the absolute entropies of pure substances  The absolute entropy of a perfect crystal at any temperature may be determined from a knowledge of the heat capacity, so long as there is no change of phase during the temperature rise.  When the temperature of a substance is raised without causing any change in the state of the substance, the motion of the molecules within the substances is increased resulting in the increase in entropy.  This process may be considered to be composed of a series of steps in which the infinitesimal change of temperature dT increases the entropy by an infinitesimal amount dS and is given by the equation: dS = Cp dT/T ------------------------------------------------------ 1 where Cp = the molar heat capacity. By integrating the equation, we get: ∆S = S2 – S1 = Cp dT2/T1 = Cp ln T2/T1 --------------------------------------- 2 Converting to common logarithm, we get: ∆S = 2.303 Cp log T2/T1 When T1 = 0, the absolute entropy is given by S = 2.303 Cp log T ------------------------------------------------ 3  WORK FUNCTION AND FREE ENERGY.  For systems of constant internal energy and volume, changes occur spontaneously in the direction of increased entropy.  To explain spontaneity for systems with constant temperature and pressure, and for systems with constant temperature and volume, two other thermodynamic functions have been introduced.  They are the Helmholtz free energy or work function (A) and the Gibbs free energy (G) or simply free energy.  The internal energy E of a system is considered to consist of 2 parts, A, consisting of isothermally available internal energy for doing work (or work function) and a second part, TS, - unavailable for doing work. i.e. E = A + TS or A = E – TS.  The driving force of any physical or chemical change resides in the component available for doing work, i.e. part ‘A’. when the system undergoes a change, the equation may be written as: ∆A = ∆E - ∆(TS) At constant temperature, ∆A = ∆E - T∆S Or ∆A = Q – W - T∆S (since ∆E = Q – W).  In a reversible process, maximum work is obtained and from the definition of entropy: ∆S = Qrev/T T∆S = Qrev Substituting this value in the above equation, we obtain ∆A = T∆S – Wmax - T∆S Or ∆A = -Wmax Or - ∆A = Wmax  The decrease of A is a measure of capacity of the system to do work and of the driving force causing the system to undergo spontaneous change in the direction of diminishing its work capacity and function.  For systems at constant temperature and volume, spontaneity of a reaction is indicated when ∆A has a negative value, indicating a loss in the capacity of the system to do work.  If a system at constant temperature and under constant atmospheric pressure increases the volume, the capacity of the system for doing work is diminishing by an amount equivalent to work of expression (P∆A) against the atmospheric pressure. The net work is then given by: Wnet = Wmax - P∆A  GIBBS FREE ENERGY. The total heat content may be divided into – isothermally available energy or free energy (G) and isothermally unavailable energy, (TS). Thus, H = G + TS This on rearrangement becomes: G = H – TS Or G = E + PV – TS (since H = E + PV) When a system undergoes a change, the change of free energy, ∆G is given by: ∆G = ∆E + ∆(PV) - ∆(TS) At constant pressure and temperature: ∆G = ∆E + P∆V - T∆S Substituting Q – W for ∆E and Q for T∆S for a reversible reaction: ∆G = Q –W + P∆V – Q (where Q = Wmax for a reversible process) ∆G = - Wmax + P∆V Multiplying by minus (-), the equation becomes: -∆G = Wmax -P∆V Or -∆G Wnet (Wnet = Wmax - P∆V).  Hence, the decrease in free energy (-∆G) is a measure of maximum net work at constant temperature and pressure available from the system.  It gives a measure both of the net capacity of the system for doing useful work and of the driving force for the spontaneous change in the direction of diminishing its free energy. If the system undergoes no change in volume, Wnet equals Wmax and ∆A and ∆G must be equal. If the system undergoes a decrease in volume, the term ∆V becomes negative and WQnet is larger than Wmax. Read up The Clapeyron equation Clasius- Clapeyron equation. THERMOCHEMISTRY. Thermochemistry is concerned with heat changes involved during isothermal chemical reactions. These processes usually occur at constant atmospheric pressure, and the heat absorbed is equal to the increase in heat content represented by Qp = ∆H. HEAT CONTENT AND HEAT OF REACTION. Generally, a substance possesses a characteristic internal energy which is due to its structure and physical state. This is known as its heat content or enthalpy. Each mole of a substance has a characteristic mass. In a chemical reaction, heat energy is either evolved or absorbed because of the difference of heat content of the substances involved in the reaction. The energy change accompanying a chemical reaction is termed as the heat of reaction. It is a constant for a given reaction carried out under a fixed set of conditions. STANDARD HEAT OF REACTION. It is defined as the amount of heat evolved or absorbed when a chemical reaction occurs between molar quantities of the substances as represented in the equation of reaction under standard conditions. Also where the reaction occurs at constant volume, then Qv = ∆E, in which the P∆V terms are not considered. However, this close approximation does not hold for reactions involving gases. In the reaction with the thermochemical equation given below C(s) + O(g) = CO2 (g) --------------------------------------------------- 1 ∆H0 250C = -94,054 cal. The subscripts in the equation represents the physical states: (s) = solid, (g) = gas, and other symbols (l) = liquid and (aq) = dilute aqueous solution. ∆H0 250C = the standard heat of reaction for the process at 250C. The negative sign accompanying the value for ∆H signifies that heat is evolved; i.e. the reaction is exothermic. Equation 1 states that when one mole of solid carbon (graphite) reacts with one mole of gaseous oxygen to produce one mole of gaseous carbon dioxide at 25 0C, 94,052 cal are liberated. This means that the reactants contain 94,052 cal in excess of the product so that this quantity of heat is evolved during the reaction. When this reaction is reversed and CO2 is converted to carbon and oxygen, the reaction would be endothermic. That means there will be absorption of 94,052 cal of heat and ∆H will also have a positive value. When the pressure is not stated, it implies that the reaction is carried out at 1 atm. Heat of reaction encompasses many different types of reactions like formation, combustion, neutralization etc. HEAT OF FORMATION. The standard heat of formation of carbon dioxide from its elements is given in equation 1. The heat content of 1 mole of carbon dioxide is 94,052 cal less than the heat content of its elements in the standard or reference states of 250C and 1 atm pressure. The heat contents of all elements in their standard states are arbitrarily assigned values of zero. So, the heat involved in the formation of a compound from its elements or the energy liberated or absorbed when one mole of a compound is formed in their standard states from its constituent element is the heat of formation of the compound. The heat of formation of carbon dioxide is -94,052 cal. HEAT OF COMBUSTION The heat of combustion is the heat involved in the complete oxidation of 1 mole of a compound at 1 atm pressure or the energy liberated when one mole of a given substance is completely oxidized. The compound is burned in the presence of oxygen in a sealed calorimeter to convert it completely to carbon dioxide and water. The combustion of methane is given below in the thermochemical equation: CH4(g) + 2O2(g) = CO2(g) + 2H2O(l) ---------------------------------------------------- 2 ∆H0 250C = -212.8 Kcal.  HESS’S LAW OF CONSTANT HEAT OF SUMMATION. This law shows that, since ∆H depends only on the initial and final states of a system, thermochemical equations for several steps in a reaction can be added and subtracted to obtain the heat of the overall reaction. This principle is known as Hess’s law of constant heat of summation and is used to obtain thermodynamic data like the heats of reactions that cannot be measured directly or by simple method. It is a consequence of the law of conservation of energy. Hess’s law of constant heat of summation states that the total enthalpy of a chemical reaction is always constant regardless of the route by which the chemical change occurs provided that initial and final conditions are the same. This is concerned with obtaining the heat of the overall heat of reaction from heats of formation of reactants and products and so Hess’s law provides that: ∆Hreaction = ∑∆Hproducts - ∑∆Hreactants --------------------------------------------- 3 And is another means of getting some results as obtained by the use of Hess’s law. According to the National Bureau of standards data, the heat of formation of CH4(g) = -17.889 CO2(g) = -94.052 H2O(l) = -68.317 Kcal/mole at 250. For elements = zero and so for oxygen O2(g) = 0. Employing the thermochemical equation 2 i.e. CH4(g) + 2O2(g) = CO2(g) + 2 H2O(l) ∆H0 250C = -212.8 Kcal Then substituting the values in Hess’s law equation 3 i.e. ∆Hreaction = ∑∆Hproducts - ∑∆Hreactants = ∑(-230.686 + 17.889) = -212.797. HEAT OF REACTION CHANGES IN SOLUTION. a. Heats of Neutralization Sodium hydroxide and hydrochloric acid are strong electrolytes and when dissolved in water dissociate in ions of Na+ and Cl-. In solution the ions gather about them in sphere of water molecules, called the hydration sphere. The positive ions attract the negatively charged (oxygen) ends of water molecules. In very dilute solution, the ions are separated by many layers of water molecules and each one behaves as if the other ions do not exist. This usually happens at infinite dilution and so warrants putting the latter’s aq in brackets after the symbol, OH(aq) or H+(aq). Actually in mixing sodium hydroxide and hydrochloric acid, caution should be exercised. Never attempt to mix concentrated solutions of acids and alkalis as so much heat can be liberated that the mixture can be showered over one. So the equation for the reaction is NaOH(aq) + HCl(aq) NaCl(aq) + H2O(l) However, the reaction is really between the hydroxide ions and the hydrogen ions and is the true source of the heat. H+(aq) + OH- H2O(l) The enthalpy changes when one mole of hydrogen ions is completely neutralized under standard conditions is called the standard heat of neutralization. ∆Hnꙫ. The value is -57.9 Kj mol-1 i.e. H+(aq) + OH-(aq) H2O(l); ∆Hnꙫ = -57.9 Kj mol-1 Definition The standard heat of neutralization is the enthalpy change that takes place when one mole of hydrogen ion from an acid is completely neutralized by an alkali to give an infinitely dilute solution. NOTE – The definition does not talk about one mole of an acid as sulphoric acid H2SO4 can give more than one mole of hydrogen ions. Acids and alkalis that are strong electrolytes have almost the same value for their heat of neutralization as reaction taking place when they neutralize one another is just the reaction between the hydrogen ions and hydroxide ions. Acids and alkalis that are weak electrolytes have very different heats of neutralization. Solutions of weak electrolytes contain molecules of the substance as well as ions. For instance, in a dilute solution of ethanoic acid CH 3COOH, only about 4% of the ethanoic acid molecule exist as ions. CH3COOH(aq) =========== CH3COO-(aq) + H+(aq) Mainly molecules a very few ions. b. Hydration Energies If a molecule of gaseous hydrogen ions is plunged in water and the resulting solution was infinitely dilute, leads to liberation of large amount of heat. The heat change for the process is called the hydration energy ∆Hhꙫ of the hydrogen ions; H+(g) + water H+(aq) ∆Hhꙫ = -1075Kj mol-1 The hydration energy, ∆Hhꙫ, is generally defined as the heat change that takes place where one mole of a gaseous ion dissolves in water to give an infinitely dilute solution. c. Heat of Solution When chemicals dissolve in a solvent, there may be either liberation of heat (exothermic reaction) or absorption of heat (endothermic reaction). For instance, sodium hydroxide dissolves exothermically while sodium nitrate dissolves endothermically. The energy liberated or absorbed when one mole of a given substance is completely dissolved in a large volume of solvent i.e. to infinite dilution is termed the heat of solution. When a mole of a solute is dissolved, the heat absorbed or liberated is not a constant quantity but varies with the concentration of the solution. No appreciable change in concentration results when the solute is added, and the heat change is thus obtained at the specified concentration. Differential heat of solution can be defined, in an equivalent way, as the heat change when an infinitely small amount of solute is dissolved in a definite quantity of solution. Since the amount of solute is infinitesimal, no change in concentration would result. The integral heat of solution is the effect obtained when 1 mole of solution is dissolved in a definite quantity of pure solvent like 100 g of water, to yield a solution. Usually, differential and integral heats of solution are not equal. In differential heat the concentration does not change when the solute is added. In integral heat, both the solute and the solvent are affected during the process. Heat of hydration is generally calculated from integral heat of solution. For instance, anhydrous sodium sulphate dissolves in water as the heat of hydration disintegrate the crystal with liberation of heat. Whereas, hydrated Na2SO4.10H2O, dissolves with the absorption of heat as no enough hydration energy is produced to overcome the crystal energy. BOND ENERGY. Energy changes in chemical reaction are due to the forming and breaking of bonds. The strength of a particular bond is usually measured by the amount of energy required to separate the atoms or ions held together by the bond. This is termed as the bond energy. This is the energy that is absorbed when a particular bond is broken, or evolved when it I formed. Thus, it is the amount of energy with a particular bond in an element or compound. For instance, H2(g) 2H(g) ∆H = + 435 Kjmol-1 H(g) + H(g) H2(g) ∆H = - 435 Kjmol-1 It can be seen that 435 Kj mol-1 of energy are absorbed when 1 mole of hydrogen bond is broken in gaseous state. The same amount of the energy is also evolved when 1 mole of hydrogen bonds is formed from gaseous hydrogen atoms. Thus, the bond energy evolved when a bond is formed and that absorbed when a bond is broken are the same only for homo-nuclear molecules e.g. H H, Cl Cl; they are slightly different for hetero-nuclear molecules e.g. H Cl. The average amount of energy associated with making or breaking 1 mole of a particular bond, in its gaseous state, is its bond energy. Bond energies indicate the strength of the bonds. The higher the bond energy, the stronger the bond. Stable substances have high energy bond while reactive substances have low energy bonds. We can use bond energies to calculate the heat change or enthalpy change of a reaction. LATTICE ENERGY. Bond energies help us to understand the stability of molecular or covalent substances. Similarly, lattice energies help us to understand the stability of ionic crystals. The lattice energy of an ionic crystal is the heat of formation of one mole of ionic compound from widely separated gaseous ions under standard conditions. Lattice energies, like bond energies, cannot be measured exponentially, they can only be determined theoretically. CHEMICAL EQUILIBRIUM. In a chemical reaction, chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. The chemical equilibrium is achieved when the rate of forward reaction is same as the reverse reaction. Since the rates are equal, there are no net changes in the concentrations of the reactant(s) and product(s). this state is known as dynamic equilibrium.  [Products] Concentration [Reactants] Time Laws of Chemical Equilibrium and Equilibrium Constant K C  Chemical equilibrium can easily be understood by understanding chemical kinetics. It is the study of the rate of the reactions under various conditions.  Another concept that forms the basis is the law of mass action. The law of mass action states that the rate of a chemical reaction is directly proportional to the product of the concentrations of the reactants raised to their respective stoichiometric coefficients.  So, given a reaction:  aA(g) + bB(g) ↔cC(g) + dD(g) using the law of mass action, forward reaction rate = K+[A]a[B]b and backward reaction rate = K-[C]c[D]d. where, [A], [B], [C] and [D] being the active masses and K + and K- are rate constants.  At equilibrium, forward and backward rates are equal and the ratio of the rate constants is a constant and is known as an equilibrium constant.  In a reaction mixture at equilibrium, the concentrations of the reactants and products are related by an equilibrium constant. For the reaction aA(g) + bB(g) ↔ cC(g) + dD(g) forward reaction rate = Backward reaction rate. K+[A]a[B]b = K-[C]c[D]d We have KC = K+/K- KC = [C]c.[D]d/[A]a.[B]b CHEMICAL KINETICS. o Rate processes are of fundamental concern to everyone connected to pharmaceutics, the science of dosage form design and formulation from the manufacturer to the patient. The manufacturer must clearly demonstrate that the drug or the dosage form that he produces I sufficiently stable that it can be stored for reasonable length of time without changing to an inactive or toxic form. The  pharmacist must be aware of the potential instability of the drugs which he handles, and the physician and patient must be assured that drug prescribed will eventually reach its locus of action in sufficient concentration to elicit the desired effect. o Kinetics provides the principles and related rate processes that are implied for providing solutions. o Kinetics deals with the rate at which changes occur in a system and factors that influence such change. o Kinetics helps us to understand the mechanism of change involved and also helps to predict the degree of change after a given time has elapsed. o Kinetics is important in pharmacy in the following ways: a. The study of drugs instability and incompatibility i.e. drugs decompose or are lost because of their conversion to a less favorable physical or chemical form. b. Solution study: the ease with which the solvent acts on a particular dosage form of drug in solution before they can be absorbed across biological membranes. c. Absorption, distribution and elimination are processes which are associated with the rate of absorption of a drug into the body, the rate of distribution of drug in the body and the rate at which this drug is removed depends on factors such as metabolism, storage within body organs or fats and by excretory routes. d. Drug action at the molecular level for which a convenient model may be constructed, assuming that the generation of the response by a drug is a rate process. CHEMICAL KINETIC THEORY. Chemical kinetic theory is based on the law of mass action. The law of mass action states that the rate of chemical reaction at a particular temperature is proportional to the product of the masses/molar concentration of the reactants each raised to a power equal to the number of molecule of the substance undergoing a reaction. xA + yB Product -------------------------------------------- 1 Rate a[A] [B] = K[A]x[B]y x y Where A and B are reactants; x and y = number of molecules of A and B. K = proportionality/rate constant of rate of reaction. RATES, ORDERS AND MOLECULARITY OF REACTIONS. 1. Rates of Reaction The rate of a reaction can be expressed as either a) A decrease in concentration per unit time of one of the reactants or b) As an increase in concentration per unit time of one of the products. Assume C0 = initial concentration at time t = 0 C = concentration after time t X = concentration of the reactant that has reacted after time t X = C0 – C Rate of reaction = rate of decrease in the concentration of reactant at a resultant time t. Rate = dC/dt – C because of decrease -------------------------------- 2  C = C0 – X Rate = d(C0 – X )/dt During increase, Rate = dX/dt. 2. Order of Reaction The rate of change in concentration of a product depends solely on the concentration of the reactant. Hence the order of a reaction expresses experimentally determined dependence of rate upon concentration of reactant. For instance, in a reaction xA + yB zC The rate of accumulation of C depends on A and B. dC/dt a [A]x[B]y ---------------------------------------------- 3 dA/dt a [A]x[B]y in the expression for rate of a chemical reaction, the order of a reaction is the sum of power of the concentration terms involved in the rate equation and this is called the order of the reaction. The order of reaction is x + y. The order of the reactant with respect to A is x, the order of reaction with respect to B is y, the order of the overall reactant with respect to A and B is x + y. The overall order is three. This reaction would be described as first order in A and second order in B. 3. Molecularity of Reaction The number of molecules involved in forming the activated complex in a step of chemical reaction. Reactions are said to be unimolecular, bimolecular or trimolecular according to whether 1, 2, 3 molecules are involved. The molecularity of a reaction is equal to the number of molecules or atoms that must collide simultaneously to form the product of reaction. If one molecule each of the two reactant in a reaction is combined, the molecularity of the reactant is two (2), the reaction can be referred to as bimolecular reaction. Molecularity can be represented mathematically from the equation Ax + By Product as -------------------------------------- 4 Molecularity = x + y. UNITS OF BASIC RATE CONSTANTS. In order to arrive at units for the rate constants appearing in zero, first, and second order rate laws, the equation expressing the law is rearranged to have the constant expressed in terms of the variables of the equation. Zero order reaction: dA/dt = K = moles/litre/seconds = moles/litre – second = moles L-1Sec-1 ------------------------------------------------ 5 First order reaction: dA/dt 1/A = K = moles/litre/second – mole litre = 1/second = Sec-1 ------------------------------------------------------- 6 Second order reaction: dA/dt 1/A2 = K = moles/litre/second(mole/litre)2 = litre/mole-second = litre second-1 mole-1  ZERO ORDER REACTIONS These are reaction in which the rate of reaction is independent of concentration of any of the reactants and it is not usually dependent with time (t). That is the rate of change of reactant is constant. dX dt = K where X = C0 – C i.e. dX/dt = dC0 – C /dt = K ------------------------------------- 7 Or C0 – C = Kt or C0 – Kt = C. C0 C Slope = -K Time The rate order equation for zero order reaction is A0 – A = Kt. Half- life for zero order is the time required for half of the material to disappear. It is the time at which A has decreased to half (1/2A) so t1/2 = 1/2A0. APPARENT ZERO ORDER KINETICS. In suspensions the concentration of solution depends on drug solubility, as the drug decomposes in solution, more drug is released from the suspended particles so that the concentrations of the drug remain constant with the solution despite its decomposition time. For an ordinary solution of drug with no reservoir from which the drug decomposing would be replaced, obeys first order equation – d[A]/dt = K[A] when the concentration (A) is rendered constant as in suspension then the first order law becomes apparently a zero order equation, and the rate of zero order reaction is d[A]/dt = K0. FIRST ORDER REACTION. These are reactions in which the rate of change in concentration of product and reactant is proportional to the first power of the concentration of the single reactant. Rate = K[A] or Rate = K[B]. For first order reaction, the rate equation is - dC/dt = KC ----------------------------------------------------------- 8  where C is the concentration remaining at time t and K is the first-order velocity constant. Integrating this equation between concentration C0 at time t = 0 and concentration C at some later time (t), it becomes ∫ ∫ lnC – lnC0 = -K (t -0) lnC = lnC0 – Kt converting to common logarithms gives log C = log C0 – Kt/2.303 or K = log 2.303/t log C0/C ------------------------------------------------- 9 The equation may also be written K = log 2.303/t log a/(a-x) -------------------------------------------------- 10 Where ‘a’ is used instead of C0, x is the decrease of concentration in time t and (a-x) = C. Therefore, first order rate formula K = 2.303/t log C0/C or 2.303/t log a/(a-x) C0 = concentration at time t = 0 = a C = concentration at time = (a-x) Plot of logarithm of concentration against time gives average for K. It can be seen that log C = log C0 – Kt/2.303 is the form y = mx + c with a negative slope. Log C0 log C slope = -K/2.303 Time (t) Expressing the rate equation as a function of log C0  Slope = - K/2.303 logC0/C Time (t) Once the constant is known, the concentration of reactant remaining at a definite time may be computed. Example 1: The decomposition of ammonium carbonate was followed by the liberation of ammonia and the volume was measured. The concentration of ammonium carbonate remaining after 50 minutes was 7.40 from an initial concentration of 44.54. a. calculate K b. how much ammonium carbonate remained undecomposed after 20 minutes? Answer: The equation to use I K = 2.303/t log C0/C a. K = log 2.303/50 log 44.54/7.40 0.461 X 0.7795 = 0.03590 min-1 b. 0.03590 = 2.303/20 log 44.54/C i.e. 0.03590 = 0.1152 x 1.6488 – 0.1152 log C 0.03590 – 0.1899 = - 0.1152 log C - 0.154 = - 0.1152 log C Log C = - 0.154/- 0.1152 = 1.1828 Therefore, C = antilog 1.1828 C = 15.23. Example 2: A solution of aspirin containing 300 mg per ml was prepared, after a period of 30 days it was analyzed and found to contain 180 mg per ml. compute, at what time will the aspirin have decomposed to 100 mg, assuming it is first order. Answer. The equation to use is K = 2.303/t log C0/C K = 2.303/30 log 300/180 0.0768 x 0.2218 = 0.0170 d-1 b) t = 2.303/0.0170 log 300/100 135.47 x 0.4771 = 64.63 days PSEUDO FIRST ORDER REACTION. This is the case in which a large excess of one of the reactants in a 2 nd order reaction is present throughout the reaction. Its concentration remains virtually constant, and the rate of concentration change follows the first order law. For instance, in the hydrolysis reaction of dilute aqueous solution, a reaction of substance A undergoing hydrolysis may appear to be proportional only to (A) because the amount of water present is so large. A+B Product Rate = K [A][B] concentration of B is fairly constant = Rate = K1[B] where K1 = K[A]. Pseudo-first order kinetic is one in which the rate of reaction is proportional to the concentration of only one of the reactants even though the reaction involves several reactants. SECOND ORDER REACTION. It can be determined by the rate of reactant A and B or one of the reactants raised to power 2. The rate of the second order reactants depends on the concentration of both reactants (A+B) with each term raised to the first power. Rate K[A][B] or depends on one (1) of the reactants raised to power 2. Rate = K[A]2 or Rate = K[B]2 For second order reaction Kt = x/a(a-x) --------------------------------------------------------------- 11 Or K = i/at (x/a-x) This is true only if the same number of molecules of a reactant are involved. A = B, but when A is not equal to B, Kt = 2.303/a – b log b/a (a-x)/(b-x) Or K = 2.303/t(a-b) log b(a-x)/a(b-x) ------------------------------------- 12 A common example of AB is the hydrolysis of chlorbutol in the presence of sodium hydroxide. HALF LIFE. The period of time required for a drug to decompose or for the concentration of a reactant to decrease to one half the original concentration. Half-life = Time at which Concentration X =a/2 or X = C0/2 Half-life t1/2 for the first order reaction: Log C0/2/C0 = -Kt/2.303 Log C0/2 ÷ C0 = -Kt/2.303 Log C0/2 x 1/C0 = -Kt/2.303 Log ½ = -Kt/2.303 -0.3010 = -Kt1/2/2.303 -0.6932 = Kt1/2 t1/2 = 0.6932/K -------------------------------------------------------------- 13 for second order t1/2 = 1/KC0 ------------------------------------------------------------------- 14 for zero order t1/2 = C0/2K = a/2K ---------------------------------------------------------- 15 Example: The combustion of methane is given in the thermochemical equation CH4(g) + 2O2(g) CO2(g) + 2H2O(l) The initial concentration of both methane and oxygen in the mixture were 0.05M. The change in concentration of oxygen during 30 minutes was 0.008 mole, the remaining concentration of the mixture is 0.042M. Compute (a) the rate constant and (b) the half-life of the reaction. Answer: Using the equation K = 1/at(x/a-x) K = 1/0.05 x 30 0.008/0.042 = 0.1269 litre-1min-1 0.6666 x 0.1904 = 0.1269 litre mol-1min-1 b) the half-life of a second-order reaction is t1/2 = 1/Ak Here the initial concentrations of the reactants are identical and so t 1/2 can be computed for the reaction. t1/2 = 1/0.05 x 0.1269 = 157.60 min. FACTORS THAT AFFECT REACTION RATES. A number of factors affect the reaction velocity. Among these are – concentration of reactants, temperature, solvents, catalysts, light, nature of reactants and surface area of reactants. 1. Concentration of Reactants Increase in concentration always increases the rate of chemical reaction. With high concentration there is more collision among the particles and hence an increase in the rate of chemical reaction e.g. Na2S2O3(aq) + 2HCl(aq) 2NaCl + H2O + SO2 + S 2. Temperature The speed of many reactions increases by about 2 to 3 times with each 10 0C rise in temperature. When the temperature is increased the reacting particles acquire more kinetic energy and they tend to travel at a greater speed. The effect of temperature on reaction rate constant (K) is given by Arrhenius equation K = Ae-Ea/RT Or log10 K = log A – Ea/2.303RT ------------------------------------------------ 16 Where K = the specific reaction rate A =a constant known as the frequency factor in thermodynamics Ea = the energy of activation R = the molar mass gas constant, 1.98 calories/deg mole. T = absolute temperature E = natural logarithm.  The mathematical prediction of shelf life is based on the application of the Arrhenius equation.  The equation holds for zero and first order reactions.   The constants, A and Ea can be evaluated by determining K at several temperatures by plotting 1/T against log K.  The slope of the line so obtained is -Ea/2.303R and the intercept on the vertical axis of the line is log A form which Ea and A can be obtained. Log A Log K slope = -Ea/2.303R K- - - - - - - - - 1/Temp. in Kelvin K = 273 + 00C. If the slope of this line is determined from the results of accelerated test at high temperature, it is possible to determine the value of the rate constant at other temperatures by extrapolation. Substitution of this value of K into the appropriate order of the reaction allows the amount of decomposition after a given time to be calculated. If a plot of logt1/2 on the vertical axis and 1/T is made, then t 1/2 can be related with Arrhenius equation, the half-life for first order reaction is related to K by the equation t1/2 = 2.303/K log C0/C and t1/2 = 0.693/K and in logarithm form log K = log 0.693 – log t1/2 and substituting this equation into log K = log A - Ea/2.303 1/RT gives log t1/2 = log 0.693 – log A + Ea/2.303 1/RT Or log t1/2 = Ea/2.303 1/RT + constant and Ea/2.303R is obtained as the slope of the line resulting from plotting log t1/2 against 1/T. Ea can also be obtained by writing Log K = log A - Ea/2.303 1/R, for temperatures T2 and T1 For temperature T2 Log K2 = log A - Ea/2.303R 1/T2 For temperature T1 Log K1 = log A - Ea/2.303R 1/T1 Substituting T1 from T2 gives Log K2/K1 = Ea/2.303R T2 - T1/T2T1 ----------------------------------------------- 17 Example The rate constant K1 for the decomposition of caffeine (1,3,7- trimethylxanthine) at 2600C (5330K) is 2.541 hr-1 and K2 at 3000C (5730K) is 10.414 hr-1. Compute the activation energy Ea in Kcal/mole for the breakdown of 1,3,7- trimethylxanthine within this temperature range. Answer Equation to be used is Log K2/K1 = Ea/2.303R T2 – T1/T2T1 Where K2 = 10.414 hr-1 K1 = 2.541 hr-1 Ea =? T1 = 5330K T2 = 5730K R = 1.98 Log 10.414/2.541 = Ea/2.303 x 1.987 x 573 – 533/573 x 533 0.6125 = Ea/24.576 x 20/305409 = 20 Ea/1397551.584 0.6125 x 1397551.584 = 20 Ea Ea = 856000.345/20 = 21400.00 cal/mol Ea = 21.40 Kcal/mol. 3. Light For photo chemical reaction Ea activation energy is provided by light, therefore, the reactions are dependent on light not temperature. It is difficult to separate photo-chemical reaction from thermal reaction, since thermal effect may be involved in subsequent chain reaction that often follows photochemical. This may arise from the heating effect produced by the increase in kinetic energy of molecules that have collided with others that have absorbed some light energy. Sunlight increase the reaction of many substances e.g. substitution reaction of methane is influenced by using ultra-violent radiation.  uv CH4 + Cl2 CH3Cl + HCl 4. Solvent The effects of solvent on the rate constant are complex. Solvent usually influence the ability of the solute to dissociate into ions, this affects the reactions that involves these ions. Usually polar solvent dissolve polar solute and non-polar solvent dissolve non-polar solute. “Like dissolve like”. 5. Catalyst A substance that increases the rate of chemical reaction without itself undergoing any permanent chemical change when present in a chemical reaction. Generally, the term catalyst is used for a substance that increases reaction rate (a positive catalyst). In the presence of a positive catalyst more reactant particles are able to react when they collide because energy barrier of a chemical reaction is lowered. Some reactions can be slowed down by negative catalysts, most catalysts are also highly specific in the type of reaction they catalyze, particularly enzymes in biochemical reactions. Catalysts are divided into two: a. Homogenous catalyst: when a catalyst dissolves in a reaction medium it is called a homogenous catalyst. Examples are enzymes in biochemical reactions or transition- metal complexes used in the liquid phase for catalyzing organic reactions. The effect of pH on hydrolysis of compounds, under this situation pH changes affect the situation based on degradation. Another example is ergometrine B.P adjusted to pH (3) for maximum stability by addition of maleic acid, maleic acid I a component of ergometrine. b. Heterogeneous catalyst: in this case the catalyst remains as solid to the surface which reactants are adsorbed. The presence of other compounds or substances which may also be adsorbed strongly to the catalyst surface by competitive exclusion of the reactant. The adsorption weakens bonds of reactant molecules and lowers the activation energy. The activated molecules then can react, and the products diffuse away from the surface in metals or oxides used in many industrial gas reactions. The catalyst provides an alternative pathway by which the reaction can proceed, in which the reaction comes to equilibrium, although it does not alter the position of the equilibrium. 6. Nature of Reactant The rate of chemical reaction is always determined by the chemical nature of reactants as different substances have different energy content. The higher the element in the series the faster the rate. For instance, if a piece of iron is placed in dilute hydrochloric acid, there is a slow evolution of hydrogen gas but with a piece of zinc the evolution of hydrogen is rapid. 7. Surface Area of Reactants Solids in powdered form due to their large surface area relative to volume react faster in solution than solids in forms or chips or lumps. For instance, the rate of evolution of carbon (V) oxide is faster when marble (CaCO3) powder is reacted with dilute HCl than when marble lumps are used.