Chemical Kinetics - CHM-203- Slides-Compiled

Download more books at 9jabaz.ng for free!  Chemical Kinetics Reaction Rate and Mechanism  INTRODUCTION Two questions are of interest to a chemist when considering a chemical reaction – Is the reaction feasible or not? and – What is the timescale of occurrence of the reaction? The f...

Download more books at 9jabaz.ng for free!  Chemical Kinetics Reaction Rate and Mechanism  INTRODUCTION Two questions are of interest to a chemist when considering a chemical reaction – Is the reaction feasible or not? and – What is the timescale of occurrence of the reaction? The first concern is addressed by thermodynamics which – predicts the spontaneity or otherwise of the chemical process as well as – the equilibrium conditions of the reaction. The time scale and the possible pathways in the transformation of the reactants to products can only be determined through the knowledge of REACTION KINETICS. In summary, thermodynamics addresses the why of a chemical process while chemical kinetics addresses the how of the reaction.  Rate of Reaction rate at which reactants are used up, or equivalently the rate at which products are formed. – units of concentration per unit time, mol dm-3 s-1 (alternative units of concentration are often used such as units of pressure – Torr, mbar or Pa, for gas phase reactions). Reaction rate is measured by monitoring the concentration of one of the reactant or products with time Consider the reaction: – The stochiometric coefficients of N2, H2 and NH3 are 1, 3, and 2 respectively. – We could determine the rate of this reaction in any one of three ways, by monitoring the changing concentration of N2, H2 or NH3.  Assuming x moles of N2 are consumed within a given time, then 3x moles of H2 will be consumed and 2x moles of NH3 formed within the same given time – this then could mean that the following would be obtained d  N2    x mol dm 3 s 1 dt d H2    3 x mol dm 3 s 1 dt d  NH 3    2 x mol dm 3 s 1 dt Clearly, the same reaction cannot have three different rates, so there appear to be a problem. the reaction rate is actually defined as the rate of change of the concentration of a reactant or product divided by its stochiometric coefficient. For the above reaction, the rate (usually given the symbol ν) is therefore  d  N2  1 d H2  1 d  NH 3  ν   dt 3 dt 2 dt – Note that a negative sign appears when we define the rate using the concentration of one of the reactants. – This is because the rate of change of a reactant is negative (since it is being used up in the reaction), but the reaction rate needs to be a positive quantity. In summary, for an hypothetical reaction: aA + bB ⟶ cC + dD, the rate in terms of all the species involved is  Illustrations 4NO(g) + O2(g) ⟶ 2N2O3(g) – For the reaction above, express the rate in terms of the change in concentration with time for each substance – How fast is [O2] decreasing when [NO] is decreasing at a rate of 1.60×10-4 mol/L·s? SOLUTION 1 d  NO  d O2  1 d  N 2O3     4 dt dt 2 dt d O2  1 d  NO  1     1.6  104  dt 4 dt 4  4.0 105 mol dm 3 s 1 [O2] is decreasing 0.25 times as fast as [NO]  For a particular reaction, the reaction rate in terms of the change in concentration with time for each substance is SOLUTION The balanced equation is 4 NH 3  g   5O2  g  4 NO  g   6 H 2O  g  Can you try this?  Rate Law Expression The major theme of any kinetic study is what is known as the rate law or rate equation. It expresses the rate as a function of concentrations and temperature The rate law is based on experiment For a general reaction: aA + bB ⟶ cC + dD, Experimentally, the rate is proportional to the concentrations of the reactants in the form Rate   A  B  x y Expressing the expression above mathematically by introducing a constant of proportionality k, gives the rate law as Rate  k  A  B  x y k is known as the rate constant that is specific to the particular reaction of interest at a given temperature and does not change as the reaction proceeds  The exponents x and y are the order of the reaction with respect to the concentrations of A and B in the reaction. – The order with respect to a reactant is a measure of how the rate of the reaction depends on the concentration of the particular reactant. – The coefficients of A and B in the balanced equation do not necessarily related to the exponents x and y in the rate law expression – The terms in the rate law expression are all experimentally determined parameters The various parameters in the rate law expression can be obtained by measuring the concentrations of the reactants to determine the initial rate – (REMEMBER THAT YOU WERE TAUGHT INITIAL RATE METHOD IN PART ONE)  Some Methods of Determining the Initial Rate Spectrometric methods measure the concentration of a component that absorbs (or emits) characteristic wavelengths of light. – For example, in the reaction of NO and O3, only NO2 has a color (an indication that NO2 absorbs some wavelength of visible light): – NO(g, colorless) + O3(g, colorless) ⟶ O2(g, colorless) + NO2(g, brown) – Known amounts of reactants are injected into a tube of known volume within a spectrometer, which is set to measure the wavelength and intensity of the color. – The rate of NO2 formation is proportional to the increase in that intensity over time.  Conductometric methods rely on the change in electrical conductivity of the reaction solution when nonionic reactants form ionic products, or vice versa. – Consider the substitution reaction between a haloalkane, such as 2-bromo-2-methylpropane, and water: – (CH3)3C-Br(l) + H2O(l) ⟶ (CH3)3C-OH (l) + H+(aq) + Br-(aq) – The HBr that forms is a strong acid, so it dissociates completely in the water. – As time passes, more ions form, so the conductivity of the reaction mixture increases. – There are other examples in which the conductivity decreases  Manometric methods employ a manometer attached to a reaction vessel of fixed volume and temperature. The manometer measures the pressure change in the vessel due to a reaction that involves a change in the number of moles of gas. – Consider the decomposition reaction of hydrogen peroxide: – 2H2O2(aq) ⟶ 2H2O(l) + O2(g) – The rate is directly proportional to the increase in pressure as O2 gas forms.  Determination of the Order of a Reaction The rate as a function of reactants’ concentration – Initial Rate Method: (Please read this up from CHM 101) The rate as a function of time - Integrated Rate Law Method – We will proceed by first looking at reactions involving a single reactant: 𝑎𝐴 ⟶ 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠 d  A Rate    k  A x dt Consider the integrated rate laws individually for the cases x = 1 (first order), x = 2 (second order), and x = 0 (zero order). First order Rate Law: Consider the reaction  Experiment showed that the rate law is Integration of the above rate law with boundary conditions: At time t  0 :  N 2O5    N 2O5 0 At time t  t :  N 2O5    N 2O5 t The rate law yields OR kt log10  N 2O5 t    log10  N 2O5 0 2.303 Generally for the reaction: 𝑎𝐴 ⟶ 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠, where the order of the reaction is 1 with respect to [A] d  A Rate    k  A dt  The integrated rate law is ln  At  kt  ln  A0 Things to note about this equation  The unit of the rate constant for a first order reaction can be obtained from the last equation by making k the subject of the expression 1   A0  k  ln     A t  t   Unit of k is therefore the reciprocal of time such as s 1 , min 1 , hr 1 etc ILLUSTRATION Problem At 1000°C, cyclobutane (C4H8) decomposes in a first- order reaction, with the very high rate constant of 87 s-1, to two molecules of ethylene (C2H4). – (a) The initial C4H8 concentration is 2.00 M. What is the concentration after 0.010 s? – (b) How long will it take for 70.0% of the C4H8 to decompose? SOLUTION  From the unit of the rate constant, the reaction is first order with respect to [cyclobutane] ln C4 H 8 t 0.010 s   kt  ln C4 H 8 0  ln C4 H 8 t 0.010 s  87 s 1  0.010 s  ln  2.00   0.1769 C4 H 8 t 0.010 s  exp  0.1769   0.84 M For 70% decomposition, it means C4 H 8 t  0.30 C4 H 8 0 1   A0  1   A0   t  ln   ln   k   A t  87 s 1  0.30  A     0   0.0138s PRACTISE QUESTIONS FACTORS AFFECTING RATE OF REACTION  A number of factors influence the rate of reactions. Such factors include: Concentration of reactants Nature of reactants and products formed Presence of catalyst, and Temperature All of these factors were taught in detail during CHM 101 and so, be advised to revisit your CHM 101 note. However, we will be looking more into the effect of catalyst as well as, temperature on the rate of reaction  Effect of Catalyst – A catalyst is a substance that speed up the rate of reaction by providing an alternative pathway of low activation energy for the reaction. – The catalyst is not used up in the course of the reaction, it can however, be transformed from one form of existence to another – Nature is the master designer and user of catalysts: every organism relies on protein catalysts, known as enzymes, to speed up life-sustaining reactions, and even the simplest bacterium employs thousands of them. Basics of Catalytic Activity A catalyst provides a different reaction mechanism with a lower activation energy, which in turn makes the rate constant larger and, thus, the reaction rate higher: Catalyst ⟹ lower Ea ⟹ larger k ⟹ higher rate  For each reaction a certain energy barrier must be surmounted.  Catalysts are classified as homogeneous or heterogeneous. – A homogeneous catalyst is one that is present in the same phase (physical state) as the reacting molecules. – A heterogeneous catalyst exists in a different phase, usually as a solid.  EFFECT OF TEMPERATURE ON RATE OF REACTION AND THE COLLISION THEORY Atoms, molecules, or ions must collide with one another for a reaction to occur – this is the basis of the COLLISION THEORY – This assumption can explain the concentration dependence of reaction rates. Increase in concentration increases the number of molecules colliding and the frequency of collision increases, thus, the rate of the reaction increases It is found that the rate of reaction is much smaller than the calculated collision frequency in a given collection of gas particles. This must mean that only a small fraction of the collisions produces a reaction. Why? – To answer this question, Svante Arrhenius proposed the existence of a threshold energy, called the activation energy, that must be overcome to produce a chemical reaction.  Consider the gas phase reaction In this reaction two Br-N bonds must be broken and one Br-Br bond must be formed. Breaking a Br-N bond requires considerable energy (243 kJ/mol), which must come from somewhere. The collision model postulates that the energy required to break the bonds comes from the kinetic energies possessed by the reacting molecules before the collision. This kinetic energy is changed into potential energy as the molecules are distorted during a collision, breaking bonds and rearranging the atoms into the product molecules.  the fraction of collisions with the required activation energy increases dramatically. When the temperature is doubled, the fraction of effective collisions much more than doubles. The fraction of effective collisions increases exponentially with temperature. This agrees with the Only collisions with energy greater observation that rates of than the activation energy are able reactions increases to react (get over the barrier). exponentially with temperature.  There is, however, another complication. – Experiments show that the observed reaction rate is considerably smaller than the rate of collisions with enough energy to surmount the barrier. – This means that many collisions, even though they have the required energy, still do not produce a reaction. Why not? The answer lies in the molecular orientations during collisions. In summary, two requirements must be satisfied for reactants to collide successfully (to rearrange to form products): – The collision must involve enough energy to produce the reaction; that is, the collision energy must equal or exceed the activation energy. – The relative orientations of the reactants must allow formation of any new bonds necessary to produce products. Taking these factors into account, the rate constant can be represented as – where z is the collision frequency (the total number of collisions per second). – The factor p in this expression, called the steric factor, reflects the fraction of collisions with effective orientations. – The factor e−Ea RT represents the fraction of collisions with sufficient energy to produce a reaction.  The expression above can be written is the well-known form as – The Arrhenius Equation Taking logarithm of both sides of the equation: One of the most important applications of this equation is in the determination of the activation energy Ea for a reaction. – This involves measuring the rate constant k at several temperatures and then plotting ln(k) versus 1/T, – Ea can also be calculated from the values of k at only two temperatures using a formula that can be derived as follows from the equation above. At temperature T1, the rate constant is k1; thus – At temperature T2, the rate constant is k2; thus  Subtracting the last two equations from one another gives Therefore, the values of k1 and k2 measured at temperatures T1 and T2 can be used to calculate Ea.  PRACTISE QUESTIONS 1 2 3 MECHANISM OF REACTIONS  Mechanism a sequence of single reaction steps that sum to the overall equation. a series of elementary steps by which a chemical reaction occurs – An elementary step - a step involving a one-, two-, or three-entity collision that cannot be explained by simpler reactions To understand a chemical reaction completely, we must know its mechanism. One of the main purposes for studying reaction rates is to learn as much as possible about the steps involved in a chemical reaction. For example – the rate law expression was determined experimentally as d  NO g   2   k  NO g   dt  The question, therefore, is what mechanism can explain this experimentally observed rate law equation? The following elementary steps were proposed – Since the first step is the slowest step in the reaction mechanism, the overall reaction must proceed at the rate of the first step. – That is, the products of the overall reaction, gaseous nitrogen dioxide and carbon dioxide, can be produced only as fast as the slowest step, Step 1. – Step 1 is therefore called the rate-determining step, which is the step in a reaction mechanism that determines the rate of the overall reaction. – A careful look at the proposed mechanism shows that NO3 is produced in the step 1 but consumed in step 2. Such a specie is called a REACTION INTERMEDIATE  The sum of the two elementary steps above gives the overall balanced equation. For a plausible reaction mechanism, the following conditions must be obeyed: – Summing the elementary steps in the reaction mechanism must give the overall balanced equation for the reaction. – The elementary steps must be reasonable. They should generally involve one reactant particle (unimolecular) or two (bimolecular). – The reaction mechanism must agree with the experimentally determined rate law. For example, consider the reaction between nitrogen dioxide and fluorine is another example of such a reaction: 2NO2(g) + F2(g) ⟶ 2NO2F(g) The experimental rate law is first order in NO2 and in F2, Rate = k[NO2][F2]  The proposed mechanism is (1) NO2(g) + F2(g) ⟶ NO2F(g) + F(g) [slow] (2) NO2(g) + F(g) ⟶ NO2F(g) [fast] Is the mechanism a plausible one or not? SOLUTION The slow step is rate determining, therefore, Rate = k1 NO2 F2 = k NO2 F2. This in agreement with the experimentally determined rate law Summing the two elementary steps above also gives the overall reaction. – Note that elemental F is produced in the step 1 and consumed in step 2 – reaction intermediate – and did not feature in the rate law The two elementary steps involved 2 molecules in each step, so each step is bimolecular. Hence the three criteria for a plausible mechanism is obeyed by the proposed mechanism  Example 2 The following elementary steps are proposed for a reaction mechanism: (1) Cl2(g) ⥫⥬ 2Cl(g) (2) Cl(g) + CHCl3(g) ⟶ HCl(g) + CCl3(g) [Slow] (3) Cl(g) + CCl3(g) ⟶ CCl4(g) If the observed rate law is Rate = k[Cl2]1/2[CHCl3] Show that the mechanism is consistent with this rate law SOLUTION The rate law from the slow step is 𝑅𝑎𝑡𝑒 = 𝑘2 𝐶𝑙 𝐶𝐻𝐶𝑙3 From the elementary steps in the mechanism, Cl and CCl3 are INTERMEDIATES and so should not feature in the rate law Step 1 is an equilibrium step such that:  Cl  2 k K eq  1  k1 Cl2  k1 Cl2   Cl   2 k1  k Cl   12  Cl    1 2   k1  Substitute for [Cl] in the rate law to obtain 𝑅𝑎𝑡𝑒 = 𝑘2 𝐶𝑙 𝐶𝐻𝐶𝑙3  k1 Cl2   12 12  k1  Rate  k2       CHCl3  12  CHCl3 k 2  Cl 2  k1   k1   k Cl2  CHCl3  12 12  k  where k  k2  1   k1  and k1 , k -1 are the rate constant for the forward and reverse reactions in step 1 The above is in agreement with the observed rate law. Also, the sum of the elementary steps is Cl2(g) + CHCl3(g) ⟶ HCl(g) + CCl4(g)  In a study of nitrosyl halides, a chemist proposes the following mechanism for the synthesis of nitrosyl bromide: If the rate law is rate = k[NO]2[Br2], is the proposed mechanism valid? If so, show that it satisfies the three criteria for validity. Experiment shows that the rate of formation of carbon tetrachloride from chloroform, CHCl3(g) + Cl2(g) ⟶ CCl4(g) + HCl(g) is first order in CHCl3, 1/2 order in Cl2, and 3/2 order overall. Show that the following mechanism is consistent with the rate law:  6. CHEMICAL THERMODYNAMICS AND ENERGETICS 1. THERMODYNAMICS 4.2 Intensive Properties It is the study of flow of energy. It encompasses the study of flow Such properties of a system which depends on concentration of heat as well as mass. and does not depend on the mass or the total number of particles in the system are categorized as Intensive properties. eg. Pressure, Density, Reractive Index. 2. THERMODYNAMICS TERMINOLOGY 5. STATE AND PATH FUNCTIONS 2.1 System Such thermodynamic functions which depend only on the initial and final states of the system and not on the path followed are The part of the Universe under observation is called system. called state functions eg. Internal energy, Enthalpy and the functions which depend on the path followed while changing 2.2 Surrounding from one state to the other are called path functions. eg. work heat. The part of the Universe not under observation is called Surrounding. 6. THERMODYNAMIC EQUILIBRIUM System + Surrounding = Universe A system is said to be under thermodynamic equilibrium when 2.3 Types of system none of the state variables are changing and it satisfies the three equilibriums. 2.3.1 Open System : A system in which both flow of mass and heat is possible. 6.1 Mechanical Equilibrium 2.3.2 Closed System : A system in which flow of heat is possible but flow of mass is not possible. There is no mechanical motion and the pressure and the volume of the system is not changing. 2.3.3 Isolated System : A system in which neither heat nor mass can flow in or out. 6.2 Thermal Equilibrium 3. STATE OF A SYSTEM There is no flow of heat and the temperature of the system does not change with time. The state variables (P, V, T, n) describes the condition of a system. On changing any one or more of these variables the state of the 6.3 Chemical Equilibrium system changes. If any chemical reaction is taking place in the system then the rate 4. PROPERTIES OF SYSTEM of forward reaction is equal to the rate of backward reaction which means that the overall moles of the system is constant. All the properties of a system can be categorized into one of the following two types : 7. INTERNAL ENERGY 4.1 Extensive Properties It is the sum total of the components of energy of the system due to the internal factors. It is denoted by U (sometimes by E). Since Such properties of a system which depends on the mass or the the system under observation is an ideal gas thus the internal total number of particles in the system are categorized as Extensive energy of the system is dependent only on the kinetic energy of Properties. eg. Total Energy, volume. the gas and therefore is only a function of temperature. U v T. Since internal energy depends only on temperature thus, it is a state function. Lakshya Educare 180 CHEMICAL THERMODYNAMICS AND ENERGETICS 8. MODES OF ENERGY TRANSFER 11. EXPANSION WORK It is the work done due to the volume changes of the gas. The There are two methods to alter the internal energy of a system viz. Heat and work. mathematical expression for the expansion work is w ³  Pex dV. 8.1 Heat Always remember, be it expansion or compression we always take the external pressure as the driving force. For a reversible process, Heat is the energy transferred due to temperature difference between the system and the surrounding. On heating, the kinetic Pex | PGAS and w ³  PGASdV. energy of the molecules increases and therefore the internal energy increases. If we draw a process curve between P and V then the work done is represented by the area covered under the P-V graph as shown 8.2 Work in Fig. Work is the energy spent to overcome an external force. When the system does work against an external pressure (expansion) it tends to reduce the internal energy and on the other hand when the system contracts due to the external pressure it tends to increase the internal energy. 9. FIRST LAW OF THERMODYNAMICS The first law of Thermodynamics states that Energy can neither be created nor destroyed. 'U = q + w Conventions : In the above system if work is done by the system then w is negative and if work is done on the system then w is NOTE positive. Also, if heat flows into the system then q is positive and Sign of w : If the volume of the system is increasing if heat flows out of the system then q is negative. then the sign of w is –ve and if volume is decreasing w is +ve 10. REVERSIBILITY Sign of 'U : If the temperature of the system is A process whose direction can be changed by an infinitesimal decreasing or the product pressure and volume (PV) is change to the system or surroundings and which can be reversed reducing then the sign of 'U is –ve else, the sign of by retracing the original path and the system is restored to the 'U is +ve. initial state. The driving force of a reversible process is very-very Sign of q : The sign of q needs to be determined using small and such a process is extremely slow. For a process to be the first law of thermodynamics. reversible there must not be any dissipative forces and also the system should be in a Quasi Static State. 12. CYCLIC PROCESS 10.1 Quasi Static State A cyclic process is one which comes back to its initial state. The A quasi static state means that the system seems to be static at all graph of a cyclic process is always a closed graph. For a cyclic intervals of time but actually is not. The motion is so slow that it process, 'Unet = 0 and qnet = – wnet. is almost impossible to detect the motion and the system seems to be in equilibrium with the surroundings at all instants of time. Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 181 13. ENTHALPY (H) 15.1 Isothermal Process Enthalpy is another thermodynamic function (like internal energy, These processes are the ones in which the temperature is constant work and heat) which we study in various thermodynamic throughout the process. processes. It is also a state function like internal energy. It is 'U = 0; 'H = 0 defined as the sum of the energy stored in the system and the energy used in doing work. Mathematically, H = U + PV. At constant w = –2.303 nRT log10(V2/V1) = – 2.303 nRT log10 (P1/P2) pressure 'H = qp and at constant volume 'U = qV. q = +2.303 nRT log10 (V2/V1) = + 2.303 nRT log10 (P1/P2) 14. HEAT CAPACITY (C) 15.2 Adiabatic Process The heat capacity of the system is the amount of heat needed to These processes are the ones in which the heat exchanged with the raise the temperature of the system by 1ºC or 1K. surroundings is zero. Such processes are defined by the equation 'T. C = q/' TVJ–1 = constant, TJP1–JJ = constant, PVJ = constant. q=0 Ÿ w = 'U 14.1 Molar Heat Capacity 'U = nCV'T = (P2V2 – P1V1)/(J–1). = (nR'T)/(J – 1) The molar heat capacity of a system (CM) is the amount of heat 'H = nCP'T needed to raise the temperature of one mole gas by 1ºC or K. 15.3 Isochoric Process q CM n'T These processes are the ones in which the volume remains constant. The molar heat capacity of a system at constant pressure (Cp) is Since the change in volume is zero therefore we can say that the amount of heat needed to raise the temperature of one mole w=0 gas the system by 1ºC at constant pressure. 'U = nCV'T = qV 'H = nCP'T 'T. Cp = qp /n' The molar heat capacity of a system at constant volume (CV) is 15.4 Isobaric Process the amount of heat needed to raise the temperature of one mole gas by 1ºC at constant volume. These are the processes in which the pressure remains constant. 'T. CV = qV /n' 'V = – nR' w = – P' 'T Thus, we can say that : 'H = nCP'T and 'U = nCV'T and 'U = nCV'T CP = CV + R. 'H = nCP'T NOTE All these processes are happening on a system containing an ideal gas therefore we can apply PV = nRT at any stage that we find suitable. 15. TYPES OF THERMODYNAMIC PROCESSES There are four important types of processes to be studied in this chapter. The basic meanings and difference of these four processes are : Lakshya Educare 182 CHEMICAL THERMODYNAMICS AND ENERGETICS 4. A cyclic process should remain cyclic whichever graph we NOTE make. Although the graph of isothermal and adiabatic processes are similar in nature it should be noted that the P-V graph of an adiabatic process is steeper than that of an isothermal process. 16. GRAPH TRANSFORMATION When a thermodynamic process is plotted in terms of two state variable it can be transformed into a graph involving the other state variable by doing the following : Note : From the given P–V graph. 1. Identify the type of curve given, whether it is P-V, V-T or P-T graph. Process 1o2 is isothermal expansion; 2o3 adiabatic expansion; 2. Then, Identify every step of the process 3o4 isothermal compression & 4o1 adiabatic compression. 3. Then one by one convert every step into the required graph bearing in mind critical points like, an expansion process will remain an expansion process and so on. Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 183 17. IRREVERSIBLE PROCESS 22. CONCEPT OF ENTROPY For an irreversible process the work done is given by “ Matter has a natural tendency to get disordered or randomised W=– ³P EXT dV. We cannot take the external pressure to be “ Energy has a tendency to become disordered or dispersed. equal to the pressure of the gas in these processes. It was concluded that any such process in which the total 18. FREE EXPANSION randomness of the universe (system + surrounding) increases is a spontaneous process. Entropy is a measure of randomness or If the external pressure of the gas is zero that is the gas is expanding disorder. It is a state function represented by S. We can safely against vaccum then the work done is always zero, this is called say that in a spontaneous process there is a tendency for increase the case of free expansion. In this process the gas does no work in entropy. Hence the statement of second law : as there is no effort put in expansion process. If no heat is supplied The entropy of an isolated system/Universe tends to increase to the gas then there is no change in temperature too. That is why OR In a spontaneous process the entropy of the Universe such a process is both Isothermal and Adiabatic. increases. 19. POLYTROPIC PROCESS 'S = qrev/T. It is a generalized form of any thermodynamic process which has 'STOTAL 'SSYSTEM  'SSURROUNDING ! 0 a form PVn = constant where n is a real number. For an isothermal (for a sponataneous change) process n = 1 and for an adiabatic process n = J. The heat capacity Thus, In a reversible process the entropy of the Universe remains of a polytropic process can be calculated using the first law of thermodynamics and comes out to be : constant i.e. 'STotal = 0 C = CV – R/(n – 1). 22.1 Entropy changes in a Thermodynamic Process 20. NEED FOR SECOND LAW The entropy changes in an thermodynamic process can be mathematically calculated by the equation : The first law talks about the conservation of energy in a process but does not speak of the feasibility of a process. It does not tell Ÿ'S Ÿ' = nCV ln (T2/T1) + nR ln (V2/V1). This expression can be whether a process will happen on its own i.e. whether the process simplified for the four processes studied earlier as : is spontaneous or not. A spontaneous process is one which Isothermal process : 'S = nR ln (V2/V1) happens on its own. Example, heat always flows spontaneously from higher temperature to lower temperature : Nothing in the first Isochoric process : 'S = nCV ln (T2/T1) law mentions that the opposite process cannot happen. According to first law any process where energy remains conserved is feasible. For isobaric process : 'S = nCP ln (T2/T1) But we need some other basis for feasibility of a process. This is Adiabatic process : 'S = 0 (qrev = 0) where the second law is important. 22.2 Important points to Remember 21. TYPES OF PROCESSES 1. Entropy of a system remains constant in a reversible adiabatic 21.1 Spontaneous processes process. Therefore, it is also known as “isentropic process”. 2. Entropy of an ideal gas will always increase in isothermal Spontaneous processes have a natural tendency to take place expansion. and no external work is needed to carry out these processes. All 3. In a reversible adiabatic process the entropy of both system natural processes are spontaneous. and surroundings remains the same and there is no overall 21.2 Non-Spontaneous processes change in entropy as well. 'SSYSTEM = 'SSURROUNDINGS = 'STOTAL = 0 They are driven by external work and cannot take place naturally. Lakshya Educare 184 CHEMICAL THERMODYNAMICS AND ENERGETICS 4. In a reversible isothermal expansion the entropy of surroundings will always decrease to balance the increase in 25.1 Enthalpy of Formation 'Hf0 system’s entropy to make the overall entropy constant. 5. In free expansion the entropy of the system always increases It is the heat absorbed or released when one mole of a compound and that of surrounding remains constant. Free expansion is is formed from its constituent elements under their standard elemental forms. The enthalpy for formation of the following both isothermal and adiabatic and is irreversible. substances is taken to be zero under 1 bar pressure and 298 K. 23. GIBB’S FREE ENERGY 'Hf0 (O2, g) = 0 'Hf0 (S, Rhombic) = 0 'Hf0 (C, graphite) = 0 'Hf0 (P, white) = 0 Gibb’s Free energy function gives us a very convenient parameter 'Hf0 (Br2, l) = 0 'Hf0 (H+, aq) = 0 to judge the spontaneity of a process from system’s perspective. At a constant temperature and pressure, 'G = –T'STOTAL and for 25.2 Enthalpy of Combustion a process to be spontaneous, 'G < 0. The change in Gibb’s free energy can also be represented in terms of the system parameters It is the heat released or absorbed when one mole of a substance as : undergoes combustion in presence of oxygen. 'GSYS = 'H – T'Ssys at a constant temperature. CH4 (g) + 2O2 (g) o CO2 (g) + 2H2O (g) 'H = – 890 kJ/mol 24. THERMOCHEMICAL EQUATION 25.3 Enthalpy of Solution A chemical equation which gives us all the information like energy It is the heat released or absorbed when 1 mole of a compound is changes associated with a chemical reaction and phases of various dissolved in excess of a solvent (water). reactants and products is called Thermochemical Equation. MgSO4 (s) + H2O (excess) oMg o 2+ (aq) + SO42– (aq) All reactions can be categorized into one of the following two 'H0Sol = – 91.211 kJ/mol categories : 25.4 Enthalpy of Hydration 24.1 Endothermic Reactions It is the energy released or absorbed when 1 mole of anhydrous Are those chemical reactions which absorb energy. or partially hydrated salt undergoes hydration by the addition of ('H = positive) water of crystallisation. e.g. CuSO4(s) + 5H2O(l) o CuSO4. 5H2O(s) 24.2 Exothermic Reactions 'HHYD = –78.9 kJ/mol Are those chemical reactions which release energy. 25.5 Enthalpy of Neutralization 'H = negative) For a chemical reaction, 'HREACTION = HPRODUCTS – HREACTANTS It is the heat released or absorbed when one equivalent of an acid The change in enthalpy during a chemical reaction occurs due to undergoes neutralisation with one equivalent of a base. e.g. breaking and making of bonds. H+ (aq) + OH– (aq) o H2O (l ) Also, 'H = 'U + 'ngRT. 'HR = – 57.3 kJ/mol 25. ENTHALPY OF REACTIONS 26. HESS LAW OF CONSTANT HEAT SUMMATION Enthalpy change can be calculated for all reactions and is Since enthalpy is a state function thus for a reaction which takes sometimes called the Heat of Reaction. Let’s take a look at various place in steps the net change in enthalpy can be calculated by types of reactions and enthalpy changes associated with them : adding the enthalpy changes of each step. This is called the Hess Law. Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 185 27. BORN HABER CYCLE 29. HEAT OF ATOMIZATION The entire thermodynamics process of formation of an ionic crystal It is defined as the energy required to convert any substance to lattice is called Born Haber cycle. An ionic compound is formed gaseous atoms. This is defined per mol of the gaseous atoms. For from its constituents through a series of steps involving example Heat of atomisation of H will be 218 kJ/mol atoms. conversion of atoms/molecules into gaseous phase for ion formation, ionisation and electron gain to form ions and then the 30. RESONANCE ENERGY reaction of gaseous ions to form solid lattice. Many compounds exhibit resonance. Due to resonance they exist 28. BOND DISSOCIATION ENTHALPY in a structure which is different from the expected one and more stable. The energy needed to break the bonds of one mole molecules is called the Bond Dissociation Enthalpy of the substance. It is Resonance energy = 'H0f (actual) – 'H0f (calculated) defined per mol of molecule. eg. Bond dissociation enthalpy of H2 is 436 kJ/mol Lakshya Educare 186 CHEMICAL THERMODYNAMICS AND ENERGETICS SOLVED EXAMPLES 20 Example : 1 – 4754 = – 2.303 × n × 8.314 × 300 log. 10 Calculate the internal energy change in each of the This given n = 2.75 moles. following cases : (i) A system absorbs 15 kJ of heat and does 5 kJ of work. Example : 4 (ii) 5 kJ of work is done on the system and 15 kJ of heat A 5-litre cylinder contained 10 moles of oxygen gas at is given out by the system. 27oC. Due to sudden leakage through the hole, all the gas escaped into the atmosphere and the cylinder got empty. If Sol. (i) Here, q = +15 kJ the atmospheric pressure is 1.0 atmosphere, calculate w = – 5 kJ the work done by the gas. (1 L atm = 101.3 J) ? According to first law of thermodynamics, 'U = q + w = 15 + (–5) = 10 kJ Sol. Vinitial = 5L, T = 27oC = 27 + 273 K = 300 K Thus, internal energy of the system increases by 10 kJ. nRT 10 u 0.0821 u 300 (ii) Here, w = + 5 kJ Vfinal = 246.3L P 1.0 q = – 5 kJ 'V = 'final – Vinitial = 246.3 – 5 = 241.3 L ? According to first law of thermodynamics, 'U = q + w = – 15 + (+ 5) = – 10 kJ wexp = – P'V = – 1 × 241.3 L atm = – 241.3 × 101.3 J Thus, the internal energy of the system decreases by 10 kJ. = – 24443.7 J. Example : 2 Example : 5 Calculate w, q and 'U when 0.75 mol of an ideal gas Two moles of an ideal gas initially at 27oC and one expands isothermally and reversibly at 27oC from a volume atmospheric pressure are compressed isothermally and of 15 L to 25 L. reversibly till the final pressure of the gas is 10 atm. Calculate q, w and 'U for the process. Sol. For isothermal reversible expansion of an ideal gas, V2 Sol. Here, n = 2 moles T = 27oC = 300 K, P1 = 1 atm, P2 = 10 atm w = – 2.303 nRT log V Putting n = 0.75 mol, V1 = 15 L, 1 P1 V2 = 25 L, T = 27 + 273 = 300 K and R = 8.314 J K–1 mol–1, w = –2.303 n RT log 2.303 u 2 u 8.314 JK 1 mol–1 × P2 we get 25 1 w = – 2.303 × 0.75 × 8.314 × 300 log 955.5 J 15 300 K × log = 11488 J 10 (–ve sign represents work of expansion) For isothermal expansion of an ideal gas, 'U = 0 For isothermal compression of ideal gas, 'U = 0 ?'U = q + w gives q = – w = + 955.5 J. Further, 'U = q + w ? q = – w = – 11488 J. Example : 3 Example : 6 Carbon monoxide is allowed to expand isothermally and 10g of argon gas is compressed isothermally and reversibly from 10 m3 to 20 m3 at 300 K and work obtained reversibly at a temperature of 27ºC from 10L to 5L. is 4.754 kJ. Calculate the number of moles of carbon Calculate q, W, ' U and ' H for this process. monoxide. R = 2.0 cal K–1 mol–1. log10 2 = 0.30 Atomic wt. of Ar = 40 V2 Sol. w = – 2.303 n RT log V V2 1 Sol. W = – 2.303 nRT log10 V1 Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 187 'U = 'H = 0 (temperature is constant) 10 5 = – 2.303 × × 2 × 300 log10 w = – nRT ln (V2/V1) 40 10 = – 5 × 8.314 × 300 × ln 5 W = 103.991 cal = – 20071.3 J 'U = 0 ; 'H = 0 (Constant temperature) q = + 20071.3 J ' q = 'U – W ?q = – W = – 103.991 cal Example : 7 Example : 9 The state of a mole of an ideal gas changed from state A A gas present in a cylinder fitted with a frictionless piston (2p, v) through four different processes and finally returns expands against a constant pressure of 1 atm from a volume to initial State A reversibly as shown below. of 2 litre to a volume of 6 litre. In doing so, it absorbs 800 J heat from surroundings. Determine increase in internal energy of process. Sol. Since, work is done against constant pressure and thus, irreversible. Given, 'V = (6 – 2) = 4 litre; P = 1 atm ? W = – 1 × 4 litre-atm = – 4 × 1.01325 × 102 J = 405.3 J Now from first law of thermodynamics q = 'U – W 800 = 'U + 405.3 ? 'U = 394.7 Joule Calculate the total work done by the system and heat absorbed by the system in the cyclic process. Example : 8 5 moles of an ideal gas at 300 K are expanded isothermally Sol. State A to State B (Isobaric expansion) from an initial pressure of 500 Pa to a final pressure of Pressure is held constant at 2p and the gas is heated 100 Pa against a constant external pressure of 100 Pa. until the volume v becomes 2v. Calculate w, q, 'U and 'H for the process. What will be the difference if the same process is carried out irreversibly ? ? W1 = – p'V = – 2p (2v – v) = – 2pv What are the values of w, q, 'U, 'H for the irreversible process ? State B to State C (Isochoric process) Volume is held constant at 2v and the gas is coolled Sol. For an isothermal irreversible expansion : until the pressure 2P reaches p. ' U = 'H = 0 ?W2 = 0 (' 'V = 0) w = – PEXT (V2–V1) State C to State D (Isobaric compression) V2 and V1 can be calculated from ideal gas equation. Pressure is held constant at p and the gas is further V1 = nRT/P1 = 5 × 8.314 × 300/500 = 24.9 m3 cooled until the volume 2v becomes v. V2 = nRT/P2 = 124.7 m3 ?W3 = – p (v – 2v) = pv Therefore, w = – 100 × (124.5 – 24.9) State D to State A (Isochoric process) = – 9980 J Volume is held constant at v and the gas is heated until the pressure p reaches 2p. q = + 9980 J ?W4 = 0 (' 'V = 0) If this process is done reversibly then the internal and external pressure should be same throughout. Lakshya Educare 188 CHEMICAL THERMODYNAMICS AND ENERGETICS Total work done by the gas = W = W1 + W2 + W3 + W4 n = 2 mol. or W = – 2pv + 0 + pv + 0 = – pv (= area ABCD) At point C : P = 0.5 atm, V = 40L As the process is cyclic 'E = 0 PV = nRT ?q = – w Ÿq = + pv 0.5 u 40 T = (0.0821) (2) = 121.8 K. where q is the heat absorbed in the cylic process. § 20 · Example : 10 WCA = – 2.303 (2) (8.314) (121.8) log10 ¨ ¸ © 40 ¹ Two mole of a perfect gas undergo the following processes: = 1404.07 J (a) a reversible isobaric expansion from Total work, W = – 2026.5 + 0 + 1404.07 (1.0 atm, 20.0L) to (1.0 atm, 40.0L). (b) a reversible isochoric change of state from = – 622.43 J (1.0 atm, 40.0L) to (0.5 atm, 40.0L) For cyclic process : 'U = 0 (c) a reversible isothermal compression from Ÿ q=–w (0.5 atm, 40.0L) to (1.0 atm, 20.0L). q = + 622.43 J (i) sketch with labels each of the processes on the same P-V diagram. Example : 11 (ii) Calculate the total work (w) and the total heat change Calculate the amount of work done in each of the following (q) involved in the above processes. cases : (iii) What will be the value of 'U, 'H and 'S for the overall process ? (i) One mole of an ideal gas contained in a bulb of 10 litre capacity at 1 bar is allowed to enter into an Sol. The overall is cyclic one, i.e., initial state is regained, thus evacuated bulb of 100 litre capacity. 'U = 0; 'H = 0 and 'S = 0. (ii) One mole of a gas is allowed to expand from a volume of 1 litre to a volume of 5 litres against the constant external pressure of 1 atm (1 litres atm = 101.3 J) Calculate the internal energy change (' 'U) in each case if the process were carried out adiabatically. Sol. (i) w = –Pext × 'V As expansion taks place into the evacuated bulb, i.e., against vacuum, Pext = 0. Henc, w = 0. For adiabatic process, q = 0 ?' 'U = q + w = 0 + 0 = 0. Now, total work W = WAoB + WBoC + WCoA (ii) 'V = V2 – V1 = 5 – 1 = 4 litres WAB = – P (VB – VA) P = 1 atm ? w = – P'V = – 1 (40 – 20) = – 20L atm = –1 × 4 litre atm = – 4 litres atm 2 = – 20 × 1.01325 × 10 J = – 4 × 101.3 J = – 405.2 J (1 L – atm = 101.3J) = – 2026.5 J The negative sign implies that the work is done by the WBC = O (Isochoric) system. VA For adiabatic process, 'U = q + w = 0 – 405.2 J = – 405.2 J. WCA = – 2.303 nRT log10 V C Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 189 Example : 12 Example : 14 An ideal gas is originally confined to a volume V1 in an 5.6 dm3 of an unknown gas at S.T.P. required 52.25 J of insulated container of volume V1 + V2. The remainder of heat to raise its temperature by 10oC at constant volume. the container is evacuated. The portion is then removed Calculate Cv, Cp and J of the gas and the gas expands to fill the entire container. If the initial temperature of the gas was T, what is the final Sol. The 22.4 dm3 of a gas at S.T.P. = 1 mol temperature. Also predict qualitatively, the entropy change of system, surroundings and the universe. 1 ? 5.6 dm3 of the gas at S.T.P. = u 5.6 0.25 mol 22.4 Thus, for 10o rise, 0.25 mol of the gas at constant volume require heat = 52.25 J ? For 1o rise, 1 mol of the gas at constant volume will 52.25 require heat = 20.9 10 u 0.25 Sol. This is a process of adiabatic free expansion of an ideal ? Cv = 20.9 J K mol –1 –1 gas. The internal energy does not change thus the temperature also stays the same, i.e., the final temperature Now, Cp = Cv + R = 20.9 J K–1 mol–1 + 8.314 J K–1 mol–1 = is still T. 29.214 J K–1 mol–1 V1  V2 Cp 29.214 'Ssystem = nR ln V1 > 0 ' V 1 + V 2 > V1 ? J = 1.4 Cv 20.9 'Ssurr = 0 ' qsurr. = 0 Example : 13 Ÿ' 'Suniv > 0 A heated copper block at 130oC loses 340 J of heat to the Example : 15 surroundings which are at room temperature of 32oC. Calculate 1.0 mol of an ideal gas, initially present in a 2.00 L insulated cylinder at 300 K is allowed to expand against vacuum to (i) the entropy change of the system (copper block) 8.00 L. Determine W, 'E, 'H, 'Suniv and 'G. (ii) the entropy change in the surroundings (iii) the total entropy change in the universe due to this process Sol. Assume that the temperature of the block and the surroundings remains constant. W = – pext 'V = 0, q = 0, Ÿ'E = 0 = 'H Sol. Tsystem = 130o C = 130 + 273 K = 403 K, Tsurr = 32oC = 32 + 273 K = 305 K qsystem = – 340 J, qsurr = + 340 J Ÿ Tf = 300 K q system 340J V2 (i) 'Ssystem Tsystem 403K -0.84 JK -1 'Ssys = R ln V = R ln 4 = 11.52 JK–1 1 'Ssurr = 0 ' qsys = qsurr = 0 q surr 340 J (ii) 'Ssurr -1 +1.11 J K Tsurr 305K Ÿ'Suniv = 11.52 JK–1 (iii) 'Stotal or 'Suniverse = 'Ssystem + 'Ssurr = – 0.84 + (+1.11) Ÿ'G = – T'Suniv = – 300 × 11.52 = – 3456 J/mol. J K–1 = 0.27 J K–1 Lakshya Educare 190 CHEMICAL THERMODYNAMICS AND ENERGETICS Thus, 0.04 mole of H+ ions will combine with 0.04 mole Example : 16 of OH– ions to from 0.04 mole of H2O and 0.01 mole of The heat of combustion of benzene in a bomb calorimeter H+ ions will remain unreacted. (i.e., constant volume) was found to be 3263.9 kJ mol–1 at ? Heat evolved when 1 mole of H+ ions combine with 25oC. Calculate the heat of combustion of benzene at constant pressure. 1 mole of OH– ions = 57.1 kJ. ? Heat evolved when 0.04 mole of H+ ions combine Sol. The reaction is: with 0.04 mole of OH– ions = 57.1 × 0.04 = 2.284 kJ 1 C6 H 6 A  7 O 2 g  o 6CO 2 g  3H 2 O A 0.2 2 (ii) 200 cm3 of 0.2 M H2SO4 u 200 mole of H2SO4 1000 In this reaction, O2 is the only gaseous reactant and CO2 = 0.04 mole of H2SO4 = 0.08 mole of H+ ions is the only gaseous product. 0.5 400 cm3 of 0.5 M KOH u 400 mole of KOH = 1 1 3 1000 ? 'n g np  nr 67 1  2 2 2 0.2 mole of KOH = 0.2 mole of OH– ions Also, we are given 'U (or qv) = – 3263.9 kJ mol–1 Thus, 0.08 mole of H+ ions will neutralize 0.08 mole of T = 25oC = 298K OH– ions. (out of 0.2 mole of OH– ions) to form 0.08 mole of H2O. 8.314 R = 8.314 J K–1 mol– = kJ K 1 mol 1 Hence, heat evolved = 57.1 × 0.08 = 4.568 kJ 1000 In case (i), heat produced = 2.284 kJ = 2284 J 'H (or qp) = 'U + 'ng RT = – 3263.9 kJ mol–1 + Total volume of the solution = 500 + 200 = 700 mL § 3 · § 8.314 · Assuming density of solution = 1g/mL ¨  mol ¸ ¨ kJ K 1 mol 1 ¸ 298K © 2 ¹ © 1000 ¹ So mass of solution = 700g Specific heat = 4.18 J K–1 g–1 = – 3263.9 – 3.7 kJ mol–1 = – 3267.6 kJ mol–1. Q 2284 Q m u C u ' T ? 'T = 0.78ºC muC 700 u 4.18 Example : 17 In case (ii), heat produced = 4.568 kJ = 4568 J Calculate the amount of heat evolved when Total mass of the solution = 200 + 400 = 600 g (i) 500 cm3 of 0.1 M hydrochloric acid is mixed with 200 cm3 of 0.2 M sodium hydroxide solution Q 4568 ? 'T = 1.82ºC 3 (ii) 200 cm of 0.2 M sulphuric acid is mixed with 400 mu C 600 u 4.18 cm3 of 0.5 M potassium hydroxide solution. Example : 18 Assuming that the specific heat of water is 4.18 J K–1 g–1 and ignoring the heat absorbed by the container, Calculate the enthalpy change accompanying the thermometer, stirrer etc., what would be the rise in transformation of C (graphite) to C(diamond). Given that temperature in each of the above cases ? the enthalpies of combustion of graphite and diamond are 393.5 and 395.4 kJ mol–1 respectively. 0.1 Sol. (i) moles of HCl = u 500 = 0.05 = 0.05 mole of H+ ions 1000 Sol. Remember, enthalpy of combustion is always negative 0.2 we are given 200 cm3 of 0.2 M NaOH u 200 1000 (i) C (graphite) + O2 (g)  o CO2(g) ; 'c Ho = – 393.5 kJ mol–1 mole of NaOH = 0.04 mole = 0.04 mole of OH– ions (ii) C (diamond) + O2 (g)  o CO2(g) ; 'c Ho = – 395.4 kJ mol–1 Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 191 In order to get this thermochemical equation, multiply We aim at C(graphite)  o C(diamond), 'trans Ho = ? eqn. (iii) by 2 and add it to eqn. (ii) and then subtract Subtracting eqn. (ii) from eqn. (i), we get eqn. (i) from their sum. We get : C + 2 H2  o CH4, C(graphite – C(diamond)  o 0; 'H = – 393.4 + 2 (–285.7) 'r Ho = – 393.5 – (–395.4) = + 1.9 kJ – (–890.2) kJ mol–1 = –74.6 kJ mol–1 or C(graphite)  o (diamond) ; 'trans H = + 1.9 kJ Hence, the heat of formation of methane is : 'f H = –74.6 kJ mol–1 Example : 19 Example : 21 Calculate the enthalpy of hydration of anhydrous copper sulphate (CuSO 4) into hydrated copper sulphate Calculate the heat of formation of KCl from the following (CuSO4.5H2O). Given that the enthalpies of solutions of data : anhydrous copper sulphate and hydrated copper sulphate are –66.5 and + 11.7 kJ mol–1 respectively (i) KOH (aq) + HCl (aq)  o KCl (aq) + H2O A , 'H = – 57.3 kJ mol–1 Sol. We are given 1 (i) CuSO4 (s) + aq  (ii) H2 (g) + O 2 g  o H 2O A , o CuSO4 (aq) ; 2 1 'sol H1 = – 66.5 kJ mol–1 'H = 286.2 kJ mol 1 1 (ii) CuSO4.5H2O (s) + aq  o CuSO4 (aq) ; (iii) H 2 g  Cl 2 g  aq  o HCl aq , 2 2 'sol H2 = + 11.7 kJ mol–1 'H = – 164.4 kJ mol–1 1 1 We aim at CuSO4 (s) + 5H2O (l)  o CuSO4.5H2O(s) ; (iv) K (s)  O 2 g  H 2 g  aq o 2 2 'hyd H = ? KOH (aq), 'H = – 487.4 kJ mol–1 'H = 'H1 – 'H2 = – 66.5 – (+11.7) = – 78.2 kJ/mol o KCl (aq), 'H = + 18.4 kJ mol–1 (v) KCl (s) + aq  1 Example : 20 Sol. We aim at : K(s) + Cl 2 g  o KCl s , ' f H ? 2 Calculate the enthalpy of formation of methane, given that In order to get this thermochemical equation, we follow the enthalpies of combustion of methane, graphite and the following two steps : hydrogen are 890.2 kJ, 393.4 kJ and 285.7 kJ mol–1 Step 1. Adding Eqns. (iii) and (iv) and subtacting Eq. (v). respectively. we have 1 1 Sol. Remember : Enthalpy of combustion is always negative. K (s) + Cl2 g  H 2 g  O2 g  o KCl s  2 2 We are given : HCl (aq) + KOH (aq) – KCl (aq) 'H = – 487.4 + (–164.4) – (18.4) = – 670.2 kJ mol–1... (vii) o CO2 + 2 H2O, 'H = – 890.2 kJ mol–1 (i) CH4 + 2O2  Step 2. To cancel out the terms of this equation which do not appear in the required equation (vi), add eqn. (i) to eqn. o CO2, 'H = – 393.4 kJ mol–1 (ii) C + O2  (vii) and subtract eqn. (ii) from their sum. This gives 1 1 (iii) H 2  O2  o H 2 O, 'H 285.7 kJ mol1 K s  Cl 2 g  o KCl s ; 2 2 o CH4, 'H = ? We aim at : C + 2H2  'fH = 670.2  57.3  286.2 = – 441.3 kJ Lakshya Educare 192 CHEMICAL THERMODYNAMICS AND ENERGETICS Example : 22 We aim at : C2H4 (g) + 3O2 (g)  o 2CO2 (g) + 2H2O(g) The combustion of 1 mole of benzene takes place at 298 K 'H = – 'H3 + 2'H1 + 2'H2 and 1 atm. After combustion, CO2 (g) and H2O (l) are produced and 3267.0 kJ of heat is liberated. Calculate = –1322.9 kJ mol–1 the standard enthalpy of formation, 'f Ho of benzene. Standard enthalpies of formation of CO2 (g) and H2O (l) Example : 24 are – 393.5 kJ mol–1 and –285.83 kJ mol–1 respectively. Given the following thermochemical equations: Sol. o C6H6 A , 'H = ? Aim : 6 C (s) + 3 H2 (g)  o SO2 (g), 'H (i) S (rhombic) + O2 (g)  ' = – 297.5 kJ mol–1 Given : o SO2 (g), 'H = – 300.0 kJ mol–1 (ii) S(monoclinic) + O2  15 Calculate 'H for the transformation of one gram atom of (i) C6H6 (l) + o 6 CO2 (g) + 3 H2O (l), 'H O (g)  rhombic sulphur into monoclinic sulphur. 2 2 = –3267.0 kJ mol–1 Sol. We aim at S (rhombic) o S (monoclinic), 'H = ? o CO2 (g), 'H = – 393.5 kJ mol–1 (ii) C (s) + O2 (g)  Equation (i) o Equation (ii) gives 1 S(rhombic) – S(monoclinic)  o 0, (iii) H2 (g) + o H2O (l), 'H = –285.83 kJ mol–1 O 2 g  'H = 297.5 – (–300.0) = 2.5 kJ mol–1 2 or o S(monoclinic), 'H = + 2.5 kJ mol–1 S(rhombic)  In order to get the required thermochemical equation, Thus, for the transformation of one gram atom of rhombic multiply Eq. (ii) by 6 and Eq. (iii) by 3 and subtract Eq. sulphur into monoclinic sulphur, 2.5 kJ mol–1 of heat is (i) from their sum, i.e. operating 6 × Eqn. (ii) + 3 × Eqn. absorbed. (iii) – Eqn. (i), we get Example : 25 6C (s) + 3H2 (g)  o C6H6 (l) ; 'H) for BaCl2.2H2O and BaCl2 are Enthalpy of solution (' 'H = 6 (–393.5) + 3 (–285.83) – (–3267.0) 8.8 and –20.6 kJ mol–1 respectively. Calculate the heat of = – 2361 – 857.49 + 3267.0 = 48.51 kJ mol–1 hydration of BaCl2 to BaCl2.2H2O. Thus, the enthalpy of formation of benzene is Sol. We are given 'f H = – 48.51 kJ mol–1 o BaCl2(aq),'sol 'H1o = 8.8 kJ mol–1 (i) BaCl2.2H2O (s) + aq  Example : 23 o BaCl2(aq), 'sol 'H o2 = – 20.6 kJ mol–1 (ii) BaCl2 (s) + aq  Calculate the enthalpy of combustion of ethylene (gas) to We aim at form CO2 (gas) and H2O (gas) at 298 K and 1 atmospheric pressure. The enthalpies of formation of CO2, H2O and C2H4 o BaCl2.2H2O (s), 'hyd Ho = ? BaCl2 (s) + 2H2O  are –393.5, –241.8, + 52.3 kJ per mole respectively. 'H = 'H2 – 'H1 = – 29.4 kJ mol–1 Sol. We are given : Example : 26 (i) C (s) + O2 (g)  o CO2 (g), 'H1o = – 393.5 kJ mol–1 Calculate the enthalpy of hydrogenation of ethylene, given that the enthalpy of combustion of enthylene, hydrogen 1 and ethane are – 1410.0, – 286.2 and – 1560.6 kJ mol–1 (ii) H 2 g  O 2 g  o H2O (g), 'H o2 = – 241.8kJ mol–1 2 respectively at 298 K. Sol. We are given (i) C2H4 (g) + 3O2 (g)  o 2CO2(g) + 2H2O (iii) 2C (s) + 2H2 (g)  o C2H4(g), 'H 3o = + 52.3 kJ mol–1 (l) , 'H = – 1410 kJ mol–1 Mahesh Tutorials Science  CHEMICAL THERMODYNAMICS AND ENERGETICS 193 890.3 kJ of heat is produced from 1 mole of CH4, i.e., 1 12 + 4 = 16 g of CH4 o H2O (l), 'H = – 286.2 kJ mol–1 (ii) H2 (g)  O 2 g  2 ? 445.15 kJ of heat is produced from 8 g of CH4 1 (b) From the given equation, when 890.3 kJ of heat is (iii) C2H6 (g) + 3 O2 g  o 2CO 2 g + 3H2O (l), 'H evolved, CO2 formed = 1 mole = 44 g 2 = – 1560.6 kJ mol–1 ? When 445.15 kJ of heat is evolved, CO2 formed = 22 g (c) From the equation, O2 used in the production of We aim at: o C2H6 (g), 'H = ? C2H4 + H2(g)  890.3 kJ of heat = 2 moles = 2 × 22.4 litres at STP Equation (i) + Equation (ii) – Equation (iii) gives = 44.8 litres at STP Hence, O2 used in the production of 445.15 kJ of heat = C2H4 (g) + H2 (g)  o C2H6 (g), 22.4 litres at STP. 'H = – 1410.0 + (–286.2) – (1560.6) = – 135.6 kJ mol–1 Example : 29 Example : 27 From the thermochemical equation, The thermite reaction used for welding of metals involves the reaction 1 C6 H 6 A  7 O 2 g  o 3H 2 O A  6CO 2 (g), 2 2 Al (s) + Fe2O3 (s)  o Al2O3 (s) + 2Fe (s) 'cH = – 3264.64 kJ mol–1, What is 'Ho at 25oC for this reaction ? Given that the calculate the energy evolved when 39 g of C6H6 are burnt standard heats of formation of Al2O3 and Fe2O3 are in an open container. – 1675.7 kJ and –828.4 kJ mol–1 respectively. Sol. From the given equation, Sol. We aim at 2Al (s) + Fe2O3(s)  o Al2O3 (s) + 2Fe When 1 mole of C6H6 (78 g of C6H6) is burnt, heat evolved (s), 'r Ho = ? = 3264.64 kJ 'r H = Sum of 'f Ho of products – Sum of ? When 39 g of C 6H 6 is burnt, heat evolved = 'f Ho of reactants 3264.64 u 39 = 1632.32 kJ 78 =['f Ho (Al2O3) + 2 × 'f Ho (Fe)] – [2 × 'f Ho (Al) + 'f Ho (Fe2O3)] Example : 30 = [–1675.7 + 0] – [0 + (–828.4)] = – 847.3 kJ mol–1 The thermochemical equation for solid and liquid rocket fuel are given below : Example : 28 1 The heat evolved in the combustion of methane is given by 2Al s  1 O 2 g  o Al2 O3 s ; 2 the equation : 'H 1667.8 kJ mol1 o CO2 (g) + 2H2O (l), 'H = – 890.3 kJ mol CH4 (g) + 2O2 (g)  –1 1 H 2 g  O 2 g  o H 2O 1 ; (a) How many grams of methane would be required to 2 produce 445.15 kJ of heat of combustion ? 'H 285.9 kJ mol1 (b) How many grams of carbon dioxide would be formed (a) If equal masses of aluminium and hydrogen are used, when 445.15 kJ of heat is evolved ? which is a better rocket fuel ? (c) What volume of oxygen at STP would be used in the (b) Determine 'H for the reaction : combustion process (a) or (b) ? 1 Al2O3(s)  o 2Al (s) + 1 O2 g. 2 Sol. (a) From the given equation. Sol. (a) From the first given equation, Lakshya Educare 194 CHEMICAL THERMODYNAMICS AND ENERGETICS 2 moles of Al (i.e., 2 × 27 g = 54 g) on combustion give Writing the reverse reaction, we have heat = 1667.8 kJ 1667.8