Pharmaceutical Analytical Chemistry 1 PDF

Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) Uncertainty of measurements Significant figures Mole Dr. Mahmoud A. Tantawy Assoc.Prof. of Analytical Chemistry...

Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) Uncertainty of measurements Significant figures Mole Dr. Mahmoud A. Tantawy Assoc.Prof. of Analytical Chemistry Faculty of Pharmacy Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) 1- What is the international system of units (SI units)? What are the SI derived units? International system of units: (SI Units) - It has seven base quantities & base units from which all other quantities can be derived: Quantity Unit Symbol Length Meter m Mass Kilogram kg Time Second s Temperature Kelvin K Amount of substance mole mol Electric current ampere A Luminous intensity candela cd Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) Derived SI units :- Quantity Symb. Expression SI unit Area A Length2 m2 Volume V Length3 m3 Density D Mass/volume kg.m-3 Velocity v Distance / time m.s-1 Acceleration A Velocity/ time m.s-2 Force F Mass x kg.m.s-2(newton N) acceleration pressure P Force/area kg.m-1.s-2(pascal pa) Energy E Force x kg.m2.s-2(joule J) distance Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) 2-What are the SI prefixes? Multiple Prefix Symbol 106 mega M 103 kilo k 102 hecto h 101 deka da 10-1 deci d 10-2 centi c 10-3 milli m 10-6 micro µ 10-9 nano n 10-15 femto f Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) 3-Some common non-SI units: 1- Volume in litre (L) 1mL (cm3) = 10-6 m3 1 L = 1000 mL = 1000 cm3 = 1000 x (10-2m)3= 10-3 m3 2- Temperature in centigrade (oC) ToK = toC + 273 toC = ToK- 273 3- Pressure in atmosphere (atm) 1 atm = 101.325 K.pascal 4- Energy in calories (cal) 1 cal = 4.148 J Pharmaceutical Analytical Chemistry 1 Units and measurements (SI units) 4-Example on Units and Measurement Example 1: A substance has a mass = 250 g and its volume = 1500 mL. Calculate the density in g/cm3 & SI units. Soln D=mass / volume = 250 g / 1500 mL = 0.166 g/mL = 0.166 g/cm3 -In SI units: Mass :250 g = 250 x 10-3 = 0.25 kg Volume 1500 cm3 = 1500 x 10-6 = 0.0015 m3 Therefore D = 0.25 / 0.0015 = 166.66 kg / m3 Pharmaceutical Analytical Chemistry 1 Uncertainty of measurements - No physical quantity can be measured with perfect certainty. - There are always errors (uncertainty) in any measurement. - Each measurement should be reported with some certain digits plus one uncertain digit. 40.51, 40.52, 40.53 Pharmaceutical Analytical Chemistry 1 Uncertainty of measurements Uncertainty depends on accuracy and precision of the measuring device and person using it. Accuracy - It is the measure of exactness of an analytical method or the closeness of agreement between the measured value and the value that is accepted either as a conventional , true value or an accepted reference value. - For example, if you used a balance to find the mass of a known standard 100.00 g mass, and you got a reading of 78.55 g, your measurement would not be accurate. Pharmaceutical Analytical Chemistry 1 Uncertainty of measurements Precision - Precision refers to how close together a group of measurements. - Precision is sometimes referred to as repeatability or reproducibility. Pharmaceutical Analytical Chemistry 1 Uncertainty of measurements N.B. One important distinction between accuracy and precision is that: - Accuracy could be determined from one measurement. - Precision needs multiple measurements. Pharmaceutical Analytical Chemistry 1 Significant figures Significant Figures - The number of significant figures is the number of figures (numbers) that are known with some degree of reliability. - The number 13.2 is said to have 3 significant figures. The number 13.20 is said to have 4 significant figures. Pharmaceutical Analytical Chemistry 1 Significant figures Rules for Significant Digits: 1 All non-zero digits are considered significant. - e.g. 91 has two significant digits (9 and 1), while 123.45 has five significant digits (1, 2, 3, 4 and 5). 2 Zeros; there are three classes of zeros: a. Leading zeros: - Not count as significant figures. - E.g. In the number 0.0034, there are only two significant figures. b. Captive zeros: - Are zeros between nonzero digits. - Count as significant figures. - E.g. 1.007 has 4 significant figures. Pharmaceutical Analytical Chemistry 1 Significant figures c. Trailing zeros: - are zeros at the right end. - They are significant only if the number contains a decimal point. - E.g. The number 100 has only one significant figure. - 1.00 has three significant figures. Pharmaceutical Analytical Chemistry 1 Significant figures Rules for rounding off numbers: (1) If the digit to be dropped is greater than 5, the last retained digit is increased by one. - e.g. 12.6 is rounded to 13. (2)If the digit to be dropped is less than 5, the last remaining digit is left as it is. - e.g. 12.4 is rounded to 12. (3) If the digit to be dropped is 5, and if any digit following it is not zero, the last remaining digit is increased by one. - e.g. 12.51 is rounded to 13. (4) If the digit to be dropped is 5 and is followed only by zeroes, the last remaining digit is increased by one if it is odd, but left as it is if even. - e.g. For example, 11.5 is rounded to 12, 12.5 is rounded to 12. Pharmaceutical Analytical Chemistry 1 Significant figures Questions 1-How many significant figures are there in each of the following values: Answer a- 6.07 x10-2 a-3 b- 0.003840 b-4 2-Round off each of the following numbers to specified significant figures: Answer a- 0.00034159 to three digits A-0.000342 b-103.351 x 102 to four digits b-103.4 x 102 c-17.9915 to five digits c-17.992 d-3.365 x 105 to three digits d-3.36 x 105 e- 4.348 to two digits e- 4.3 Pharmaceutical Analytical Chemistry 1 The Mole and molar mass The Mole - Is the quantity of a given substance that contains as many molecules or formula units as the number of atoms in exactly 12 g carbon. e.g. number of ethanol molecules in 1 mole ethanol = number of carbon atoms in 12 g carbon (which is called Avogadro's number NA). NA = 6.022 X 1023 - A mole of a substance contains Avogadro's number (6.022 X 1023) of molecules (or formula units). - E.g. a mole of O atoms contains 6.022 X 1023 O atoms. A mole of O2 molecules contains 6.022 X1023 molecules, that is 2 X 6.022 X 1023 O atoms. Pharmaceutical Analytical Chemistry 1 The Mole and molar mass - The molar mass of a substance is the mass of one mole of the substance. - Carbon has molar mass of 12 g / mol. - For all substances, the molar mass (formula weight or molecular weight) in grams (g) is numerically equal to the weight in atomic mass units (amu). - E.g. ethanol has molecular weight of 46.1 amu (or g) and a molar mass of 46.1 g / mol. Pharmaceutical Analytical Chemistry 1 The Mole and molar mass To Calculate the Number of moles (n): - n = N / NA n= number of moles. N= number of molecules, ions or atoms. NA= Avagadro’s number (6.022 x 1023) - Or n = m / M.wt n= number of moles. m= mass in grams. M.wt= molecular weight. Example 1: How many grams of Cu are there in 2.55 mole of Cu? Solution: 1 mole Cu = 63.5 g Cu , so 2.55 mole Cu contains 162 g Cu Pharmaceutical Analytical Chemistry 1 The Mole and molar mass Example 2: What number of moles of aluminum is present in 125 g of Al ? Solution: 1 mole Al = 27.0 g Al ….. Mole = 125 g = 4.63 mole Example 3: How many carbon atoms are there in a 1.0 carat diamond? Diamond is pure carbon and one carat is exactly 0.2 g. Solution: 1 mole C = 12.01 g C …. Mole = 0.2 g 0.01665 mole 1 mole C = 6.022 X 1023 atoms C 0.01665 mole = ……….. Atoms 1.003 X 1022 atoms C Pharmaceutical Analytical Chemistry 1 Percentage composition of compounds Empirical formula and Molecular formula Stoichiometry Limiting reactant %Yield General problems Dr. Mahmoud A. Tantawy Associate Professor Faculty of Pharmacy Pharmaceutical Analytical Chemistry 1 Percentage composition of compounds I- Percentage composition of compounds - Suppose that A is part of something. - We define the mass percentage of A as follows: - Mass% A = (mass of A in the whole / mass of the whole) × 100 Example 1: What is the percentage of Fe in Fe2 O3? Solution: One mole of Fe2 O3 contains: 2 mole Fe = 2 X 55.8 g Fe = 111.6 g Fe 3mole O = 3 X16.0 g O The sum of the masses of one mole = 159.6 g Fe2 O3 The percentage of Fe in Fe2 O3 is = 69.92% Fe Pharmaceutical Analytical Chemistry 1 Empirical formula and Molecular formula II- Empirical formula and molecular formula - The simplest formula is called the empirical formula. - The multiple of the empirical formula of a compound is the molecular formula. - Molecular weight = n × empirical formula weight - n = molecular weight / empirical formula weight Pharmaceutical Analytical Chemistry 1 Empirical formula and Molecular formula Benzene Cyclohexane Molecular formula: C6H6 Molecular formula: C6H12 Empirical formula: CH Empirical formula: CH2 C6H6 = 6 (CH) or (CH) 6 C6H12 = 6 (CH2) or (CH2) 6 Acetic acid CH3COOH Molecular formula: C2H4O2 Empirical formula: CH2O C2H4O2 = 2 (CH2O) or (CH2O) 2 Pharmaceutical Analytical Chemistry 1 Empirical formula and Molecular formula Example 2: A 1.26 g sample of pure caffeine contains 0.624 g C, 0.065 g H, 0.364 g N and 0.208 g O. What is the empirical formula of Caffeine, and if the molecular weight of caffeine is 194 g. What is the molecular formula? Pharmaceutical Analytical Chemistry 1 Empirical formula and Molecular formula Solution: We calculate the number of moles of each element present. ? mole C = 0.624 g C x 1 mol C / 12.0 g C = 0.052 mol C ? mole H = 0.065 g H x 1 mol H / 1.0 g H = 0.065 mol H ? mol N = 0.364 g N x 1 mol N / 14.0 g N = 0.026 mol N ? mole O = 0.208 g O x 1 mol O / 16.0 g O = 0.013 mol O - Division of each of these values by the smallest value (0.013) gives the ratio. 4 mole C : 5 mole H : 2 mole N : 1 mol O The empirical formula of caffeine is C4H5N2O. The empirical formula weight indicated by C4H5N2O is 97. n=2  The molecular formula of caffeine is C8H10N4O2 Pharmaceutical Analytical Chemistry 1 Stoichiometry III- Stoichiometry - The quantitative relationships between reactants and products in a balanced equation are known as stoichiometry. - The chemical equation for the combustion of ethanol, C2H5OH C2H5OH + 3O2 2CO2 + 3H2O 1 mole C2H5OH + 3 mole O2 2mole CO2 + 3 mole H2O - In this example, we can write six equivalencies: 1 mole C2H5OH  3 mole O2. 1 mole C2H5OH  2 mole CO2. 1mole C2H5OH  3 mole H2O. 3 mole O2  2 mole CO2. 3 mole O2  3 mole H2O. 2 mole CO2  3 mole H2O. Pharmaceutical Analytical Chemistry 1 Stoichiometry Example 3: a) How many moles of O2 are needed to burn 1.8 mole C2H5OH? b) How many moles of water will form when 3.66 mole CO2 are produced during the combustion of C2H5OH? Solution: a) 1 mole C2H5OH  3 mole O2 O2 needed for combustion = 1.8 x 3/1 = 5.4 mole O2 b) 2 mole CO2 ~ 3 mole H2O H2O formed = 3.66 x 3/2 = 5.49 mole H2O Pharmaceutical Analytical Chemistry 1 Limiting reactant IV- Limiting reactant - Often reactants are added to a reaction vessel in different amounts. - In that case, only one of the reactant is completely consumed at the end of the reaction. - The limiting reactant is the reactant that is entirely consumed when a reaction goes to completion. - Other reactants are in excess and remain partially un-reacted. - The number of moles of a product is always determined by the number of moles of a limiting reactant. Pharmaceutical Analytical Chemistry 1 Limiting reactant Example 4: How many moles of H2 can be prepared from 4 moles of Fe and 5 moles of H2O? 3 Fe + 4H2O Fe3 O4 + 4H2 Solution: - The first step is to determine which reactant limits the reaction. - 3 moles Fe = 4 moles H2 - 4 moles Fe = X moles H2 - X = 5.33 moles H2 - For H2O, 4 moles = 4 moles H2 - 5 moles = Y moles H2 - Y = 5 moles H2 , water is the limiting reactant. Pharmaceutical Analytical Chemistry 1 %Yield V- Theoretical and percentage yields - Frequently, the quantity of a product actually obtained from the reaction is less than the amount calculated. - The percent yield relates the amount of product that is actually obtained (the actual yield) to the amount of product from theoretical calculation (the theoretical yield). - Percent yield = (actual yield / theoretical yield) × 100 Pharmaceutical Analytical Chemistry 1 %Yield Example 5: How many grams of N2F4 can theoretically be prepared from 4.0 g of NH3 14.0 g of F2? The chemical equation for the reaction is: 2 NH3 + 5 F2 N2F4 + 6 HF , and if 4.8 g of N2F4 is obtained from the experiment, what is the percent yields? (Mol.Wt. NH3 = 17 g, F2 = 38 g) Solution: ? mole NH3 = 4.0/ 17 = 0.235 mole NH3 ? mole F2 = 14.0 / 38.0 = 0.368 mole F2 The F2, therefore is the limiting reactant. 2NH3 + 5F2 N2F4 + 6HF 190 104 14 ? ? g N2F4 = 14 x 104 / 190 = 7.66 g N2F4 (theoretical yield) Percent yield = (4.8 / 7.66) × 100 = 62.7% Pharmaceutical Analytical Chemistry 1 General problems Problem 1 What is the empirical formula of a compound which has a mass % composition of 50.05% S and 49.95% O? Step 1: Find the atomic masses of the elements present Sulfur (S) : 32.066 gmol-1 Oxygen (O) : 16.000 gmol-1 Pharmaceutical Analytical Chemistry 1 General problems Step 2: Determine the number of moles of each element present Since we are dealing with percentages, we can express the mass % as grams if we assume we have 100g of the compound. Therefore, 100g of our compound contains 50.05g of sulfur and 49.95g of oxygen. Convert number of grams to number of moles Number of mol Sulfur = mass of sulfur in sample (g) atomic mass of sulfur (gmol-1) = 50.05g = 1.56 mol. 32.066 gmol-1 Similarly, the no. of mol of Oxygen is found to be 3.12mol Step 3: Determining the ratios of elements Sulfur: 1.56mol Ratio 1.56 : 3.12 Oxygen: 3.12mol Ratio must be in whole numbers. Here we must divide across by 1.56 Therefore, we have a ratio of 1:2 giving an empirical formula of SO2 Pharmaceutical Analytical Chemistry 1 General problems Problem 2 Benzene (C6H6) reacts with Bromine to produce bromobenzene (C6H6Br) and hydrobromic acid. If 30. g of benzene reacts with 65 g of bromine and produces 56.7 g of bromobenzene, what is the percent yield of the reaction? C6H6 + Br2 ------> C6H5Br + HBr 30.g 65 g 56.7 g 78g/mol 160.g/mol 157g/mol 30.g(1mol/78g) 65g(1mol/160g) 0.38 mol 0.41 mol (If Br2 limiting) 0.41 mol 0.41 mol (If C6H6 limiting) 0.38 mol 0.38 mol 0.38mol(157g/1mol) = 60.g 56.7g/60.g(100)=94.5%=95% Pharmaceutical Analytical Chemistry 1 CHEMICAL EQUILIBRIUM Classification of reactions Law of mass action The relation between Kc and Kp Predicting the direction of a reaction Le Chatelier’s principle Dr. Mahmoud A. Tantawy Associate Professor Faculty of Pharmacy Pharmaceutical Analytical Chemistry 1 Classification of reactions I- Classification of reactions 1- Homogenous reactions Only one phase is present. e.g. in gas phase H2(g) + I2(g) 2 HI(g) 2- Heterogeneous reactions The mixture is not uniform e.g. C2H5OH(L) Alumina C2H4(g) + H2O(L) Pharmaceutical Analytical Chemistry 1 Classification of reactions I- Classification of reactions 1-Irreversible reaction Reactions proceed in one direction. i.e from reactants to products A+B AB 2- Reversible reactions Reactions proceed in either direction forward or backward e.g. N2(g) + 3H2(g) 2NH3(g) Pharmaceutical Analytical Chemistry 1 Classification of reactions N2 + 3H2 2NH3 The double arrow tells us that this reaction can go in both directions N2 + 3H2 2NH3 1) Reactants react to become products, N2 + 3H2 f 2NH3 (‘forward’ reaction) while simultaneously, 2) Products react to become reactants N2 + 3H2 b 2NH3 (‘reverse’ or backward reaction) Pharmaceutical Analytical Chemistry 1 Classification of reactions For a reversible chemical reaction: An equilibrium state is attained when the rate at which the forward reaction is proceeding equals the rate at which the reverse reaction is proceeding. Pharmaceutical Analytical Chemistry 1 Law of mass action II- Law of mass action Consider the following reaction jA (g) + kB (g) mC(g) + nD(g) rate(f)= Kf [A] j [B] k rate(b) = Kb [C] m [D] n At equilibrium rate(f) = rate(b)  Kf [A] j [B] k = Kb [C] m [D] n m n Kf [C ] [ D] = = K (Kc) equilibrium Kb j k constant [ A] [ B ] Pharmaceutical Analytical Chemistry 1 Law of mass action - The numerical value of K varies with temperature. - It dose not vary with changes in the amounts of substances used to establish the equilibrium, changes in pressure, or the presence of catalyst. - N.B. If Kc is a large number, the forward reaction is fairly complete. If Kc is a small number, the reverse reaction is fairly complete. Pharmaceutical Analytical Chemistry 1 Law of mass action If the reaction occurs in more than one step: e.g. H2S H+ + HS- K1 = [ H ][ HS ] [H 2S ] [ H  ][ S  ] HS- H+ + S-2 K2 = [ HS ] Kc= K1 K2 Pharmaceutical Analytical Chemistry 1 Law of mass action If equilibrium involves more than one phase it is called heterogeneous equilibrium The concentration of pure solid or pure liquid is constant and do not appear in the expression for equilibrium constant. Ex : CaCO3(s) CaO(S) + CO2(g) Kc = [CO2] Ex : 3Fe(S) + 4 H2O(g) Fe3O4(s) + 4H2(g) [ H 2 ]4 Kc = [ H 2 O]4 Pharmaceutical Analytical Chemistry 1 Law of mass action Example : For reaction N2O4(g) 2NO2(g) The concentration of substances present in an equilibrium mixture at 25oc are [N2O4] = 4.27 x 10-2 mol/L, [NO2] = 1.4x10-2 mol/L. What is the value of Kc for this reaction? Solution [ NO2 ] 2 (1.4x10 2 mole/L) 2 Kc =  2  4.66x10-3 mole/L [ N 2 O4 ] (4.27x10 mole/L) Pharmaceutical Analytical Chemistry 1 The relation between Kc and Kp III- The relation between Kc and Kp For reactions involving gases: The partial pressures of the reactants and products are proportional to their molar concentrations Example: N2(g) + 3H2(g) 2NH3(g) Kc = [NH3]2 Kp = P2NH3 [N2] [H2]3 PN2.P3H2 Kp = P2(NH3) P(N2) P3(H2) Kp = Kc (RT)n(g) Where: n(g) = (no of moles of gaseous products – no of moles of gaseous reactants.) Pharmaceutical Analytical Chemistry 1 The relation between Kc and Kp Example: 2SO3(g) 2 SO2(g) + O2(g) at 11000 k Kc is 0.0271 mol / L what is Kp at the same temperature? Solution n = 3 – 2 = 1 Kp = Kc (RT)+1 = 0.0271 mol/L x 0.0821 L.atm / k.mol x 1100 k = 2.45 atm Pharmaceutical Analytical Chemistry 1 Predicting the direction of a reaction IV- Predicting the direction of a reaction Reaction Quotient (Q) For reaction PCl5(g) PCl3(g) + Cl2(g) at 2500C Is the reaction in equilibrium ? Will it go forward or backward ? To know this: a value obtained by substituting with the initial concentration into the expression of equilibrium constant called (reaction quotient) (Q) For this example Q = [PCl3]o [Cl2]o [PCl5]o Pharmaceutical Analytical Chemistry 1 Predicting the direction of a reaction To know the direction compare between the value of Q & Kc: 1- If Q < Kc The reaction will move from left to right, i.e forward reaction to approach equilibrium. 2- If Q = Kc  The system is in equilibrium 3- If Q > Kc  The reaction will move from right to left i.e backward reaction Pharmaceutical Analytical Chemistry 1 Predicting the direction of a reaction Example : For the reaction 2SO2(g) + O2(g) 2SO3(g) at 827oc Kc = 36.9 mol/L If 0.05 mol of SO2(g), 0.03 mol of O2(g) & 0.125 mol of SO3(g) are mixed in 1 liter container at 827oc. In what direction will the reaction proceed? Solution [ SO3 ]2 (0.125mol/L) 2 Q = [ SO ]2[O ](0.05mol/L) 2(0.03mol / L) 208 Lmol /L / mol 2 2 Since Q (208 mol/L) > Kc(3 6.9 mol/L) The reaction will proceed from right to left. i.e SO3 will dissociate (i.e backward reaction). Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle V- Le Chatelier’s principle It predicts how a system in equilibrium will respond to changes in experimental conditions (such as temperature or pressure). “A system in equilibrium reacts to a stress in a way that counteracts the stress and establishes a new equilibrium state.” Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle 1- Concentration changes If a component is added to a system at equilibrium, the equilibrium position will shift in the direction that lowers the concentration of that component. If a component is removed the opposite effect occurs. Example H2(g) + I2(g) 2 HI(g) So increase H2 or I2 will shift the reaction towards the formation of HI (Forward reaction) Removal of H2 & I2 leads to decomposition of HI (backward reaction) Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle 2- Pressure changes Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas. Ex SO2(g) + 2 O2(g) 2 SO3(g) 3 moles 2 moles If pressure increased (or volume decreased), the position of equilibrium is shifted to the right & vise versa. Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle N.B - For reactions in which n = 0, pressure changes have no effect on the position of equilibrium. Ex N2(g) + O2(g) 2NO(g) - For heterogeneous equilibrium; the effect of pressure is predicated by counting the number of moles of gas indicated on each side of the equation. Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle 3- Temperature changes For exothermic reactions: Eg: N2 + 3H2 2NH3 H = -92 KJ Since H is -ve, the reaction will evolve heat. N2(g) + 3H2(g) 2NH3(g) + 92KJ Inc temp will shift the reaction from right to left. i.e backward For endothermic reactions: CO2(g) + H2(g) CO(g) + H2O(g) H = + 41.2 KJ CO2 + H2 + 41KJ CO + H2O Inc temp will shift the reaction from left to right. i.e foreword Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle Increasing the temp will favor the endothermic reaction & decreasing the temp will favor the exothermic reaction N.B The numerical value of K will change by change in temp. Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle 4- Effect of catalyst The addition of a catalyst causes a system to achieve equilibrium faster but dose not alter the position of equilibrium. 5- Addition of an inert gas No effect on position of equilibrium. Example: Predict how the value of K changes as the temperature is increased: 1- N2(g) + O2(g) 2NO (g) Ho = 181 KJ 2- 2SO2 (g) + O2(g) 2SO3 (g) Ho = -198 KJ Answer: 1-the reaction shifts right and K increases 2-the reaction shifts to the left and K decreases Pharmaceutical Analytical Chemistry 1 Le Chatelier’s principle Problem: How the following changes affect the equilibrium position of the following reaction: N2O4(g) 2NO2(g) Ho = 58 KJ Change Shift Addition of N2O4 (g) Right Addition of NO2 (g) Left Removal of N2O4 (g) Left Removal of NO2 (g) Right Addition of He (g) None Decrease container volume Left Increase container volume Right Increase temperature Right Decrease temperature Left Pharmaceutical Analytical Chemistry 1 Problem No1. At 500 K, 1.0 mol of NOCl(g) is introduced into a one liter container. At equilibrium the NOCl(g) is 9.0% dissociated: 2 NOCl(g) 2 NO(g) + Cl2(g) Calculate the value of Kc for equilibrium at 500 K. Pharmaceutical Analytical Chemistry 1 Problem 2 Pharmaceutical Analytical Chemistry 1 Problem No2. Pharmaceutical Analytical Chemistry 1 Problem Pharmaceutical Analytical Chemistry 1 Problem Pharmaceutical Analytical Chemistry 1 Chemical Kinetics Reaction rates & their measurements Factors affecting the rate of chemical reaction Rate laws Half life time Dr. Mahmoud A. Tantawy Associate Professor Faculty of Pharmacy Pharmaceutical Analytical Chemistry 1 Reaction rates & their measurements Chemical kinetics is the study of REACTION RATES and their relation to the way the reaction proceeds. I- Reaction rates & their measurements Reaction Rate is the change in concentration of a reactant or product with time. The rate is usually expressed in moles/L.s. Consider the following reaction: NO + O3 NO2 + O2  [ NO]  [O3 ] Rate of the reaction =  t t Rate of the reaction =  [ NO2 ] [O2 ]  t t Pharmaceutical Analytical Chemistry 1 Reaction rates & their measurements Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. ∆𝐴 Rate = -½. = ∆𝐵/∆𝑡 ∆𝑡 Example 1: 4NH3(g) + 5O2(g) 4 NO(g) + 6 H2O(g) Suppose that at particular moment during reaction, ammonia reacts at the rate of 0.24 mole.L-1.s-1. a.What is the rate at which O2 react ? b. What is the rate at which H2O is formed ? Pharmaceutical Analytical Chemistry 1 Factors affecting the rate of chemical reaction II- Factors affecting the rate of chemical reaction 1) The nature of the reactants and products (e.g.) - White phosphorus ignites when exposed to oxygen in the air. By contrast, red phosphorus can be kept in open containers for long periods 2) Surface area of the reactants (S.A) - Powdered chalk (mostly calcium carbonate, CaCO3) reacts rapidly with dilute HCl because it has a large total surface area. A stick of chalk has a much smaller surface area, so it reacts much more slowly. Rate  S.A.  1/ particle size 3) Temperature Rate  temperature 4) Catalyst The presence of the catalyst will increase the rate of the chemical reaction. 5) The concentrations of the reacting species - Rate of chemical reaction is  concentration. - As the concentration increases, the collision between molecules increases, so the reaction goes faster. Pharmaceutical Analytical Chemistry 1 Rate laws III- Rate Laws - Generally, if we follow a chemical reaction over a period of time, we find that its rate gradually decreases as the reactants are consumed. - the rate is nearly always directly proportional to the concentration of the reactants raised to some power. AB Rate  [A]x Where, x is called the order of the reaction. r = K [A]x r : is the rate of the reaction K: is the rate constant that is affected by temp. Pharmaceutical Analytical Chemistry 1 Rate laws 1st order 2nd order 3rd order zero order reaction reaction reaction reaction If x = 1 If x = 2 If x = 3 If x = 0 r1 = K [A] r1 = K [A]2 r1 = K [A]3 r1 = K [A]0 i.e r1 = K if we double r2 = K[2A]1 = 2 r2 = K [2A]2 = 4 r2 = K [2A]3 = 8 r2 = r1 = K the concn. KA K A2 K A3 of A ,i.e., r2 = 2 r1 r2 = 4r1 r2 = 8 r1 becomes 2A The rate is also The rate The rate The rate of the doubled becomes 4 becomes 8 reaction is not times the times the affected by initial one initial one concn. Pharmaceutical Analytical Chemistry 1 Rate laws If A B Rate  [A]x x = 1 in 1st order, 2 in 2nd order, 0 in zero order Rate = K [A]x K: Rate constant affected by temp. If A+B+C D If the order of the reaction of A = x If the order of the reaction of B = y If the order of the reaction of C = z  r = K AX BY CZ & the overall order of the reaction = x + y + z N.B. The values of x and k can only be determined from experiment. Pharmaceutical Analytical Chemistry 1 Rate laws Example 2: Below some data collected in a series of experiments on the reaction of nitric oxide with Br2 2 NO(g) + Br2(g) 2NOBr(g) at 273oc Initial concentration mol/L Rates mol.L- Experiment 1.s-1 NO Br2 1 0.1 0.1 12 2 0.1 0.2 24 3 0.1 0.3 36 4 0.2 0.1 48 5 0.3 0.1 108 Determine the rate law for the reaction and compute the value of the rate constant. N.B. r (rate law) cannot be determined from the equation alone. Pharmaceutical Analytical Chemistry 1 Rate laws Solution Rate law of the reaction r = K [NO]x [Br2]y r1 = K[NO]1x [Br2]1y r2 = K[NO]2x [Br2]2y To know the value of x from the table, chose 2 experiments such that the conc of Br2 is equal and NO is different r4 0.2 x 48 4 2 For NO ( )   ( )  ( ) 2  2x = 4 = 22 x=2 r1 0.1 12 1 1 To know the value of y from the table, chose 2 experiments such that the conc of NO is equal and Br2 is different For Br2 r2 0.2 y 24 ( )  2y = 21 y=1 r1 0.1 12 N.B. Always the larger r in the numerator & the smaller in the denominator r = K [NO]2.[Br2] To know the value of K substitute in any experiment 12mol.L1.s 1 4 2  2 1 K= 3 3 3 1.24x10 L.mol.s 1x10 mol.L Pharmaceutical Analytical Chemistry 1 Rate laws Example 3: The following data were collected for the reaction of t-butyl bromide with hydroxide ion at 35oC (CH3)3 C Br + OH- (CH3)3 C OH + Br- Initial concentration mol/L Rates mol.L-1.s-1 Experiment (CH3)3 C Br OH- 1 0.1 0.1 0.001 2 0.2 0.1 0.002 3 0.3 0.1 0.003 4 0.1 0.2 0.001 5 0.1 0.3 0.001 What is the rate law and rate constant of this reaction? Pharmaceutical Analytical Chemistry 1 Rate laws Solution Rate = K [(CH3)3CBr]x [OH]y r2 0.2 x 0.002 ( )  For (CH3)3C Br: r1 0.1 0.001 x=1 i.e the reaction is 1st order with respect to [(CH3)3C Br] In experiments 1, 4, 5 the concentration of [(CH3)3C Br] is the same, but the concentration of [OH] is changing and has no effect on the rate. y=0 Therefore the reaction is zero order with respect to [OH]  r = K [(CH3)3C Br]1 [OH]0 r = K [(CH3)3CBr] 0.001 mol.L-1.s-1 = (K) (0.1) (mol / L).  K = 0.001 / 0.1 = 0.01 s-1 Pharmaceutical Analytical Chemistry 1 Rate laws Example 4: For the reaction I- + CIO- OH- IO- + Cl- In order to use the initial rate method to obtain the rate law for this reaction, the following experiments were run: Initial concentration mol/L Rates mol.L-1.s-1 Experiment I- ClO- OH- 1 0.01 0.01 0.01 6.1 x 10-4 2 0.02 0.01 0.01 12.2 x 10-4 3 0.01 0.02 0.01 12.2 x 10-4 4 0.01 0.01 0.02 3 x 10-4 What is the rate law & rate constant for this reaction? Pharmaceutical Analytical Chemistry 1 Rate laws Solution l- From exp l & 2 r2 0.02 x 12.2 x10 4 ( )   21 r1 0.01 6.1x10 4 x=1 i.e the reaction is 1st order for I- 2- From exp 1& 3 r3 0.02 y 12.2 x10 4 ( )   21 r1 0.01 6.1x10 4 y=1 i.e the reaction is 1st order for C1O- r4 0.02 z 3x10 4 1 3- From exp l & 4 ( )   r1 0.01 6.1x10 4 2  z = -1 i.e the reaction is inversely proportional for (OH) = [OH]-1 K (0.01mol/ L)(0.01mol/ L) 6.1x10-4mol.L-1.s-1 = 0.01mol / L  K=6.1x10-2 s-1 The overall order of the reaction = 1 + 1-1 = 1 Pharmaceutical Analytical Chemistry 1 Half life time IV- Concentration & time: (half lives) Half life: t1/2 “It is the length of time required to decrease the concentration of the reactant to 1/2 its initial value.” i.e when t = t1/2 [C] = 1/2[C]o Order of the Rate law Relation of t1/2 K units reaction concentration & time Zero order r=K mol / L.s First order r = K [C] s-1 Second order r = K [C]2 L/mol.s Where C0 is the initial concentration, C is the concentration at time t N.B. t1/2 in first order reaction is not affected by concentration. Pharmaceutical Analytical Chemistry 1 Half life time Example 5: Considering the reaction: A B The initial concentration of A was 5 x 10-3 mol.L-1, and after 30 min the concentration was 5 x 10-4 mol.L-1. Assuming that reaction is first order, calculate: a) Half life time of the reaction b) Concentration of A after 2 hr c) When will the concentration of A reach 1 x 10-4 mol.L-1 d) When will the concentration of B reach 2.5 x 10-3 mol.L-1 Pharmaceutical Analytical Chemistry 1 Half life time Solution 4.8 × 10-7 mol/L 50.8 min Pharmaceutical Analytical Chemistry 1 Half life time Example 6: Considering the reaction: X Y The initial concentration of X was 8 x 10-2 mol.L-1, and after 20 min the concentration was 6 x 10-2 mol.L-1. Assuming that reaction is zero order, calculate: a) Half life time of the reaction b) Concentration of X after 1 hr c) When will the concentration of X reach 1 x 10-2 mol.L-1 d) When will the concentration of Y reach 1 x 10-2 mol.L-1 Pharmaceutical Analytical Chemistry 1 Half life time Solution