*Measuring how fast reactions occur. Lecture 2: Stoichiometry & Rate Laws

Reaction Rates and Stoichiometry To generalize, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t = Reactants decrease, so the rate is negative. Products increase, so the rate is positive.

Chemical Kinetics Reaction Rate - Homework 3 bread + 2 tomato + 2 meat + 1 cheese  4 sandwiches 3.Write the rate in terms of each reactant and product

Reaction Rate - Example

4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(g) 6.a. Write the formulas for rate in terms of each reactant and product: If, in this reaction, water is formed at a rate of 36 M/min: b.at what rate is the ammonia used? c.at what rate is the oxygen used? d.at what rate is the nitrogen dioxide formed?

Answers: (a) 8.4  10 –7 M/s, (b) 2.1  10 –7 M/s PRACTICE EXERCISE 14.3 Relating Rates at Which Products Appear and Reactants Disappear The decomposition of N 2 O 5 proceeds according to the following equation: If the rate of decomposition of N 2 O 5 at a particular instant in a reaction vessel is 4.2  10 –7 M/s, what is the rate of appearance of (a) NO 2, (b) O 2 ?

Rate Laws Differential rate laws express the relationship between the concentration of reactants and the rate of the reaction. Usually just called “the rate law.”

Differential Rate Laws – Effect of Concentration on Rate Rate = k [A] m [B] n m and n are the order of reaction for each reactant k is the rate constant The overall order of the reaction = m + n Note: coefficients are NOT part of the rate law equation. aA + bBcC + dD

Rate Law – Effect of Concentration on Rate Rate = k [A] m [B] n The exponent (order) indicates how the concentration of each reactant affects the rate. First order: m=1 –Ex. If [A] triples (x 3), the rate triples (x 3 1 ) Second order: m=2 –Ex. If [A] doubles (x 2), the rate quadruples (x 2 2 ) aA + bBcC + dD

Rate Laws Rate laws are always determined experimentally. The initial rate is measured for varying concentrations of reactants. By determining how the rate changes, the order of the reactants can be determined.

Rate Law – Effect of Conc. A+ B  C Problem 1 – Consider the initial rate data for the reaction: Experiment[A][B] Initial Rate (M·s -1 ) a.What is the order of reactant A? b. What is the order of reactant B? c. What is the overall order of the reaction? Pick two experiments where only [A] changes ([B] is constant). In experiments 2 &3, [A] doubles (x2) but the rate increases by x8 (x2 3 ). Therefore the order with respect to A is third (m=3). Pick two experiments where only [B] changes ([A] is constant). In experiments 1&2, [B] doubles (x2) and the rate doubles (x2 1 ). Therefore the order with respect to B is first (n=1). m + n = = 4

Rate Law – Effect of Conc. A+ B  C Problem 1 – Consider the initial rate data for the reaction: Experiment[A][B] Initial Rate (M·s -1 ) d. Write the rate equation for the reaction. e. Calculate the value of the specific rate constant, k. Rate = k[A] 3 [B] 1 Use data from any trial and plug it into the rate law equation: 2.0 M/s = k(0.1 M) 3 (0.1 M) 1 k = 20,000 M -3 s -1

Rate Law – Effect of Conc. 3X+ 2Y  Z Problem 2– Consider the initial rate data for the reaction: Experiment[X][Y] Initial Rate (M·s -1 ) a.What is the order of reactant X? b. What is the order of reactant Y? c. What is the overall order of the reaction? Pick two experiments where only [X] changes ([Y] is constant). In experiments 1 &3, [S] triples (x3) and the rate doesn’t change (x3 0 ). Therefore the order with respect to X is zero (m=0). (Rate is not affected by changes in [X]. Pick two experiments where only [Y] changes ([X] is constant). In experiments 1&2, [Y] triples (x3) and the rate changes by x9 (x3 2 ). Therefore the order with respect to Y is second (n=2). m + n = = 2

Rate Law – Effect of Conc. 3X+ 2Y  Z Problem 2– Consider the initial rate data for the reaction: Experiment[X][Y] Initial Rate (M·s -1 ) d. Write the rate equation for the reaction. e. Calculate the value of the specific rate constant, k. f. Calculate the rate when the [X] and [Y] = 0.25 M. Rate = k[X] 0 [Y] 2 = k[Y] 2 Use data from any trial and plug it into the rate law equation: 1.0 M/s = k(0.1 M) 2 k = 100 M -1 s -1 Rate = (100 M -1 s -1 ) (0.25M) 2 = 6.3 M·s -1

Rate Law – Units of Rate Constants The units of the rate constant depend on the overall reaction order. The units of RATE are always M/s (or mol/s). The units of the rate constant have to cancel out the concentration units to make the rate have the correct unit.

Rate Law – Units of Rate Constants

Rate Law – Effect of Conc. 2 NO(g) + Cl 2 (g)  2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO] [Cl 2 ] Initial Rate (M·s -1 ) x x x x 10 -6

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L) Initial Rate (M·s -1 ) x x x x In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. R = k[NO] x [Cl 2 ] y The rate quadruples, so the reaction is second order with respect to [NO]  R = k[NO] 2 [Cl 2 ] y

Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: Experiment[NO](mol/L) [Cl 2 ] (mol/L) Initial Rate (M·s -1 ) x x x x R = k[NO] 2 [Cl 2 ] y In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ]  R = k[NO] 2 [Cl 2 ]

Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO](mol/L) [Cl 2 ] (mol/L) Initial Rate (M·s -1 ) x R = k[NO] 2 [Cl 2 ] 1.43 x x M/s = k(0.250 M) 2 (0.250 M) 1.43 x k = 1.43 x M -2 s -1

Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2+1 = 3  The reaction is 3 rd order