University:
Seattle Pacific University
Course:
CHM 1211 | General Chemistry I
Academic year:

2020
Views:

232

Pages:

3
Author:

rehinhea0c

CHEM RATE OF REACTIONS 1. The table presents data for the reaction: K1 2H2(g) + 2NO(g) → 2H2O(g) + N2(g) The temperature of the reaction is constant. The initial rate is in arbitrary units. a) What happens to the initial rate, when the concentration of NO is doubled and the concentration of H2 is constant? III Î IV, concentration doubled, rate quadrupled (x4) b) What happens to the initial rate, when the concentration of H2 is doubled and the concentration of NO is constant? I Î II, concentration doubled, rate doubled (x2) c) Which of the following equations shows the correct relationship between the initial rate and the concentrations of H2 and NO – use the information from your answers to a and b to represent the right level of proportionality (this equation is what we call the rate law for this reaction)? (A) rate = k [H2] [NO] (C) rate = k [H2]2 [NO] (B) rate = k [H2]2 [NO]2 (D) rate = k [H2] [NO]2 d) What is the value of the proportionality constant k (which is called the rate constant)? k = 12 /{(6.0·10-3)(2.0·10-3)2} = 5.0·108 1 CHEM 2. The reaction below was studied and produced the results in the table. 2 ClO2 + 2 OH- → ClO3- + ClO2- + H2O a) Determine the rate law for the reaction. 2 => 1, when [ClO2] is tripled and [OH-] is held constant, the rate increases by a factor of 9 (0.0248/0.00276). 3x = 9; x = 2. It is second order in ClO2. 2 => 3, when [OH-] is tripled and [ClO2] is held constant, the rate increases by a factor of 3 (0.00828/0.00276). 3x = 3; x = 1. It is first order in OH-. Rate = k1 [ClO2]2 [OH-] b) Calculate the rate constant. k = 0.0248 M/s-1 / {(0.060)2(0.030)} = 2.3·102 M-2·s-1 c) Calculate the rate when [ClO2] = 0.010 M and [OH-] = 0.025 M. Rate = (2.3·102)(0.010)2(0.025) = 5.7·10-4 M/s-1 2 CHEM 3. The following data were obtained for the reaction A + B + C → products: Experiment [A]0 [B]0 [C]0 Initial rate, v0 (mol L-1 s-1) 1 2 3 4 5 1.25 x 10-3 M 2.50 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 3.01 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 3.02 x 10-3 M 3.02 x 10-3 M 1.00 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 1.25 x 10-3 M 3.75 x 10-3 M 1.15 x 10-3 M 0.0087 0.0174 0.0508 0.457 ? a) Write the rate law for the reaction. Explain your reasoning in arriving at your rate law. 1 => 2, when [A] is doubled and [B] and [C] are held constant, the rate increases by a factor of 2 (0.0174/0.0087). 2x = 2; x = 1. It is first order in A. 1 => 3, when [B] is increased by a factor of 2.4 and [A] and [C] are held constant, the rate increases by a factor of 5.8 (0.0508/0.0087). 2.4x = 5.8; x = 2. It is second order in B. 3 => 4, when [C] is increased by a factor of 3 and [A] and [B] are held constant, the rate increases by a factor of 9 (0.457/0.0508). 3x = 9; x = 2. It is second order in C. Rate = k1 [A][B]2 [C]2 b) Determine the value of the rate constant. k = (0.0087) / {(1.25·10-3)(1.25·10-3)2(1.25·10-3)2} = 2.85·1012 L4mol-4s-1 c) Use the data to predict the reaction rate for experiment 5. Rate = (2.85·1012 L4mol-4s-1) (3.01·10-3)(1.00·10-3)2(1.15·10-3)2 = 0.0113 mol·L-1s-1 3