Question

C. Given the following data for the reaction A+B products and assuming that the order of each reactant does not change with an increase in temperature, answer the following questions. B] 0.100 MO rate (moL-hr) 0.0375 0.0375 0.0750 0.1120 Expt. Temp 2 3 4 25°C 0.100 M 25 C 25°C 45°C 0.100 M 0.200 M 0.100 M 0.500 M 0.100 M 0.200 M 1. How long will it take to use up half of A? 2. At what concentrations of A and B will the reaction rate be 0.186 moVL-hr? 3. What percent of A will be used up after 25 minutes? 4. What is the activation energy of the reaction?

Answer 1

Rate = K[A]^x [B]^y implies K = Rate/[A]^x [B]^y From Expt (1) & (2) 0.0375/(0.1)^x (0.1)^y = 0.0375/(0.1)^x (0.1)^y implies (0.5/0.1)^y = 1 implies 5^y = 1 = 5^0 implies y = 0 order w. r. t b = 0 from Expt (1) & (3): 0.0375/(0.1)^x (0.1)^y = 0.0750/(.02)^x (0.1)^y implies (0.2/0.1)^x = 0.0750/0.0375 implies 2^x = 2 = 2^1 implies x = 1 order w. r. t A = 1 therefore total order = 1 + 0 = 1 therefore It is a 1^st order reaction Rate = K[A]^1 [B]^0 implies K = 0.0375 mol/L - hr/(0.1)^1(0.1)^0 mol/L = 0.375 hr^- 1 t_1/2 = 0.693/K implies t_1/2 = 0.693/0.375 hr^- 1 = 1.848 hrs because Rate = K[A]^1 [B]^0 implies [A]^1 times 1 implies [A] = 0.186/0.375 = 0.496 mol/L therefore [A] = 0.496 mol/L & [B] = 0.1 M K = 2.303/t log ([A]_0/[A]_t) implies 0.375 = 2.303/0.4167 log (0.1/[A]_t = 0.4167 hr) t = 25 min = 0.4167 hrs implies log(0.1/[A]_t) = 0.0678