please answer PART B. thanks :) A reaction has a rate constant of 1.24 times 10^
ID: 1008896 • Letter: P
Question
please answer PART B. thanks :)
A reaction has a rate constant of 1.24 times 10^-4 s^-1 at 29 degreeC and 0.227 s^-1 at 76 degreeC. You may want to reference (pages 576 - 582) section 15.6 while completing this problem. Determine the activation barrier for the reaction. Express your answer in units of kilojoules per mole. E_a = 140 kJ/mol The activation energy E_a is an energy barrier or hump that must be surmounted for the reactants to be transformed into products. The activation energy was found using the following equation: ln (k_2/k_1) = E_a/R (1/T_1 - 1/T_2) E_a = R ln (k_2/k_1)/(1/T_1 - 1/T_2) E_a = 8.314 J/mol-K ln (0.227s^-1/1.24 times 10^-4s^-1)/(1/302 K - 1/349 K) times 1 kJ/1000 J E_a = 140 kJ/mol What is the value of the rate constant at 17 degreeC? Express your answer in units of inverse seconds.Explanation / Answer
Part B:
ln(K2/K1) = [Ea / R][1/T1 - 1/T2]
Where Ea is activation energy
T1 and T2 are temperatures
K1 and K2 are rate constants at T1 and T2
ln 1.24 x 10-4/ x = 140 / 0.008314 ( 1/ 290K - 1 / 302)
ln 1.24 x 10-4/x = 2.31
1.24 x 10-4 /x = e2.31
1.24 x 10-4 /x = 10.07
x = 1.231 x 10-5 s-1
Rate constant at 170C = 1.231 x 10-5s-1
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