Question

1. A reaction has a rate constant of 1,11 [1/M-) at 998 K and activation ceroy of 1401 J/mol (a) What is the rate constant, k, [1/ Mes) at 898 K (b) What is the value of the rate constant at 800 K? (e) How will the activation energy and the rate constant change if a catalyst is added at 898 K?

Answer 1

(a) ln (K2/K1) = - (Ea/ R)* ((1/ T2) - (1/T1))

Rate constant (K1)  = 1.11, T1 = 998 K , Ea (activation energy) = 1.4*10+5 , T2 = 898 K ,

R, universal gas constant = 8.314 J / Kmol

ln (K2/1.11) = - (1.4*10+5/8.314) ( (1/898) - ( 1/998)) = - 1.8789

K2/1.11 = e- 1.8789 = 0.152

K2 = 1.11*0.152 = 0.169 1/ M*s

(b) ln (K2/K1) = - (Ea/ R)* ((1/ T2) - (1/T1))

Rate constant (K1)  = 1.11, T1 = 998 K , Ea (activation energy) = 1.4*10+5 , T2 = 800 K ,

R, universal gas constant = 8.314 J / Kmol

ln (K2/1.11) = - (1.4*10+5/8.314) ( (1/800) - ( 1/998)) = - 4.176

K2/1.11 = e- 4.176 = 0.015359

K2 = 1.11*0.15359 = 0.017 1/ M*s

(c) A catalyst will increase the rate of a reaction by proceeds the reaction through an alternate pathway with lower activation energy. So, at a constant temperature activation energy decreases with the addition of a catalyst.

As activation energy decreases, More reactant molecules will be converted to products. So, reaction rate will increase and rate constant will also increase. So, The addition catalyst increase the rate constant at a constant temperature.