Chemical EquilibriumHard
Question
In a reaction carried out at 500 K, 0.001% of the total number of collisions are effective. The energy of activation of the reaction is approximately
Options
A.15.8 kcal mol-1
B.11.5 kcal mol-1
C.12.8 kcal mol-1
D.zero
Solution
The fraction of molecules having energy equal to or greater than Ea is:
x =
= e -Ea/RT 
∴ log x =
(R = 2 cal mol-1 K-1, T = 500K)
log 10-5 =
Ea = 11.5 × 103 cal mol-1 = 11.5 kcal mol-1
x =
= e -Ea/RT ∴ log x =
(R = 2 cal mol-1 K-1, T = 500K)log 10-5 =
Ea = 11.5 × 103 cal mol-1 = 11.5 kcal mol-1
Create a free account to view solution
View Solution FreeMore Chemical Equilibrium Questions
For the following gases equilibrium, N2O4 (g) ⇋ 2NO2 (g)Kp is found to be equal to Kc. This is attained when :...X2(g) + Y2(g) ⇋ 2XY(g) reaction was studied at a certain temperature. In the beginning 1 mole of X2 was taken in a...An aqueous solution of volume 500 ml, when the reaction 2Ag+ (aq) + Cu(s) $\rightleftharpoons$Cu2+ (aq) + 2Ag(s) reached...At constant temperature, the equilibrium constant (Kp)for the decomposition reaction,N2O4 ⇋ 2NO2 is expressed by K...At 200°C, PCl5 dissociates as PCl5(g) $\rightleftharpoons$PCl3(g) + Cl2(g). It was found that the equilibrium vapours ar...