Chemical EquilibriumMedium
Question
The activation energies for the forward and reverse elementary reactions in the system A$\rightleftharpoons$B are 10.303 and 8.000 kcal, respectively at 500 K. Assuming the pre-exponential factor to be the same for both the forward and reverse steps, the equilibrium constant of the reaction at 500 K is
Options
A.1.00
B.10.0
C.100
D.0.1
Solution
$\Delta H = \varepsilon a_{f} - \varepsilon a_{b} = 10.303 - 8.000 = 2.303\text{ kcal}$
$\text{Now, }K_{eq} = \frac{A_{f}}{A_{b}}.e^{- \Delta H/RT} = 1 \times e^{- \frac{2.303 \times 10^{3}}{2 \times 500}} = 0.1$
Create a free account to view solution
View Solution FreeMore Chemical Equilibrium Questions
Which one of the following has lowest melting point :-...At chemical equilibrium, the...The value of ᐃH for the reactionX2(g) + 4Y2(g) ⇋ 2XY4(g) is less than zero. Formation of XY4(g) will be favo...The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is...NO2 is involved in the formation of smog and acid rain. It is formed importantly as:NO(g) + O3(g) $\rightleftharpoons$ N...