Chemical EquilibriumHard
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
The system PCl5(g) ⇋ PCl3(g) + Cl2(g) attains equilibrium. If the equilibrium cocentrateion of PCl3(g) is doubled,...When two reactants $A$ and $B$ are mixed to give products $C$ and $D$, the reaction quotient $Q$ at the initial stage of...If for the equilibria NH2COONH4(s) $\rightleftharpoons$N2 + H2 + CO + O2, the value of KP at 800 K is 27 × 2λ/2 and the ...Rate of disappearance of the reactant ‘A’ in the reversible reaction A$\rightleftharpoons$B at two temperatures is given...The reaction ZnO(s) + CO(g) $\rightleftharpoons$Zn(g) + CO2(g) has an equilibrium constant of 1 atm at 1500 K. The equil...