Question

For the given equilibrium reaction, $2 A ( g ) \rightleftharpoons 2 B ( g )+ C ( g )$ the equilibrium constant $\left(K_{c}\right)$ at $1000 \,K$ is $4 \times 10^{-4}$. Calculate $K_{p}$ for the reaction at $800\, K$ temperature.

• A. 0.044
• B. 0.026
• C. 0.33
• D. 1

Answer 1

Answer: (B)

$2 A(g) \rightleftharpoons 2 B(g)+C(g)$
$K_{p}=K_{C} \cdot(R T)^{\Delta n_{g}}$
Given, $K_{C}=4 \times 10^{-4}$
$T=1000 \,K$
$\Delta n_{g}=$ moles of product $-$ moles of reactant
$=3-2=1$
$K_{p}=4 \times 10^{-4}(8.314 \times 800)^{1}$
$K_{p}=0.026$