Question

At $700$K, the equilibrium constant, $K_p$ for the reaction $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ is $1.80\times {10}^{-3}kpa.$What is the numeriacal value in moles per litre of $K_c$ for this reaction at the same temperaure?

• A. $3.09\times {10}^{-7}$ mol $L^{-1}$
• B. $3.09\times {10}^{-6}$ mol $L^{-1}$
• C. $3.09\times {10}^7$ mol $L^{-1}$
• D. $3.09\times {10}^6$ mol $L^{-1}$

Answer 1

Answer: (A)

$3.09\times {10}^{-7}$ mol $L^{-1}$

Given: $2S{O}_3\rightleftharpoons 2S{O}_2+O_2$

1 atm=10${}^5$ pa

$\Delta n=n_{products}-n_{reactants}$

$K_p=1.8\times {10}^{-3}KPa=1.8Pa=1.8\times {10}^{-5}atm$

$T=700K$

$R=0.0821lit-atm$

$K_p=K_c(RT{)}^{\Delta n_g}$

$K_c=\frac{K_p}{RT}$

$K_c=\frac{1.8\times {10}^{-5}}{700\times 0.0821}=0.031\times {10}^{-5}=3.1\times {10}^{-7}$ $mol{L}^{-1}$