Question

At $700K$, 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 $1.8\times {10}^{-3}kPa$. What is the numerical value on moles per $d{m}^3$ of $K_c$ for the reaction at the same temperature ?

• A. $3.09\times {10}^{-4}$
• B. $3.09\times {10}^{-7}$
• C. $3.09\times {10}^{-10}$
• D. $3.09\times {10}^{-12}$

Answer 1

Answer: (B)

Given reaction scheme

$2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$

∆n_g= n_products-n_reactants $=3-(2+1)=3-2=1$

Formula: $K_p=K_c(RT{)}^{\Delta n}$

K_p is equilibrium constant from partial pressures = 1.8 kpa = $1.8\times {10}^{3}pa$

K_c is the equilibrium constant from concentrations =?

R is gas constant = 8314 L pa K^-1 mol^-1

T is temperature = 700K

On substituting we get

$K_c=\frac{1.8\times {10}^{3}pa}{8314Lpa{K}^{-1}mo{l}^{-1}\times 700K}$

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

Hence option B is correct.