Question

At $700K$, the equilibrium constant, $K_P$, for the reaction ${2SO}_3\left(g\right)\leftrightharpoons {2SO}_2\left(g\right)+O_2\left(g\right)$ is $1.8\times {10}^{-3}$ kPa. The value of $K_c$ for the above reaction at the same temperature in moles per m${}^3$ would be:

• A. $1.1\times {10}^{-7}$
• B. $3.1\times {10}^{-4}$
• C. $6.2\times {10}^{-7}$
• D. $9.3\times {10}^{-7}$

Answer 1

The correct option is B $3.1\times {10}^{-4}$

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

As we know that,

$K_p=K_c{\left(RT\right)}^{\Delta n}$

whereas,

Given that:-

$K_p=1.8\times {10}^{-3}kPa=1.8\left(\text{Given}\right)$

$T=700K\left(\text{Given}\right)$

$R=8.314\times {10}^{-3}{m}^{3}kPa{K}^{-1}{mol}^{-1}$

$K_c=?$

$\Delta n=$ change in no. of moles = no. of moles of product- no. of moles of reactant

$\Delta n=n_p-n_r=(2+1)-\left(2\right)=1$

$\therefore 1.8\times {10}^{-3}=K_c\times {(8.314\times {10}^{-3}\times 700)}^1$

$\Rightarrow K_c=\frac{1.8\times {10}^{-3}}{8.314\times {10}^{-3}\times 700}=3.1\times {10}^{-4}{m}^{-3}mol$