Question

$\frac{K_p}{K_c}$ for the following reaction will be:
$CO\left(g\right)+O_2\left(g\right)\rightleftharpoons C{O}_2\left(g\right)$

• A. $RT$
• B. $\frac{1}{(RT{)}^{\frac{1}{2}}}$
• C. $\frac{1}{{\left(RT\right)}^2}$
• D. $\frac{RT}{2}$

Answer 1

Answer: (B)

$\frac{1}{(RT{)}^{\frac{1}{2}}}$

Solution:-

As we know that,

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

For the reaction-

${CO}_{\left(g\right)}+\frac{1}{2}{O_2}_{\left(g\right)}⟶{C{O}_2}_{\left(g\right)}$

$\Delta n_g=n_P-n_R=1-(1+\frac{1}{2})=-\frac{1}{2}$

Therefore,

$K_p=K_c{\left(RT\right)}^{-\frac{1}{2}}$

$\Rightarrow \frac{K_p}{K_c}=\frac{1}{{\left(RT\right)}^{\frac{1}{2}}}$

Hence for the given reaction-

$\frac{K_p}{K_c}=\frac{1}{{\left(RT\right)}^{\frac{1}{2}}}$