Question

For a reversible reaction $\frac{K_p}{K_c}={\left(RT\right)}^{\Delta n}$ at $300K$. If $\frac{K_p}{K_c}=14941.5$ (in litre atm unit) for a given reaction then $\Delta n$ for the reaction is :

[Given that $log\frac{K_p}{K_c}=4.17$ and $log\left(24.93\right)=1.39$]

Answer 1

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

$log\frac{K_p}{K_c}=\Delta n\times log\left(RT\right)$

$log14941.5=\Delta n\times log(0.0821\times 300)$

$4.17=\Delta n\times log24.93$

$4.17=\Delta n\times 1.39$

$\Delta n=3$