Chemistry - Thermodynamics - I Question with Solution | TestHub

Heat of reaction at constant volume = $\Delta E$.

Heat of reaction at constant pressure = $\Delta H$.

Given $\Delta E-\Delta H=1200cal$.

We know $\Delta H=\Delta E+P\Delta V$

or $\Delta H=\Delta E+\Delta n_{g}RT$.

So, $\Delta n_g=\frac{\Delta H-\Delta E}{RT}=\frac{-(\Delta E-\Delta H)}{RT}=\frac{-1200}{2\times 300}=-2$.

Now, $K_P=K_C(RT{)}^{\Delta n_{}}$.

$\frac{K_p}{K_C}={\left(0.821\times 300\right)}^{-2}=1.648\times 10^{-3}$