Chemistry - Thermodynamics - 1 Question with Solution | TestHub

$\frac{{\text{K}}_{\text{p}}}{{\text{K}}_{\text{c}}}={\left(\text{RT}\right)}^{{\text{Δn}}_{\mathrm{g}}}$

${\left(\text{RT}\right)}^{{\text{Δn}}_{\mathrm{g}}}=\text{24.63 }\mathrm{L}\mathrm{atm}{mol}^{-1}$

${\left(\text{0.0821}\times 300\right)}^{\text{Δ}{\text{n}}_{\text{g}}}=\text{24.63}$

$\therefore \text{Δ}{\text{n}}_{\text{g}}=1$

$\text{Δ}\text{H}=\text{Δ}\text{E}+\text{Δ}{\text{n}}_{\text{g}}\text{RT}$

$\therefore \text{Δ}\text{E}=\text{Δ}\text{H}-\text{RT}=\text{98.8}-2\times 10^{-3}\times 300=\text{98.2 kcal}$