**Answer: -321.30 - 150 R**

$C_6H_5COOH (s) + \frac{15}{2} O_2(g) \longrightarrow 7CO_2(g) + 3H_2O(l)$

Since, $\Delta n_g = n_P - n_R$

$\Delta n_g$ = 7 - $\frac{15}{2} = \frac{-1}{2}$

$\Delta H = \Delta E + \Delta n_g RT$

$\Delta H = -321.30 - ( \frac{1}{2} \times R \times 300 )= -321.30 - 150 R$