JEE Main 2024 — 5 April, Shift 1 — Question 79

The combustion reaction of benzoic acid is $\mathrm{C_7H_6O_2(s) + \tfrac{15}{2}O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l)}$

The change in number of moles of gaseous species is $\Delta n_{\text{gas}} = 7 - \tfrac{15}{2} = -\tfrac{1}{2}$

The relation between heat at constant pressure and constant volume is $\Delta H = \Delta U + \Delta n_{\text{gas}}RT$

Substituting the values gives $\Delta H = -321.30 - \tfrac{1}{2}RT$

Given, $\Delta H = (-321.30 - xR)\,\mathrm{kJ}$

Comparing both expressions, $xR = \tfrac{1}{2}RT$

At $27^\circ\mathrm{C} = 300\,\mathrm{K}$ $x = \tfrac{1}{2} \times 300 = 150$.

Thus, the value of x is 150.