The difference between the reaction enthalpy change ($$\Delta_r H$$) and reaction internal energy change ($$\Delta_r U$$) for the reaction: $$$2\text{C}_6\text{H}_6(l) + 15\text{O}_2(g) \longrightarrow$$$ at $$300$$ K is ($$R = 8.314$$ J mol$$^{-1}$$ K$$^{-1}$$)

For any chemical reaction carried out at a fixed temperature $$T$$, the relation between the reaction enthalpy change $$\Delta_r H$$ and the reaction internal energy change $$\Delta_r U$$ is

$$\Delta_r H - \Delta_r U = \Delta n_g \, R T \qquad -(1)$$
where $$\Delta n_g$$ is the change in the number of moles of gaseous species (moles of gaseous products minus moles of gaseous reactants) and $$R$$ is the gas constant.

First write the complete balanced combustion equation for benzene:

$$2\,C_6H_6\,(l) + 15\,O_2\,(g) \rightarrow 12\,CO_2\,(g) + 6\,H_2O\,(l)$$

Count the gaseous moles on each side:

• Reactants: $$15$$ mol $$O_2\,(g)$$
• Products: $$12$$ mol $$CO_2\,(g)$$

Hence

$$\Delta n_g = n_{g,\;products} - n_{g,\;reactants} = 12 - 15 = -3$$

Substitute $$\Delta n_g = -3$$, $$R = 8.314 \,\text{J mol}^{-1}\text{K}^{-1}$$ and $$T = 300\,\text{K}$$ into equation $$(1)$$:

$$\Delta_r H - \Delta_r U = (-3)(8.314)(300)$$

$$\Delta_r H - \Delta_r U = -7482.6 \,\text{J mol}^{-1} \approx -7482 \,\text{J mol}^{-1}$$

Therefore, the difference between the reaction enthalpy change and the reaction internal energy change is $$-7482 \,\text{J mol}^{-1}$$.

Option D which is: $$-7482 \text{ J mol}^{-1}$$