RT $$

Here:

$$\Delta H$$ is the enthalpy change.
$$\Delta U$$ is the internal energy change.
$$\Delta n_{\mathrm{gas}}$$ is the change in the number of moles of gas during the reaction.
$$R$$ is the universal gas constant, which is approximately $$8.314 \, \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1}$$ or $$0.008314 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \, \mathrm{K}^{-1}$$.
$$T$$ is the temperature in Kelvin. Given the temperature is $$25^{\circ}\mathrm{C}$$, we convert it to Kelvin as $$25 + 273.15 = 298.15 \, \mathrm{K}$$.

The standard combustion reaction of benzoic acid is:

$$ \mathrm{C}_6\mathrm{H}_5\mathrm{COOH}(\mathrm{s}) + 7.5 \mathrm{O}_2(\mathrm{g}) \rightarrow 7 \mathrm{CO}_2(\mathrm{g}) + 3 \mathrm{H}_2\mathrm{O}(\mathrm{l}) $$

From the reaction, the change in the number of moles of gas, $$\Delta n_{\mathrm{gas}}$$, can be calculated as:

$$ \Delta n_{\mathrm{gas}} = \text{moles of gaseous products} - \text{moles of gaseous reactants} $$

So,

$$ \Delta n_{\mathrm{gas}} = 7 - 7.5 = -0.5 $$

Now, substituting these values into the equation, we get:

$$ \Delta H = \Delta U + \Delta n_{\mathrm{gas}} RT $$

$$ -2546 \, \mathrm{kJ} = \Delta U + (-0.5) (0.008314 \, \mathrm{kJ} \, \mathrm{mol}^{-1} \mathrm{K}^{-1}) (298.15 \, \mathrm{K}) $$

$$ -2546 \, \mathrm{kJ} = \Delta U - 1.238 \, \mathrm{kJ} $$

$$ \Delta U = -2546 \, \mathrm{kJ} + 1.238 \, \mathrm{kJ} $$

$$ \Delta U = -2544.762 \, \mathrm{kJ} $$

Rounding to the nearest decimal, we get:

$$ \Delta U \approx -2544.8 \, \mathrm{kJ} $$

Therefore, the correct answer is:

Option A: $$-2544.8 \mathrm{~kJ}$$

Thermodynamics

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