The standard enthalpy of formation of benzoic acid, C₆H₅COOH, is -385 kJ mol^-1 at 298 K. Calculate the standard enthalpy of combustion of benzoic acid at this temperature, given that the standard enthalpy of formation of liquid water, H₂O is -285.8 kJ mol^-1 and gaseous carbon dioxide, CO₂, is -393.51 kJ mol^-1.

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The standard enthalpy of combustion of a compound can be calculated using the standard enthalpies of formation of the reactants and products. The balanced chemical equation for the combustion of benzoic acid (C7H6O2) is:$$ C_7H_6O_2(s) + 7.5O_2(g) \rightarrow 7CO_2(g) + 3H_2O(l) $$Using the given standard enthalpies of formation ($\Delta H_f^\circ$):- $\Delta H_f^\circ$ of benzoic acid (C7H6O2) = -385 kJ/mol- $\Delta H_f^\circ$ of liquid water (H2O) = -285.8 kJ/mol- $\Delta H_f^\circ$ of gaseous carbon dioxide (CO2) = -393.51 kJ/molThe standard enthalpy of combustion ($\Delta H_c^\circ$) can be calculated using the formula:$$ \Delta H_c^\circ = \sum (\Delta H_f^\circ \text{ of products}) - \sum (\Delta H_f^\circ \text{ of reactants}) $$$$ \Delta H_c^\circ = [7(-393.51) + 3(-285.8)] - [-385] $$$$ \Delta H_c^\circ = [-2754.57 - 857.4] + 385 $$$$ \Delta H_c^\circ = -3227.0 \, \text{kJ/mol} $$Therefore, the standard enthalpy of combustion of benzoic acid at 298 K is -3227.0 kJ/mol.

The Standard Enthalpy of Formation of Benzoic Acid, C6H5COOH, Is

The Standard Enthalpy of Formation of Benzoic Acid, C₆H₅COOH, Is