Question

At \(300\,\mathrm{K}\), the enthalpies of formation of \(\mathrm{C_6H_5COOH(s)}\), \(\mathrm{CO_2(g)}\), and \(\mathrm{H_2O(l)}\) are \(-409\), \(-393\), and \(-286\,\mathrm{kJ\,mol^{-1}}\), respectively. 
The enthalpy of combustion of benzoic acid (in \(\mathrm{kJ\,mol^{-1}}\)) is

• A. \(-1600\)
• B. \(+1600\)
• C. \(-3200\)
• D. \(+4800\)

Hint:

For any reaction, \[ \boxed{ \Delta H_{\text{reaction}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}). } \] Remember that \[ \boxed{\Delta H_f^\circ(\mathrm{O_2})=0.} \]

Answer 1

The Correct Option is C

Step 1: Write the balanced combustion reaction. \[ \mathrm{C_6H_5COOH(s)+\frac{15}{2}O_2(g)\rightarrow7CO_2(g)+3H_2O(l)} \]

Step 2: Apply Hess's law. The enthalpy of combustion is \[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}). \] Since \[ \Delta H_f^\circ(\mathrm{O_2})=0, \] \[ \Delta H = \left[7(-393)+3(-286)\right] - \left[(-409)\right]. \]

Step 3: Calculate the value. \[ \Delta H = (-2751-858)+409 = -3200\;\mathrm{kJ\,mol^{-1}}. \] Hence, \[ \boxed{\Delta H_{\text{comb}}=-3200\;\mathrm{kJ\,mol^{-1}}.} \] Therefore, the correct option is \(\boxed{(C)}\).