### Related Formula Enthalpy of reaction from enthalpy of formation data: Delta H_textreaction = sum Delta H_f(textProducts) - sum Delta H_f(textReactants)$\Delta H_{\text{reaction}} = \sum \Delta H_f(\text{Products}) - \sum \Delta H_f(\text{Reactants})$ ### Core Logic Write out the balanced thermochemical equation for the combustion of benzene (C_6H_6$C_6H_6$): C_6H_6(l) + frac152O_2(g) rightarrow 6CO_2(g) + 3H_2O(l)$C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)$ Given parameters: - Delta H_c = -3267.0mathrm\ kJ/mol$\Delta H_c = -3267.0\mathrm{\ kJ/mol}$ - Delta H_f[CO_2] = -393.5mathrm\ kJ/mol$\Delta H_f[CO_2] = -393.5\mathrm{\ kJ/mol}$ - Delta H_f[H_2O] = -286.0mathrm\ kJ/mol$\Delta H_f[H_2O] = -286.0\mathrm{\ kJ/mol}$ - Delta H_f[O_2] = 0mathrm\ kJ/mol$\Delta H_f[O_2] = 0\mathrm{\ kJ/mol}$ ### Step 1: Applying Hess's Law Substitute these values into the reaction expression: -3267 = [6(-393.5) + 3(-286.0)] - Delta H_f[C_6H_6]$-3267 = [6(-393.5) + 3(-286.0)] - \Delta H_f[C_6H_6]$ -3267 = [-2361.0 - 858.0] - Delta H_f[C_6H_6]$-3267 = [-2361.0 - 858.0] - \Delta H_f[C_6H_6]$ -3267 = -3219.0 - Delta H_f[C_6H_6]$-3267 = -3219.0 - \Delta H_f[C_6H_6]$ Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48mathrm\ kJ/mol$\Delta H_f[C_6H_6] = -3219.0 + 3267.0 = 48\mathrm{\ kJ/mol}$ ### Pattern Recognition Always set up products minus reactants when using heat of formation data. Pay close attention to stoichiometric coefficients (multiply CO_2$CO_2$ by 6$6$ and H_2O$H_2O$ by 3$3$) to ensure accurate bookkeeping. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Thermodynamics