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(A) The enthalpy change of a reaction $(\Delta H_{rxn})$ is calculated using the enthalpies of formation $(\Delta H_f)$ of products and reactants:$\De

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$-605 \ kJ \ mol^{-1}$

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(A) The enthalpy change of a reaction $(\Delta H_{rxn})$ is calculated using the enthalpies of formation $(\Delta H_f)$ of products and reactants:$\Delta H_{rxn} = \sum \Delta H_f(	ext{products}) - \sum \Delta H_f(	ext{reactants})$Given:$\Delta H_f(C_2H_{2(g)}) = 230 \ kJ \ mol^{-1}$$\Delta H_f(C_6H_{6(g)}) = 85 \ kJ \ mol^{-1}$For the reaction $3 \ C_2H_{2(g)} ightarrow C_6H_{6(g)}$:$\Delta H_{rxn} = [1 	imes \Delta H_f(C_6H_{6(g)})] - [3 	imes \Delta H_f(C_2H_{2(g)})]$$\Delta H_{rxn} = [85] - [3 	imes 230]$$\Delta H_{rxn} = 85 - 690$$\Delta H_{rxn} = -605 \ kJ \ mol^{-1}$

At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?
$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$

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At $298 \ K$ and constant pressure, the heat of formation of $C_2H_{2(g)}$ and $C_6H_{6(g)}$ is $230 \ kJ \ mol^{-1}$ and $85 \ kJ \ mol^{-1}$ respectively. What will be the change in heat for the reaction?$3 \ C_2H_{2(g)} \rightarrow C_6H_{6(g)}$