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(C) The standard enthalpy change of reaction $\Delta_rH^{\circ}$ is calculated using the formula:$\Delta_rH^{\circ} = \sum \Delta_fH^{\circ}(	ext{pro

Vedclass · Accepted Answer

$-1299 \ kJ \ mol^{-1}$

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(C) The standard enthalpy change of reaction $\Delta_rH^{\circ}$ is calculated using the formula:$\Delta_rH^{\circ} = \sum \Delta_fH^{\circ}(	ext{products}) - \sum \Delta_fH^{\circ}(	ext{reactants})$For the reaction $C_2H_{2(g)} + \frac{5}{2}O_{2(g)} ightarrow 2CO_{2(g)} + H_2O_{(\ell)}$:$\Delta_rH^{\circ} = [2 	imes \Delta_fH^{\circ}(CO_2) + 1 	imes \Delta_fH^{\circ}(H_2O)] - [1 	imes \Delta_fH^{\circ}(C_2H_2) + \frac{5}{2} 	imes \Delta_fH^{\circ}(O_2)]$Since $\Delta_fH^{\circ}(O_2) = 0 \ kJ \ mol^{-1}$ (standard state of an element):$\Delta_rH^{\circ} = [2(-393) + (-286)] - [227 + 0]$$\Delta_rH^{\circ} = [-786 - 286] - 227$$\Delta_rH^{\circ} = -1072 - 227 = -1299 \ kJ \ mol^{-1}$

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