Consider the following data for the reaction | vedclass

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(B) The standard enthalpy change of the reaction is calculated as:
$\Delta H^\circ = \Sigma \Delta_f H^\circ(	ext{products}) - \Sigma \Delta_f H^\ci

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(B) The standard enthalpy change of the reaction is calculated as:
$\Delta H^\circ = \Sigma \Delta_f H^\circ(	ext{products}) - \Sigma \Delta_f H^\circ(	ext{reactants})$
$\Delta H^\circ = 2 	imes \Delta_f H^\circ(XY) - [\Delta_f H^\circ(X_2) + \Delta_f H^\circ(Y_2)]$
$\Delta H^\circ = 2(42) - (8 + 80) = 84 - 88 = -4 \ kJ \ mol^{-1} = -4000 \ J \ mol^{-1}$.
The standard entropy change of the reaction is:
$\Delta S^\circ = \Sigma S^\circ(	ext{products}) - \Sigma S^\circ(	ext{reactants})$
$\Delta S^\circ = 2 	imes S^\circ(XY) - [S^\circ(X_2) + S^\circ(Y_2)]$
$\Delta S^\circ = 2(200) - (140 + 250) = 400 - 390 = 10 \ J \ K^{-1} \ mol^{-1}$.
Using the Gibbs-Helmholtz equation:
$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
$\Delta G^\circ = -4000 \ J \ mol^{-1} - (600 \ K 	imes 10 \ J \ K^{-1} \ mol^{-1})$
$\Delta G^\circ = -4000 - 6000 = -10000 \ J \ mol^{-1} = -10 \ kJ \ mol^{-1}$.

Compound	$\Delta_f H^\circ$ $(kJ \ mol^{-1})$	$S^\circ$ $(J \ mol^{-1} \ K^{-1})$
$XY(g)$	$42$	$200$
$X_2(g)$	$8$	$140$
$Y_2(g)$	$80$	$250$

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