Consider the reaction 2H_2S(g) + 3O_2(g) righ | vedclass

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(A) The enthalpy change of the reaction is calculated using the formula: $\Delta_r H^\circ = \sum \Delta_f H^\circ(	ext{products}) - \sum \Delta_f H^

Vedclass · Accepted Answer

$1126$

Vedclass · Answer

(A) The enthalpy change of the reaction is calculated using the formula: $\Delta_r H^\circ = \sum \Delta_f H^\circ(	ext{products}) - \sum \Delta_f H^\circ(	ext{reactants})$.Substituting the given values:$\Delta_r H^\circ = [2 	imes \Delta_f H^\circ(H_2O) + 2 	imes \Delta_f H^\circ(SO_2)] - [2 	imes \Delta_f H^\circ(H_2S) + 3 	imes \Delta_f H^\circ(O_2)]$.Since $O_2$ is in its standard elemental state, $\Delta_f H^\circ(O_2) = 0 	ext{ kJ mol}^{-1}$.$\Delta_r H^\circ = [2(-286.0) + 2(-297.0)] - [2(-20.1) + 3(0)]$.$\Delta_r H^\circ = [-572.0 - 594.0] - [-40.2]$.$\Delta_r H^\circ = -1166.0 + 40.2 = -1125.8 	ext{ kJ mol}^{-1}$.The magnitude of the enthalpy change is $|-1125.8| = 1125.8 	ext{ kJ mol}^{-1}$.Rounding to the nearest integer, we get $1126 	ext{ kJ mol}^{-1}$.

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