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(B) The standard enthalpy change of reaction is given by: $\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(	ext{products}) - \sum \Delta_f H^{\circ}(	e

Vedclass · Accepted Answer

$-3100$

Vedclass · Answer

(B) The standard enthalpy change of reaction is given by: $\Delta_r H^{\circ} = \sum \Delta_f H^{\circ}(	ext{products}) - \sum \Delta_f H^{\circ}(	ext{reactants})$.For the reaction $2C_2H_6(g) + 7O_2(g) ightarrow 4CO_2(g) + 6H_2O(l)$:$\Delta_r H^{\circ} = [4 	imes \Delta_f H^{\circ}(CO_2) + 6 	imes \Delta_f H^{\circ}(H_2O)] - [2 	imes \Delta_f H^{\circ}(C_2H_6) + 7 	imes \Delta_f H^{\circ}(O_2)]$.Since $\Delta_f H^{\circ}(O_2) = 0 	ext{ kJ mol}^{-1}$ (standard state of an element):$\Delta_r H^{\circ} = [4(-390) + 6(-285)] - [2(-85) + 7(0)]$.$\Delta_r H^{\circ} = [-1560 - 1710] - [-170]$.$\Delta_r H^{\circ} = -3270 + 170 = -3100 	ext{ kJ}$.

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