The formation enthalpies,Delta H_{f}^{ominus} | vedclass

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Question

(A) The reaction for the formation of water is: $H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(g)}$; $\Delta H_f^{\ominus} = -242 \ kJ \ mol^{-1}$

Vedclass · Accepted Answer

$466$

Vedclass · Answer

(A) The reaction for the formation of water is: $H_{2(g)} + \frac{1}{2}O_{2(g)} ightarrow H_2O_{(g)}$; $\Delta H_f^{\ominus} = -242 \ kJ \ mol^{-1}$.To calculate the bond enthalpy of the $O-H$ bond,we consider the atomization of reactants:$H_{2(g)} ightarrow 2H_{(g)}$; $\Delta H = 2 	imes \Delta H_f^{\ominus}(H_{(g)}) = 2 	imes 220 = 440 \ kJ \ mol^{-1}$.$\frac{1}{2}O_{2(g)} ightarrow O_{(g)}$; $\Delta H = \Delta H_f^{\ominus}(O_{(g)}) = 250 \ kJ \ mol^{-1}$.The formation of $H_2O_{(g)}$ from gaseous atoms is: $2H_{(g)} + O_{(g)} ightarrow H_2O_{(g)}$; $\Delta H = -2 	imes (B.E._{O-H})$.Using Hess's Law:$\Delta H_f^{\ominus}(H_2O_{(g)}) = [2 	imes \Delta H_f^{\ominus}(H_{(g)}) + \Delta H_f^{\ominus}(O_{(g)})] - 2 	imes (B.E._{O-H})$$-242 = [2 	imes 220 + 250] - 2 	imes (B.E._{O-H})$$-242 = 690 - 2 	imes (B.E._{O-H})$$2 	imes (B.E._{O-H}) = 690 + 242 = 932$$B.E._{O-H} = 466 \ kJ \ mol^{-1}$.

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