Calculate Delta H^{circ} for the reaction,Na_ | vedclass

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(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.Given equations:$(A) \ Na_{(s)} + H_2

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$-581 \ kJ$

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(B) We need to find $\Delta H^{\circ}$ for the reaction: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.Given equations:$(A) \ Na_{(s)} + H_2O_{(l)} \longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)} \quad \Delta H^{\circ} = -146 \ kJ$$(B) \ Na_2SO_{4(s)} + H_2O_{(l)} \longrightarrow 2NaOH_{(s)} + SO_{3(g)} \quad \Delta H^{\circ} = +418 \ kJ$$(C) \ 2Na_2O_{(s)} + 2H_{2(g)} \longrightarrow 4Na_{(s)} + 2H_2O_{(l)} \quad \Delta H^{\circ} = +259 \ kJ$To get the target reaction,perform the operation: $2 	imes (A) + \frac{1}{2} 	imes (C) - (B)$.$2 	imes [Na_{(s)} + H_2O_{(l)}$ $\longrightarrow NaOH_{(s)} + \frac{1}{2} H_{2(g)}] \implies 2Na_{(s)} + 2H_2O_{(l)}$ $\longrightarrow 2NaOH_{(s)} + H_{2(g)} \quad \Delta H^{\circ} = 2 	imes (-146) = -292 \ kJ$$\frac{1}{2} 	imes [2Na_2O_{(s)} + 2H_{2(g)}$ $\longrightarrow 4Na_{(s)} + 2H_2O_{(l)}] \implies Na_2O_{(s)} + H_{2(g)}$ $\longrightarrow 2Na_{(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = \frac{259}{2} = 129.5 \ kJ$$-(B) \implies 2NaOH_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)} + H_2O_{(l)} \quad \Delta H^{\circ} = -418 \ kJ$Adding these: $(2Na_{(s)} + 2H_2O_{(l)} + Na_2O_{(s)} + H_{2(g)} + 2NaOH_{(s)} + SO_{3(g)})$ $\longrightarrow (2NaOH_{(s)} + H_{2(g)} + 2Na_{(s)} + H_2O_{(l)} + Na_2SO_{4(s)} + H_2O_{(l)})$Simplifying gives: $Na_2O_{(s)} + SO_{3(g)} \longrightarrow Na_2SO_{4(s)}$.$\Delta H^{\circ} = -292 + 129.5 - 418 = -580.5 \ kJ \approx -581 \ kJ$.

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