Based on the following thermochemical equatio | vedclass

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Question

(A) Let the given equations be:$(i) H_2O_{(g)} + C_{(s)} 	o CO_{(g)} + H_{2(g)}; \Delta H_1 = 131 \ kJ$$(ii) CO_{(g)} + \frac{1}{2}O_{2(g)} 	o CO_{2

Vedclass · Accepted Answer

$-393$

Vedclass · Answer

(A) Let the given equations be:$(i) H_2O_{(g)} + C_{(s)} 	o CO_{(g)} + H_{2(g)}; \Delta H_1 = 131 \ kJ$$(ii) CO_{(g)} + \frac{1}{2}O_{2(g)} 	o CO_{2(g)}; \Delta H_2 = -282 \ kJ$$(iii) H_{2(g)} + \frac{1}{2}O_{2(g)} 	o H_2O_{(g)}; \Delta H_3 = -242 \ kJ$Adding equations $(i)$,$(ii)$,and $(iii)$:$(H_2O_{(g)} + C_{(s)}) + (CO_{(g)} + \frac{1}{2}O_{2(g)}) + (H_{2(g)} + \frac{1}{2}O_{2(g)}) 	o (CO_{(g)} + H_{2(g)}) + CO_{2(g)} + H_2O_{(g)}$Canceling common terms on both sides:$C_{(s)} + O_{2(g)} 	o CO_{2(g)}$Therefore,$X = \Delta H_1 + \Delta H_2 + \Delta H_3 = 131 + (-282) + (-242) = -393 \ kJ$.

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