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(B) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.To obtain the target reaction

Vedclass · Accepted Answer

$450$

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(B) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.To obtain the target reaction $(4) \ C_{(s)} + O_{2_{(g)}} 	o CO_{2_{(g)}}$,we can add equations $(1)$,$(2)$,and $(3)$:$(1) \ H_2O_{(g)} + C_{(s)} 	o CO_{(g)} + H_{2(g)} ; \Delta H_1 = 100 \ kJ$$(2) \ CO_{(g)} + \frac{1}{2}O_{2_{(g)}} 	o CO_{2_{(g)}} ; \Delta H_2 = -300 \ kJ$$(3) \ H_{2(g)} + \frac{1}{2}O_{2_{(g)}} 	o H_2O_{(g)} ; \Delta H_3 = -250 \ kJ$Adding $(1) + (2) + (3)$:$(H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2}O_{2_{(g)}} + H_{2(g)} + \frac{1}{2}O_{2_{(g)}}) 	o (CO_{(g)} + H_{2(g)} + CO_{2_{(g)}} + H_2O_{(g)})$Canceling common terms on both sides:$C_{(s)} + O_{2_{(g)}} 	o CO_{2_{(g)}}$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = 100 + (-300) + (-250) = -450 \ kJ$.Since $\Delta H_4 = -x \ kJ$,we have $-x = -450$,which implies $x = 450$.

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