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(C) According to Hess's Law of Constant Heat Summation,the total enthalpy change of a reaction is the same whether it occurs in one step or several st

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$x = y + z$

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(C) According to Hess's Law of Constant Heat Summation,the total enthalpy change of a reaction is the same whether it occurs in one step or several steps.Given reactions:$(1) \ C_{(gr)} + O_{2_{(g)}} 	o CO_{2_{(g)}}, \Delta H = x \ kJ/mol$$(2) \ C_{(gr)} + \frac{1}{2} O_{2_{(g)}} 	o CO_{(g)}, \Delta H = y \ kJ/mol$$(3) \ CO_{(g)} + \frac{1}{2} O_{2_{(g)}} 	o CO_{2_{(g)}}, \Delta H = z \ kJ/mol$If we add reaction $(2)$ and reaction $(3)$:$(C_{(gr)} + \frac{1}{2} O_{2_{(g)}}) + (CO_{(g)} + \frac{1}{2} O_{2_{(g)}}) 	o CO_{(g)} + CO_{2_{(g)}}$Canceling $CO_{(g)}$ from both sides and combining $\frac{1}{2} O_{2_{(g)}}$ terms:$C_{(gr)} + O_{2_{(g)}} 	o CO_{2_{(g)}}$This is identical to reaction $(1)$.Therefore,the enthalpy change for reaction $(1)$ is the sum of the enthalpy changes of reactions $(2)$ and $(3)$:$x = y + z$

On the basis of the following reactions,which one is correct?
$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$
$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$
$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$

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Consider the given data $:$
$(a) \ HCl_{(g)} + 10 \ H_2O_{(l)} \rightarrow HCl \cdot 10 \ H_2O \quad \Delta H = -69.01 \ kJ \ mol^{-1}$
$(b) \ HCl_{(g)} + 40 \ H_2O_{(l)} \rightarrow HCl \cdot 40 \ H_2O \quad \Delta H = -72.79 \ kJ \ mol^{-1}$
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On the basis of the following reactions,which one is correct?$C_{(gr)} + O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = x \ kJ/mol$$C_{(gr)} + \frac{1}{2} O_{2_{(g)}} \to CO_{(g)}, \Delta H = y \ kJ/mol$$CO_{(g)} + \frac{1}{2} O_{2_{(g)}} \to CO_{2_{(g)}}, \Delta H = z \ kJ/mol$