Given H_{2(g)} + Cl_{2(g)} rightarrow 2HCl_{( | vedclass

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(C) Given reactions:$(i) H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$Dividing by $2$: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{

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$-98$

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(C) Given reactions:$(i) H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$Dividing by $2$: $\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)} \rightarrow HCl_{(g)}$; $\Delta H = -22 \, Kcal$$(ii) 2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$Dividing by $2$: $Na_{(s)} + HCl_{(g)} \rightarrow NaCl_{(s)} + \frac{1}{2}H_{2(g)}$; $\Delta H = -76 \, Kcal$Adding the two resulting equations:$(Na_{(s)} + HCl_{(g)}) + (\frac{1}{2}H_{2(g)} + \frac{1}{2}Cl_{2(g)})$ $\rightarrow (NaCl_{(s)} + \frac{1}{2}H_{2(g)}) + HCl_{(g)}$$Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$$\Delta H = (-22) + (-76) = -98 \, Kcal$

Given $H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)}$; $\Delta H = -44 \, Kcal$ and $2Na_{(s)} + 2HCl_{(g)} \rightarrow 2NaCl_{(s)} + H_{2(g)}$; $\Delta H = -152 \, Kcal$,calculate $\Delta H$ for the reaction $Na_{(s)} + 0.5 Cl_{2(g)} \rightarrow NaCl_{(s)}$ in $Kcal$.

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