Calculate the heat of reaction for the process: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$ given the following data:
$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H = -8.4 \, Kcal$
$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H = -17.3 \, Kcal$
$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H = -12.5 \, Kcal$
$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H = +3.9 \, Kcal$ (in $, Kcal$)
$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H = -8.4 \, Kcal$
$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H = -17.3 \, Kcal$
$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H = -12.5 \, Kcal$
$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H = +3.9 \, Kcal$ (in $, Kcal$)
- A$-42.1$
- B$-34.3$
- C$34.3$
- D$42.1$
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