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(A) To find the enthalpy change for the reaction $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$,we can manipulate the given equations:$(i)$ $NH_3(g) + aq \r

Vedclass · Accepted Answer

$-42.1$

Vedclass · Answer

(A) To find the enthalpy change for the reaction $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$,we can manipulate the given equations:$(i)$ $NH_3(g) + aq \rightarrow NH_3(aq)$,$\Delta H_1 = -8.4 \, Kcal$$(ii)$ $HCl(g) + aq \rightarrow HCl(aq)$,$\Delta H_2 = -17.3 \, Kcal$$(iii)$ $NH_3(aq) + HCl(aq) \rightarrow NH_4Cl(aq)$,$\Delta H_3 = -12.5 \, Kcal$$(iv)$ $NH_4Cl(s) + aq \rightarrow NH_4Cl(aq)$,$\Delta H_4 = +3.9 \, Kcal$We need the target equation: $NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$.Adding equations $(i)$,$(ii)$,and $(iii)$ gives:$NH_3(g) + HCl(g) + 2aq \rightarrow NH_4Cl(aq)$,$\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -8.4 - 17.3 - 12.5 = -38.2 \, Kcal$.Now,subtract equation $(iv)$ from this result (or add the reverse of equation $(iv)$):$NH_4Cl(aq) \rightarrow NH_4Cl(s) + aq$,$\Delta H = -3.9 \, Kcal$.Adding the equations:$NH_3(g) + HCl(g) + 2aq + NH_4Cl(aq) \rightarrow NH_4Cl(aq) + NH_4Cl(s) + aq$$NH_3(g) + HCl(g) \rightarrow NH_4Cl(s)$$\Delta H_{total} = -38.2 + (-3.9) = -42.1 \, Kcal$.

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