A, B, C and D are some compounds. The enthalpy of formation of A(g), B(g), C(g) and D(g) is 9.7,-110,81 and -393 ~kJ ~mol^-1 respectively. What is Δ r H ( . in .kJ mol^-1 ) for the given reaction? A(~g)+3 ~B(~g) longrightarrow C(~g)+3 D(~g)

Correct answer: (A) -777.7. Given, Δ H_A(~g)=9.7 ~kJ ~mol^-1 Δ H_B(~g)=-110.81 ~kJ ~mol^-1 Δ H_C(~g)=81 ~kJ ~mol^-1 Δ H_D(~g)=-393 ~kJ ~mol^-1 ~A(~g)+3 ~B(~g) arrow C(~g)+3 D(~g) H H =Δ H_product -Δ H_Reactant =[3 ×(-393)+81]-[9.7+3(-110)] =-1098+320.3 Δ_r H=-777.7 ~kJ ~mol^-1

Q10 / 160 · AP EAMCET 2024 (22 May Shift 1) · Chemistry

A, B, C and D are some compounds. The enthalpy of formation of $\mathrm{A}(\mathrm{g}), \mathrm{B}(\mathrm{g}), \mathrm{C}(\mathrm{g})$ and $\mathrm{D}(\mathrm{g})$ is $9.7,-110,81$ and $-393 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. What is $\Delta r \mathrm{H}\left(\right.$ in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ for the given reaction? $\mathrm{A}(\mathrm{~g})+3 \mathrm{~B}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{~g})+3 \mathrm{D}(\mathrm{~g})$

Correct answer
A
-777.7
B
+777.7
C
-1418.3
D
+1418.3

Solution

Given, $\begin{aligned} & \Delta \mathrm{H}_{\mathrm{A}}(\mathrm{~g})=9.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{B}}(\mathrm{~g})=-110.81 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{C}}(\mathrm{~g})=81 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \Delta \mathrm{H}_{\mathrm{D}}(\mathrm{~g})=-393 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \mathrm{~A}(\mathrm{~g})+3 \mathrm{~B}(\mathrm{~g}) \rightarrow \mathrm{C}(\mathrm{~g})+3 \mathrm{D}(\mathrm{~g}) \\ & \begin{array}{l} \mathrm{H} \\ \mathrm{H} \end{array}=\Delta \mathrm{H}_{\text {product }}-\Delta \mathrm{H}_{\text {Reactant }} \\ & \quad=[3 \times(-393)+81]-[9.7+3(-110)] \\ & =-1098+320.3 \\ & \Delta_{\mathrm{r}} \mathrm{H}=-777.7 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}$

Hence, the correct answer is option: 'A'.

Solution

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