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In the reaction:
${AB}_{2} (l) + 3 X_{2} (g) ⇌ {AX}_{2} (g) + 2 {BX}_{2} (g)$ $ΔH = - 270 kcal$ per mole. of ${AB}_{2} (l)$ . The enthalpies of formation of ${AX}_{2} (g)$ and ${BX}_{2} (g)$ are in the ratio of 4 : 3 and have opposite sign. The value of ${ΔH}_{f}^{\circ} ({AB}_{2} (l)) = + 30 kcal / mol$ Then:

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$K_{p} = K_{c} RT$ and ${ΔH}_{f}^{\circ} ({AX}_{2}) + {ΔH}_{f}^{\circ} ({BX}_{2}) = - 240 kcal / mol$

${ΔH}_{f}^{\circ} ({AX}_{2}) = - 96 kcal / mol$

${ΔH}_{f}^{b} ({BX}_{2}) = + 480 kcal / mol$

$K_{p} = K_{c} and {ΔH}_{f}^{\circ} ({AX}_{2}) = + 480 kcal / mol$

answer is C.

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Detailed Solution

$\begin{array}{l} - 270 = 4 a - 6 a - 30 \\ \Rightarrow a = 120 \\ {ΔH}_{f}^{\circ} ({AX}_{2}) = 480 \\ {ΔH}_{f}^{\circ} ({BX}_{2}) = - 360 \\ {Δn}_{g} = 0 \\ k_{P} = k_{C} \end{array}$

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