In the reaction AB2(l) + 3X2(g) = AX2(g) + 2BX2(g) AH = - 270 kcal per mol. of AB2(l), the enthalpies of formation of AX2(g) & BX2(g) are in the ratio of 4:3 and have opposite signato of 4.3 and have opposite sign. The valueof AH° (AB2(()) = + 30 kcal/mol. Then(A) AH° (AX2) = -96 kcal/mol(B) AH° (BX2) = + 480 kcal /mol(C) Kp = Kc & AH (AX2) = + 480 kcal /mol(D) Kp = Kc RT & AH1° (AX2) + AH1°(BX2) = -240 kcal /mol

In the reaction AB2(l) + 3X2(g) = AX2(g) + 2BX2(g) AH = - 270 kcal per

1.	In the reaction AB2(l) + 3X2(g) = AX2(g) + 2BX2(g) AH = - 270 kcal per mol. of AB2(l), the enthalpies of formation of AX2(g) & BX2(g) are in the ratio of 4:3 and have opposite signato of 4.3 and have opposite sign. The valueof AH° (AB2(()) = + 30 kcal/mol. Then(A) AH° (AX2) = -96 kcal/mol(B) AH° (BX2) = + 480 kcal /mol(C) Kp = Kc & AH (AX2) = + 480 kcal /mol(D) Kp = Kc RT & AH1° (AX2) + AH1°(BX2) = -240 kcal /mol
Answer» ΔH = -270 kcal/molΔH(AB₂) = 30 kcal/molTo find : ΔH(AX₂) = ?Solution :AB₂(L) + 3X₂(G) ⇔ AX₂(g) + 2BX₂(g) ΔH = -270 kcal/mol ΔH(AX₂) + 2Δ(BX₂) - ΔH(AB₂) = -27 4x - 2x = -240 -2x = -240 x = 120ΔH(AX₂) = 4×120 = 480 ΔH(AX₂) = 480 kcal/molKp = Kc×(RT)^(Δn×g) [Δn = 0] Kp = KcThe correct option is (C)