Question

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 sign
ato of 4.3 and have opposite sign. The value
of 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 1

Given :

ΔH =  -270 kcal/mol

ΔH(AB₂) = 30 kcal/mol

To 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 = 120

• ΔH(AX₂) = 4×120 = 480

      ΔH(AX₂) = 480 kcal/mol

• Kp = Kc×(RT)^(Δn×g)    [Δn = 0]

       Kp = Kc

The correct option is (C)