Question

In the reaction AB2($\mathcal{l}$) + 3X2(g) $\rightleftharpoons$ AX2(g)+ 2BX2(g) $\Delta$H = – 270 kcal per mol. of AB2($\mathcal{l}$),the enthalpies of formation of AX2(g)& BX2(g) are in the ratio of 4 : 3 and have opposite sign. The value of DHf0 (AB2(l)) = + 30 kcal/mol. Then

• A. $\Delta$Hf0 (AX2) = – 96 kcal /mol
• B. $\Delta$Hf0 (BX2) = + 480 kcal /mol
• C. Kp = Kc&$\Delta$Hf0 (AX2) = + 480 kcal /mol
• D. Kp = Kc RT &$\Delta$Hf0 (AX2) + $\Delta$Hf0(BX2) = –240 kcal /mol

Answer 1

Answer: (C)

Kp = Kc&$\Delta$Hf0 (AX2) = + 480 kcal /mol

Answer 2

AB2 ($\mathcal{l}$) + 3X2 (g) $\rightleftharpoons$ AX2 (g) + 2BX2

$\Delta$H = – 270 Kcal

$\Delta$HF⁰ (AX2) + 2 $\Delta$HF⁰ (BX2) – $\Delta$HF⁰ (AB2) = – 270

$\Delta$HF⁰ (AX2) + 2 $\Delta$HF⁰ (BX2) = – 240

$\Rightarrow$ 4x – 6x = – 240

$\Rightarrow$– 2x = – 240

$\Rightarrow$ x = 120

$\Delta$HF⁰ (AX2) = 4 × 120 = 480 Kcal/mole

Kp = Kc (RT)$\Delta$ng

$\Delta$ng = 0 So, Kp = Kc