Question

Equilibrium constant $K_P$ for ${2H}_{2}S\left(g\right)\rightleftharpoons {2H}_2\left(g\right)+S_2\left(g\right)$ is 0.0118 atm at 1065C and heat of dissociation is 42.4 Kcal.Find equilibrium constant at 1132C.

• A. $K_{P_2}=0.025atm$
• B. $K_{P_2}=1.20atm$
• C. $K_{P_2}=0.0142atm$
• D. $K_{P_2}=2.129atm$

Answer 1

Answer: (A)

$K_{P_2}=0.025atm$

As we know,
$2.303log\frac{K_{P_2}}{K_{P_1}}=\frac{△H}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
$2.303log\frac{K_{P_2}}{K_{P_1}}=\frac{42.4\times {10}^3}{2}\left[\frac{1405-1338}{1405\times 1338}\right]$
$\therefore \frac{K_{P_2}}{K_{P_1}}=2.129$
$\therefore K_{P_2}=2.129\times 0.0118=0.025atm$