Question

$H_2{S}_{\left(g\right)}$ initially at a pressure of 10 atm and a temperature of 800 K, dissociates as $2H_2{S}_{\left(g\right)}\rightleftharpoons 2H_2+S_{2\left(g\right)}$ At equilibrium, the partial pressure of $S_2$ vapor is 0.02 atm. Thus, $K_p$ is:

• A. $3.23\times {10}^{-7}$
• B. $6.45\times {10}^{-7}$
• C. $1.55\times {10}^6$
• D. $6.2\times {10}^{-7}$

Answer 1

The correct option is A $3.23\times {10}^{-7}$

$2H_2{S}_{\left(g\right)}\rightleftharpoons 2H_{2\left(g\right)}+S_{2\left(g\right)}$

Pressure

at $t=0$ $Pi$ $-$ $-$

at eqm $Pi-P$ $2P$ $P$

as $P=0.02$ thus $Pi-P=10-0.02$

$Pi=10$ $2P=0.04$

$K_P=\frac{\left[P_{S_2}\right]{\left[P_{H_2}\right]}^2}{{\left[P_{H_{2}S}\right]}^2}$

$K_P=\frac{\left[0.02\right]{[2\times 0.02]}^2}{{[10-0.02]}^2}$

$K_P=3.23\times {10}^{-7}$ atm.