Question

If 0.2 mol of $H_2\left(g\right)$ and 2.0 mol of S(s) are mixed in a 1.0 litre vessel at ${90}^{o}C$, the partial pressure of $H_{2}S\left(g\right)$ formed according to the reaction,

$H_2\left(g\right)+S\left(s\right)⟺H_{2}S\left(g\right)$; $K_p=6.8\times {10}^{-2}$ would be :

• A. 0.38 atm
• B. 0.19 atm
• C. 0.6 atm
• D. none of the above

Answer 1

The correct option is A 0.38 atm

$H_2\left(g\right)+S\left(s\right)\iff H_{2}S\left(g\right)$
0.2 2 0
0.2-x 2-x x
$\because \Delta n=0,K_p=K_c$
$K_c=\frac{\left[H_{2}S\right]}{\left[H_2\right]}=\frac{x}{(0.2-x)}=6.8\times {10}^{-2}$

On solving this, we get

$=>x=1.27\times {10}^{-2}$

$P_{H_{2}S}=\left[x\right]RT=1.27\times {10}^{-2}\times 0.0821\times (273+90)=0.38atm$