Question

If $0.2mol$ of $H_2\left(g\right)$ and $2mol$ of $S\left(s\right)$ are mixed in a $1d{m}^3$ vessel at ${90}^{o}C,$ the partial pressure of $H_{2}S\left(g\right)$ formed according to the reaction would be :
$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S;K_P=6.8\times {10}^{-2}$

• A. 0.19 atm
• B. 0.38 atm
• C. 0.6 atm
• D. 0.072 atm

Answer 1

The correct option is B 0.38 atm

We know that,

$1d{m}^3=1L;Concentration=\frac{Moles}{Volume}$

Since, volume =$1d{m}^3=1L$

so, Concentration = Moles

According to the reaction:

$H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$
$Initially:0.220$
$Equilibrium:0.2-x2-xx$

$\because △n_g=0,\therefore K_p=K_c$
$K_c=\frac{\left[H_{2}S\right]}{\left[H_2\right]}=\frac{x}{(0.2-x)}=6.8\times {10}^{-2}$

On solving this, we get
$\Rightarrow x=1.27\times {10}^{-2}$
$P_{H_{2}S}=CRT=1.27\times {10}^{-2}\times 0.0821\times (273+90)=0.38atm$