Question

At ${87}^{0}C$ the following equilibrium is established.
$H_2\left(g\right)+S\left(s\right)\leftrightharpoons H_{2}S\left(g\right);K_c=0.08$
If $0.3$ mole hydrogen and $2$mole sulphur are heated to ${87}^0$ C in a $2$L vessel, what will be the comcentration of $H_{2}S$ at equilibrium?

• A. $0.011M$
• B. $0.022M$
• C. $0.044M$
• D. $0.08M$

Answer 1

The correct option is A $0.011M$

$K_c=\frac{\left[H_{2}S\right(g\left)\right]}{\left[H_2\right(g\left)\right]}$
$\Rightarrow 8\times {10}^{-2}=\frac{x}{0.3-x}$
$\Rightarrow x=0.022$
$\left[H_{2}S\right(g\left)\right]=\frac{0.0022}{2}$
$=0.011M$