Question

A mixture of $1.0mole$ of sulphur and $0.2mole$ of hydrogen heated at ${90}^{o}C$ in a $onelitre$ flask. The equilibrium constant for the formation of hydrogen sulphide, $H_2\left(g\right)+S\left(s\right)\rightleftharpoons H_{2}S\left(g\right)$ is $6.8\times {10}^{-2}$. Calculate the pressure of $H_{2}S$ at equilibrium.

Answer 1

$Step1:$

Initial moles of $H_2=0.244$

Initial moles of $S=1$

$Kp=6.8\times {10}^{-2}$

Given equation:

$H_{2\left(g\right)}+S_{\left(s\right)}\rightleftharpoons H_2{S}_{\left(g\right)}\Rightarrow \left(1\right)$

Initial moles: $0.2$ $1$

At equilibrium : $(0.2-\alpha )$ $(1-\alpha )$ $\alpha$

Here, in the above equation, we can see that hydrogen is the limiting reagent.

$\therefore K_p=\frac{\left[H_{2}S\right]}{\left[H_2\right]}$

$\Rightarrow 6.8\times {10}^{-2}=\frac{\alpha }{0.2-\alpha }$

$1.36\times {10}^{-2}-(6.8\times {10}^{-2})\alpha =\alpha$

$\alpha +0.068\alpha =1.36\times {10}^{-2}$

$\alpha =\frac{1.36\times {10}^{-2}}{1.068}$

$\alpha =1.273\times {10}^{-2}$

So, at equilibrium moles of $H_2=0.2-\alpha =0.21.273\times {10}^{-2}=0.1873$

$Step2:$

Now, using the Ideal Gas equation,

$PV=nRT\to \left(1\right)$

Where $P=$ total pressure of the vessel

$n=$ total no. of moles $=(0.2-\alpha )+(1-\alpha )+\alpha =1.2-\alpha =1.2-1.273\times {10}^{-2}=1.1873$

$V=$ volume of vessel $=1$ litre

$R=$ Ideal gas constant $=0.082Latm{K}^{-1}mo{l}^{-1}$

$T=$ total temperature $={90}^{o}C=90+273=363K$

Substituting all the values in eq. (1), we get

$P\times 1=1.1873\times 0.082\times 363$

$\Rightarrow P=35.34atm$

$Step3:$

Thus,

The partial pressure of $H_{2}S$ at equilibrium

$=\left(molefractionof{H}_{2}S\right)\times \left(totalpressure\right)$

$=\left[\frac{1.273\times {10}^{-2}}{1.1873}\right]\times 35.34$

$=0.3789atm$

$\approx 0.38atm$