Chemistry - Chemical Equilibrium Question with Solution | TestHub

                     ${\text{NH}}_4\text{H}\text{S}\left(\text{s}\right)\rightleftharpoons \text{N}{\text{H}}_3(\text{g})+{\text{H}}_2\text{S}(\text{g})$

Initial moles       $\frac{3.06}{51}$                  0                   0

Moles at eq.    $\frac{3.06}{51}\times \frac{70}{100}$       $\frac{3.06}{51}\times \frac{30}{100}$       $\frac{3.06}{51}\times \frac{30}{\mathrm{}\mathrm{}\mathrm{}100}$

Given:

 V = 2 litre,

T = 300K,

 $\text{Δ}\text{n}=\mathrm{}\mathrm{}\mathrm{}2-0=2$

 $\therefore {\text{K}}_{\text{c}}=\left[\text{N}{\text{H}}_3\right]\left[{\text{H}}_2\text{S}\right]$

 $\begin{array}{c}{\text{K}}_{\text{c}}=\frac{3.06\times 30}{51\times 100\times 2}\times \frac{3.06\times 30}{51\times 100\times 2}\\ =8.1\times 10^{-5}{\text{M}}^2 \end{array}$

Also,

 $\begin{array}{c}{\text{K}}_{\text{p}}={\text{K}}_{\text{c}}{\left(\text{RT}\right)}^{\text{Δ}\text{n}}\\ =8.1\times {10}^{-5}{\left(0.082\times 300\right)}^2\\ =4.90\times {10}^{-2}{\text{atm}}^2\end{array}$

Addition of more  ${\text{NH}}_4\text{H}\text{S}$  on this equilibrium will cause no effect because concentration of  ${\text{NH}}_4\text{H}\text{S}$  is not involved in formula of  ${\text{K}}_{\text{p}}$ and  ${\text{K}}_{\text{c}}$ .