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Question

At $${90}^{o}C$$, the following equilibrium is established
$${H}_{2}(g)+S(s)\rightleftharpoons {H}_{2}S(g)$$ $${K}_{p}=6.8\times {10}^{-2}$$
If $$0.2$$ mol of hydrogen and $$1.0$$ mol of sulphur are heated to $${90}^{o}C$$ in a $$1.0$$ litre vessel, what will be the partial pressure of $${H}_{2}S$$ at equilibrium?


Solution

$$K_C=\cfrac {[H_2S]}{[H_2][S]}$$
$$\because K_P=K_C (RT)^{\Delta ng}$$ where $$\Delta ng$$= Difference in gaseous moles of products and reactants
Here $$\Delta ng= (1)-(1)=0$$
$$\therefore K_P=K_C=6.8\times 10^{-2}= \cfrac {[H_2S]}{0.2\times 1}$$
$$\therefore [H_2S]=1.36 \times 10^{-2}$$= Number of moles of $$H_2S$$
Partial pressure of $$H_2S=$$ (mole fraction of $$H_2S$$) $$\times$$ Total pressure
$$=X_{H_2S}\times P_t.....(1)$$
$$P_t= 1 atm$$
$$P_{H_2S}=\cfrac {1.36\times 10^{-2}}{1.36\times 10^{-2}+ 0.2+1}=0.0112 atm$$

Chemistry

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