Question

# At $$675$$K, $$H_2(g)$$ and $$CO_2(g)$$ react to form $$CO(g)$$ and $$H_2O(g)$$, $$K_p$$ for the reaction is $$0.16$$. If a mixture of $$0.25$$ mole of $$H_2(g)$$ and $$0.25$$ mol of $$CO_2$$ is heated at $$675$$K, mole $$\%$$ of $$CO(g)$$ in equilibrium mixture is:

A
7.14
B
14.28
C
28.57
D
33.33

Solution

## The correct option is B $$14.28$$We know that $$K_P=K_C(RT)^{\Delta n}$$                                     here, $$\Delta n=n_{Product}-n_{reactant}$$                                               $$=2-2=0$$$$\therefore K_P=K_C$$$$\therefore K_C=0.16$$                      $$H_2+CO_2 \longrightarrow CO+H_2O$$$$t=0$$           $$0.25$$     $$0.25$$           $$0$$          $$0$$At eqm  $$0.25-x$$  $$0.25-x$$     $$x$$          $$x$$Total number of moles at eqm= $$0.25-x+0.25-x+x+x=0.50$$now, $$K_C=\cfrac {[CO][H_2O]}{[H_2][CO_2]}$$$$\Rightarrow 0.16= \cfrac {x^2}{(0.25-x)^2}$$$$\Rightarrow 0.4= \cfrac {x}{0.25-x}\Rightarrow x=\cfrac {1}{4}$$$$\therefore$$ Moles of $$CO=\cfrac {1}{4}$$$$\therefore$$ mole % of $$CO=\cfrac {\text{moles of CO}}{\text{total moles}}\times 1000=\cfrac {1}{14\times 0.50}\times 100=14.28$$Chemistry

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