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Degree of Dissociation

At 675K, ...

Question

At $675$ K, $H_{2} (g)$ and $C O_{2} (g)$ react to form $C O (g)$ and $H_{2} O (g)$ , $K_{p}$ for the reaction is $0.16$ . If a mixture of $0.25$ mole of $H_{2} (g)$ and $0.25$ mol of $C O_{2}$ is heated at $675$ K, mole $%$ of $C O (g)$ in equilibrium mixture is:

7.14

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14.28

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28.57

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33.33

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Solution

The correct option is B $14.28$
We know that $K_{P} = K_{C} (R T)^{Δ n}$
here, $Δ n = n_{P r o d u c t} - n_{r e a c t a n t}$
$= 2 - 2 = 0$
$∴ K_{P} = K_{C}$
$∴ K_{C} = 0.16$
$H_{2} + C O_{2} ⟶ C O + H_{2} O$
$t = 0$ $0.25$ $0.25$ $0$ $0$
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, $K_{C} = \frac{[C O] [H_{2} O]}{[H_{2}] [C O_{2}]}$
$\Rightarrow 0.16 = \frac{x^{2}}{(0.25 - x)^{2}}$
$\Rightarrow 0.4 = \frac{x}{0.25 - x} \Rightarrow x = \frac{1}{4}$
$∴$ Moles of $C O = \frac{1}{4}$
$∴$ mole % of $C O = \frac{moles of CO}{total moles} \times 1000 = \frac{1}{14 \times 0.50} \times 100 = 14.28$

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Similar questions

Q. At

675 K

H_{2} (g)

and

C O_{2} (g)

react to form

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and

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, mole percentage of

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Q. A mixture of one mole of

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Q. The equilibrium constant for the reaction

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At 675K, H2(g) and CO2(g) react to form CO(g) and H2O(g), Kp for the reaction is 0.16. If a mixture of 0.25 mole of H2(g) and 0.25 mol of CO2 is heated at 675K, mole % of CO(g) in equilibrium mixture is:

At $675$ K, $H_{2} (g)$ and $C O_{2} (g)$ react to form $C O (g)$ and $H_{2} O (g)$ , $K_{p}$ for the reaction is $0.16$ . If a mixture of $0.25$ mole of $H_{2} (g)$ and $0.25$ mol of $C O_{2}$ is heated at $675$ K, mole $%$ of $C O (g)$ in equilibrium mixture is: