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Law of Mass Action

At 675 K, ...

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 $0.25$ mole of $C O_{2}$ is heated at $675 K$ , mole percentage of $C O (g)$ in equilibrium mixture is:

A

7.14

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B

14.28

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C

28.57

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D

33.33

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Solution

The correct option is B $14.28$

$H_{2}$
$C O_{2}$
$C O$
$H_{2} O$
Initial moles:
$0.25$
$0.25$
$0$
$0$
Moles at equilibrium: $0.25 - x$ $0.25 - x$
$x$
$x$
The expression for the equilibrium constant is $\frac{x}{0.25 - x} \times \frac{x}{0.25 - x} = 0.16$
Hence, $x = 0.2$
Thus, mole percentage of $C O$ in the equilibrium mixture is $\frac{0.2}{0.5} \times 100 = 14.28 %$

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Similar questions

675

H_{2} (g)

C O_{2} (g)

C O (g)

H_{2} O (g)

K_{p}

for the reaction is

0.16

. If a mixture of

0.25

H_{2} (g)

0.25

C O_{2}

675

%

C O (g)

in equilibrium mixture is:

Q. A mixture of one mole of

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attains equilibrium at a temperature of

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and a total pressure of 0.1 atm for the change

{C O}_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

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