A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some… — Solution | JEE Chemistry

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A vessel at $1000 \, \text{K}$ contains $CO_2$ with a pressure of $0.5 \, \text{atm}$ . Some of $CO_2$ is converted into $CO$ on addition of graphite. If total pressure at equilibrium is $0.8 \, \text{atm}$ , then $K_p$ is:

A
$0.18 \, \text{atm}$
B
$1.8 \, \text{atm}$
C
$0.3 \, \text{atm}$
D
$3 \, \text{atm}$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The equilibrium reaction is

CO_2 + C \rightleftharpoons 2CO

Initial pressure of $CO_2$ is $0.5 \, \text{atm}$ and total equilibrium pressure is $0.8 \, \text{atm}$ .

Find: The value of $K_p$ .

Let the decrease in pressure of $CO_2$ be $x$ . Then the equilibrium pressures are:

$CO_2 : 0.5 - x$
$CO : 2x$

Using the total pressure at equilibrium:

(0.5 - x) + 2x = 0.8

So,

0.5 + x = 0.8

x = 0.3

Detailed Equilibrium Calculation

Therefore, the equilibrium partial pressures are:

P_{CO_2} = 0.5 - 0.3 = 0.2 \, \text{atm}

P_{CO} = 2 \times 0.3 = 0.6 \, \text{atm}

Use Total Pressure Increase Directly

Now use the expression for the equilibrium constant:

K_p = \frac{(P_{CO})^2}{P_{CO_2}}

Substituting the equilibrium pressures:

K_p = \frac{(0.6)^2}{0.2} = \frac{0.36}{0.2} = 1.8 \, \text{atm}

Therefore, the correct option is B.

A quick observation is that the total pressure increases from $0.5 \, \text{atm}$ to $0.8 \, \text{atm}$ , so the net increase is $0.3 \, \text{atm}$ . Since

CO_2 + C \rightleftharpoons 2CO

one mole of gaseous reactant produces two moles of gaseous product, the net increase in gaseous pressure is exactly $x$ . Hence $x = 0.3$ directly, after which the same $K_p$ calculation follows.

Common mistakes

Taking the change in $CO$ as $+x$ instead of $+2x$ is incorrect because the stoichiometric coefficient of $CO$ is $2$ . Always use the balanced reaction to write equilibrium changes.
Using total pressure $0.8 \, \text{atm}$ directly in the $K_p$ expression is wrong because $K_p$ requires partial pressures of the gaseous species. First find equilibrium partial pressures of $CO_2$ and $CO$ .
Including graphite in the equilibrium constant is incorrect because solid carbon does not appear in the expression for $K_p$ . Only gaseous species are included here.

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Answer

Step-by-step solution

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