A vessel at 1000 K contains with a pressure of 0.5 atm. Some of is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then is: [2025]

Question

A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

Smart Achievers · Accepted Answer

(2)

$\begin{matrix} {CO}_{2} (g) & + & C (s) & ⟶ & 2 CO (g) \\ t = 0 & 0.5 atm & 0 & 0 \\ t = \infty & (0.5 - p) atm & 0 & 2 p atm \end{matrix}$

$Total pressure at equilibrium = P_{{CO}_{2}} + P_{CO} = 0.5 - p + 2 p = 0.5 + p = 0.8,$ $p = 0.3 atm$

Hence, $P_{{CO}_{2}} = 0.5 - p = (0.5 - 0.3) = 0.2 atm$

$P_{CO} = 2 p = 2 \times 0.3 = 0.6 atm$

$K_{p} = \frac{P_{CO}^{2}}{P_{{CO}_{2}}} = \frac{{(0.6)}^{2}}{0.2} = 1.8$

Solution

Q.
A vessel at 1000 K contains ${CO}_{2}$ with a pressure of 0.5 atm. Some of ${CO}_{2}$ is converted into CO on addition of graphite. If total pressure at equilibrium is 0.8 atm, then $K_{p}$ is: [2025]

1 3 atm

2 1.8 atm

3 0.18 atm

4 0.3 atm

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