Q33 / 120 · VITEEE 2011 · Chemistry

A vessel at $1000 \mathrm{~K}$ contains $\mathrm{CO}_{2}$ with a pressure of $0.5 \mathrm{~atm} .$ Some of the $\mathrm{CO}_{2}$ is converted into $\mathrm{CO}$ on the addition of graphite. The value of $\mathrm{K}$ if the total pressure at equilibrium is $0.8 \mathrm{~atm}$, is

  • Correct answer
    A
    $1.8 \mathrm{~atm}$
  • B
    $3 \mathrm{~atm}$
  • C
    $0.3 \mathrm{~atm}$
  • D
    $0.18 \mathrm{~atm}$

Solution

$\mathrm{CO}_{2}(\mathrm{~g})+\mathrm{C}(\mathrm{s}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})$ Initial $\quad 0.5 \mathrm{~atm}$ pressure At equi $(0.5-\mathrm{x})$ $\therefore$ Total no. of moles at equilibrium $$ \begin{aligned} \therefore \quad & x=0.3 \mathrm{~atm} \\ \therefore \quad & \mathrm{p}_{\mathrm{CO}_{2}}=0.5-0.3=0.2 \\ & \mathrm{P}_{\mathrm{CO}}=2 x=2 \times 0.3=0.6 \mathrm{~atm} \\ & \mathrm{~K}=\frac{\mathrm{p}_{\mathrm{CO}}^{2}}{\mathrm{p}_{\mathrm{CO}_{2}}}=\frac{(0.6)^{2}}{0.2}=1.8 \mathrm{~atm} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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