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