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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. If the total pressure at equilibrium is $0.8 \mathrm{~atm}$, the value of $\mathrm{K}$ is
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The correct answer is:
$1.8 \mathrm{~atm}$
$1.8 \mathrm{~atm}$
Total pressure at equilibrium $=0.8 \mathrm{~atm}$; Total no.of moles $=p+x$.
Therefore $\mathrm{p} \propto \mathrm{n} ; \frac{0.5}{0.8}=\frac{\mathrm{p}}{\mathrm{p}+\mathrm{x}} \Rightarrow \mathrm{x}=0.3$
$$
\mathrm{K}_{\mathrm{p}}=\frac{\mathrm{P}_{\mathrm{CO}}^2}{\mathrm{P}_{\mathrm{CO}_2}}=\frac{0.6 \times 0.6}{0.2}=1.8 \mathrm{~atm}
$$
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