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Q. A vessel at $1000 \,K$ contains $CO_2$ with a pressure of $0.5\, atm$. Some of the $CO_2$ is converted into $CO$ on the addition of graphite. The value of $K$ if the total pressure at equilibrium is $0.8 \,atm$, is

VITEEEVITEEE 2011

Solution:

$\ce{CO2(g) + C(s) <=> 2CO(g)}$
image
$= \left(0.5 - x\right) + 2x = 0.8$
$\therefore x = 0.3\, atm.$
$\therefore p_{CO_2}= 0.5 - 0.3 = 0.2$
$PCO = 2x = 2 \times 0.3 = 0.6 \,atm$
$K = \frac{p^{2}_{co}}{p_{co_2}} = \frac{\left(0.6\right)^{2}}{0.2} = 1.8\,atm$