Question

At a certain temperature, $K_P$ for dissociation of solid $Ca{CO}_3$ is $4\times {10}^{-2}atm$ and for the reaction, $C\left(s\right)+{CO}_2\rightleftharpoons 2CO$ is 2.0 atm, respectively. Calculate the pressure of CO at this temperature when solid C, CaO, CaCO${}_3$ are mixed and allowed to attain equilibrium.

• A. $P_{CO}^{\prime }=1.1$
• B. $P_{CO}^{\prime }=1.4$
• C. $P_{CO}^{\prime }=0.03$
• D. $P_{CO}^{\prime }=0.28$

Answer 1

Answer: (D)

$P_{CO}^{\prime }=0.28$

Equilibrium I:
$Ca{CO}_3\left(s\right)\rightleftharpoons CaO\left(s\right)+{CO}_2\left(g\right)$

$K_P=4\times {10}^{-2}=P_{{CO}_2}^{\prime }$

Equilibrium II:
Also $C\left(s\right)+{CO}_2\rightleftharpoons 2CO$

$K_P=\frac{{\left(P_{CO}\right)}^2}{P_{{CO}_2}^{\prime }}=2.0$

Now both are at equilibrium and thus $P_{{CO}_2}=4\times {10}^{-2}$ which should remain constant, thus
$2.0=\frac{{{(P}_{CO}^{\prime })}^2}{4\times {10}^{-2}}$
$P_{CO}^{\prime }=\sqrt{4\times {10}^{-2}\times 2}=0.28$