Question

the value of Kp for the reaction,CO2(g)+C(s)->2CO(g) is 3.0 at 1000K. if initially pCO2=0.48bar and pCO=0bar and pure graphite is present, calulate the equilibrium partial pressures of CO and CO2.

Answer 1

Answer : The partial pressure of CO and $CO_2$ is 0.665 and 0.1475 bar respectively.

Explanation:

The balanced equilibrium reaction is:

               $CO_2(g)+C(s)\rightleftharpoons 2CO(g)$

At t=0        0.48           0          0  

At $t=t_{eq}$    (0.48 - p)               2p

The expression used for $K_p$ is,

$K_p=\frac{(p_{CO})^2}{p_{CO_2}}$

C does not come in this expression because it is in solid state and does not bear any pressure.

$3=\frac{(2p)^2}{0.48-p}$

p = 0.3325 bar

Now we have to calculate the partial pressure of CO and $CO_2$  

$p_{CO}=2p=2\times 0.3325 bar=0.665 bar$

$p_{CO_2}=0.48- p=0.48-0.3325 bar=0.1475 bar$

Therefore, the partial pressure of CO and $CO_2$ is 0.665 and 0.1475 bar respectively.