Question

CO(g) + 2H2 = CH3OH , derive an expression for equilibrium constant Kp in terms of the extent of reaction ∆ and the total pressure p1 if initially 2 mol of CO and 3 mol of the H2 are mixed ?​

Answer 1

Answer : The expression for equilibrium constant Kp will be, $K_p=\frac{\alpha(5-2\alpha)^2}{(2-\alpha)(3-4\alpha)^2(P_1)^2}$

Solution :

Degree of dissociation = $\alpha$

Total pressure = $P_1$

The given equilibrium reaction is,

                       $CO(g)+2H_2(g)\rightleftharpoons CH_3OH(g)$

Initially              2          3                     0

At equilibrium  $(2-\alpha)$   $(3-2\alpha)$           $\alpha$

$\text{ Total number of moles}=2-\alpha+3-2\alpha+\alpha=5-2\alpha$

Now we have to calculate the partial pressure of $CO$ and $H_2$

$\text{ Partial pressure of }CO=\frac{\text{Moles of }CO}{\text{Total number of moles}}\times P_1=\frac{2-\alpha}{5-2\alpha}\times P_1$

$\text{ Partial pressure of }H_2=\frac{\text{Moles of }H_2}{\text{Total number of moles}}\times P_1=\frac{3-2\alpha}{5-2\alpha}\times P_1$

$\text{ Partial pressure of }CH_3OH=\frac{\text{Moles of }CH_3OH}{\text{Total number of moles}}\times P_1=\frac{\alpha}{5-2\alpha}\times P_1$

The relation between the $K_p$ and the total pressure is,

$K_p=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}$

Now put all the values of partial pressure, we get

$K_p=\frac{(\frac{\alpha}{5-2\alpha}\times P_1)}{(\frac{2-\alpha}{5-2\alpha}\times P_1)\times (\frac{3-2\alpha}{5-2\alpha}\times P_1)^2}$

$K_p=\frac{\alpha(5-2\alpha)^2}{(2-\alpha)(3-4\alpha)^2(P_1)^2}$

Therefore, the expression for equilibrium constant Kp will be, $K_p=\frac{\alpha(5-2\alpha)^2}{(2-\alpha)(3-4\alpha)^2(P_1)^2}$