Answer : The equilibrium constant for this reaction is, 0.0475
Solution : Given,
Initial pressure of [tex]N_2[/tex] = 1.42 bar
Initial pressure of [tex]H_2[/tex] = 2.87 bar
[tex]K_p[/tex] = 0.036
The given equilibrium reaction is,
[tex]CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)[/tex]
Initially 0.596 0.256 0
At equilibrium (0.596-x) (0.256-x) 2x
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{CH_2Cl_2})^2}{(p_{CH_4})(p_{CCl_4})}[/tex]
Thus, the partial pressure of [tex]CCl_4[/tex] at equilibrium = 0.218 = (0.256 - x)
That means,
(0.256 - x) = 0.218
x = 0.038 atm
The partial pressure of [tex]N_2[/tex] at equilibrium = (1.42-x) = (1.42-0.287) = 1.133 bar
The partial pressure of [tex]CH_4[/tex] at equilibrium = (0.596-x) = (0.596-0.038) = 0.558 atm
The partial pressure of [tex]CH_2Cl_2[/tex] at equilibrium = 2x = 2 × 0.038 = 0.076 atm
Now put all the values of partial pressure in above expression, we get:
[tex]K_p=\frac{(p_{CH_2Cl_2})^2}{(p_{CH_4})(p_{CCl_4})}[/tex]
[tex]K_p=\frac{(0.076)^2}{(0.558)(0.218)}[/tex]
[tex]K_p=0.0475[/tex]
Therefore, the equilibrium constant for this reaction is, 0.0475
