Respuesta :
Answer : The partial pressure of [tex]N_2,H_2\text{ and }NH_3[/tex] at equilibrium are, 1.133, 2.009, 0.574 bar respectively. The total pressure at equilibrium is, 3.716 bar
Solution : Â 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]N_2(g)+H_2(g)\rightleftharpoons 2NH_3(g)[/tex]
Initially           1.42    2.87       0
At equilibrium   (1.42-x)  (2.87-3x)    2x
The expression of [tex]K_p[/tex] will be,
[tex]K_p=\frac{(p_{NH_3})^2}{(p_{N_2})(p_{H_2})^3}[/tex]
Now put all the values of partial pressure, we get
[tex]0.036=\frac{(2x)^2}{(1.42-x)\times (2.87-3x)^3}[/tex]
By solving the term x, we get
[tex]x=0.287\text{ and }3.889[/tex]
From the values of 'x' we conclude that, x = 3.889 can not more than initial partial pressures. So, the value of 'x' which is equal to 3.889 is not consider.
Thus, the partial pressure of [tex]NH_3[/tex] at equilibrium = 2x = 2 × 0.287 = 0.574 bar
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]H_2[/tex] at equilibrium = (2.87-3x) = [2.87-3(0.287)] = 2.009 bar
The total pressure at equilibrium = Partial pressure of [tex]N_2[/tex] + Partial pressure of [tex]H_2[/tex] + Partial pressure of [tex]NH_3[/tex]
The total pressure at equilibrium = 1.133 + 2.009 + 0.574 = 3.716 bar
