Question

Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 7.0 atm NO and 3.5 atm Cl2. (Hint: Kp is relatively large; assume the reaction goes to completion then comes back to equilibrium.)
2 NO(g) + Cl2(g) --> 2 NOCl(g)Kp = 2.9 ✕ 103 at 149°C

Answer 1

Answer:  The pressures of $NO$, $Cl_2$, and $NOCl$ in an equilibrium mixture are 0.4 atm , 0.2 atm and 6.6 atm respectively.

Explanation:

Initial pressure of $NO$ = 7.0 atm

Initial pressure of $Cl_2$ = 3.5 atm

The given balanced equilibrium reaction is,

                   $2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

Initial pressure    7.0 atm   3.5 atm      0 atm

At eqm. conc.    (7-2x)atm  (3.5-x)atm   (2x) Matm

The expression for equilibrium constant for this reaction will be,

$K_p=\frac{[NOCl]^2}{[NO]^2[Cl_2]}$

$K_p=\frac{(2x)^2}{(7-2x)^2\times (3.5-x)}$

we are given : $K_p=2.9\times 10^3$

Now put all the given values in this expression, we get :

$2.9\times 10^3=\frac{(2x)^2}{(7-2x)^2\times (3.5-x)}$

$x=3.3atm$

Pressure of $NO$ at equilibrium = (7-2x) = $(7-2\times 3.3)=0.4 atm$

pressure of $Cl_2$ at equilibrium = (3.5-x) = $(3.5-3.3)=0.2 atm$

pressure of $NOCl$ at equilibrium = (2x) = $(2\times 3.3)=6.6 atm$