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11111nata11111 [884]
3 years ago
14

Calculate the pressures of NO, Cl2, and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 7.0 a

tm 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
Chemistry
1 answer:
kobusy [5.1K]3 years ago
5 0

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

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