The decomposition of NOCl to form NO and Cl₂,2 NOCl(g) $\leftrightarrows$
2 NO(g) + Cl₂(g) ,has a K_p value of 1.6 $\times$ 10^-5 at some temperature.If the initial partial pressures of NO and Cl₂ in a closed container are 0.80 atm and 0.40 atm,respectively,what is approximate equilibrium partial pressure of NOCl?

Question

The decomposition of NOCl to form NO and Cl₂,2 NOCl(g)

\leftrightarrows

2 NO(g) + Cl₂(g) ,has a K_p value of 1.6

\times

10^-5 at some temperature.If the initial partial pressures of NO and Cl₂ in a closed container are 0.80 atm and 0.40 atm,respectively,what is approximate equilibrium partial pressure of NOCl?

Quizplus · Accepted Answer

Let x be the equilibrium partial pressure of NOCl. Then, using the equilibrium constant expression:
Kp = (PNO)2(PCl2)/PNOCl2
1.6 x 10^10 = (0.8-x)^2 (0.4-x)
Since x is expected to be small, we can approximate (0.8-x)^2 and (0.4-x) as 0.64 and 0.4, respectively.
1.6 x 10^10 = 0.64(0.4-x)^2
Solving for x, we get:
x = 0.77 atm (rounded to two significant figures) 
Therefore, the answer is A.

The Decomposition of NOCl to Form NO and Cl2,2 NOCl(g) ⇆\leftrightarrows⇆

The Decomposition of NOCl to Form NO and Cl₂,2 NOCl(g) $\leftrightarrows$