# Chem II

posted by
**Rick**
.

2.5 mol NOCl(g) was placed in a 2.50 L reaction vessel at 400 degrees C. After equilibrium was established, it was found that 28% of the NOCl had dissociated according to the equation:

2NOCl(g)<-->2NO(g)+Cl2(g)

Calculate the equilibrium constant, Kc, for the reaction.

2NOCl(g)<-->2NO(g)+Cl2(g)

Kc=[(NO)^2(Cl2)]/(NOCl)^2

Starting with 2.5 mol NOCl, if 28% is dissociated, then (Cl2) must be 2.5 mol x 0.28. (NO) must be twice that, and (NOCl) must be 2.5 mol - (Cl). Plug and chug. Check my thinking.

Bob, the choices are:

a. 0.021

b. 0.039

c. 0.169

d. 26

e. 47

I keep getting 0.169

Help me with my thinking. Once I plug and chug, I should be dividing each by 2.50 L's and squaring the results of NO and NOCl before completing it, right?