Question

The equilibrium constant, Kc, for the following reaction is 1.28×10-3 at 231 K. 2NOBr(g) goes to 2NO(g) + Br2(g) .

When a sufficiently large sample of NOBr(g) is introduced into an evacuated vessel at 231 K, the equilibrium concentration of Br2(g) is found to be 0.200 M.

Calculate the concentration of NOBr in the equilibrium mixture. __M

Answer 1

Let the initial concentration of NOBr be C Molar

2 NOBr(g)       <->   2NO(g)   +   Br2(g)
C               0           0       (initial)
C-2x               2x           x       (equilibrium)

Given at equilibrium,
[Br2] = 0.200 M

So,
x = 0.200 M

So,
[NO] = 2x
= 2*0.200 M
= 0.400 M

Now use:
Kc = [NO]^2[Br2]/[NOBr]^2
1.28*10^-3 = (0.400)^2 * 0.200 / [NOBr]^2
[NOBr]^2 = 25
[NOBr] = 5.0 M

Answer: 5.0 M