Question

She introduced 0.174 moles of nobr(g) into a 1.00 liter container, she found the equilibrium concentration of br2(g) to be 1.79×10-2 m. calculate the equilibrium constant, kc, she obtained for this reaction.

Answer 1

The equilibrium reaction is

2 NOBr ⇄ 2 NO + Br₂

The initial molarity of NOBr is 0.174 mol/1 L = 0.174 M. Let's apply the ICE approach which stands for Initial-Change-Equilibrium.

2 NOBr ⇄ 2 NO + Br₂
I 0.174 0 0
C -2x +2x +x
----------------------------------------------------
E 0.174-2x 2x x

Since E for Br₂ is 1.79*10⁻² M, then that means x = 1.79*10⁻² M. We can compute E for the other compounds.

E for NOBr = 0.174 - 2(1.79*10⁻²) = 0.1382 M
E for NO = 2(1.79*10⁻²) = 0.0358 M

The expression for Kc according to the reaction is:
Kc = [NO]²[Br₂]/[NOBr]²
Kc = [0.0358]²[1.79*10⁻²]/[0.1382]²
Kc = 1.2×10⁻³