A student ran the following reaction in the laboratory at 341 K 2NO(g)Br2(g) =2NOBr(g) When she introduced 0.103 moles of NO(g) and 9.75x102 moles of Br2(g) into a 1.00 liter container, she found the equilibrium concentration of Br^(g) to be 6.21*102M Calculate the equilibrium constant, Ke she obtained for this reaction

To solve this equilibrium problem we will use an ICE Chart. We recognise 3 stages: Initial (I), Change (C) and Equilibrium (E). We complete each row with the concentration or change of concentration in that stage. Since the container is of 1.00 L, the initial concentrations are [NO] = 0.103 M and [Br₂] = 9.75 × 10⁻² M and the equilibrium concentration of Br₂ is 6.21 × 10⁻² M. Then,

2 NO(g) + Br₂(g) ⇄ 2 NOBr(g)

I 0.103 9.75 × 10⁻² 0

C -2x -x +2x

E 0.103 -2x 9.75 × 10⁻² - x 2x

Also, we know that

[Br₂]eq = 9.75 × 10⁻² - x = 6.21 × 10⁻² ⇒ x = 3.54 × 10⁻² M

We can use the value of x to find the concentrations at equilibrium:

[NO] = 0.103 -2x = 0.103 - 2 . 3.54 × 10⁻² = 3.22 × 10⁻² M

[Br₂] = 6.21 × 10⁻² M

[NOBr] = 2x = 2 . 3.54 × 10⁻² = 7.08 × 10⁻² M

We can use these concentrations in the equilibrium constant (Kc) expression.

[tex]Kc=\frac{[NOBr]^{2} }{[NO]^{2}.[Br_{2}] } =\frac{(7.08 \times 10^{-2} )^{2} }{(3.22 \times 10^{-2})^{2}.6.21 \times 10^{-2} } =77.9[/tex]