Answer:
Kc = 77.9
Explanation:
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 <em>concentration or change of concentration in that stage</em>. 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.
![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](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%20%7D%7B%5BNO%5D%5E%7B2%7D.%5BBr_%7B2%7D%5D%20%7D%20%3D%5Cfrac%7B%287.08%20%5Ctimes%2010%5E%7B-2%7D%20%20%29%5E%7B2%7D%20%7D%7B%283.22%20%5Ctimes%2010%5E%7B-2%7D%29%5E%7B2%7D.6.21%20%5Ctimes%2010%5E%7B-2%7D%20%7D%20%3D77.9)