Answer:
1. NO and Br₂
2. 77
3. 8.8
Explanation:
Let's consider the following reaction.
2 NO(g) + Br₂(g) ⇄ 2 NOBr(g)
The equilibrium constant for this reaction is:
![Kc_{1}=\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]} =1.3 \times 10^{-2}](https://tex.z-dn.net/?f=Kc_%7B1%7D%3D%5Cfrac%7B%5BNOBr%5D%5E%7B2%7D%7D%7B%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D%7D%20%3D1.3%20%5Ctimes%2010%5E%7B-2%7D)
1. At this temperature does the equilibrium favor NO and Br₂, or does it favor NOBr?
Since Kc₁ < 1, the reactants are favored, that is, NO and Br₂.
2. Calculate Kc for 2 NOBr(g) ⇄ 2 NO(g) + Br₂(g)
The equilibrium constant is:
![Kc_{2}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}} =\frac{1}{Kc_{1}} =77](https://tex.z-dn.net/?f=Kc_%7B2%7D%3D%5Cfrac%7B%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D%7D%7B%5BNOBr%5D%5E%7B2%7D%7D%20%3D%5Cfrac%7B1%7D%7BKc_%7B1%7D%7D%20%3D77)
3. Calculate Kc for NOBr(g) ⇄ NO(g) + 1/2 Br₂(g)
The equilibrium constant is:
![Kc_{3}=\frac{[NO][Br_{2}]^{1/2} }{[NOBr]} =\sqrt{\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}} =\sqrt{Kc_{2}} =8.8](https://tex.z-dn.net/?f=Kc_%7B3%7D%3D%5Cfrac%7B%5BNO%5D%5BBr_%7B2%7D%5D%5E%7B1%2F2%7D%20%7D%7B%5BNOBr%5D%7D%20%3D%5Csqrt%7B%5Cfrac%7B%5BNO%5D%5E%7B2%7D%5BBr_%7B2%7D%5D%7D%7B%5BNOBr%5D%5E%7B2%7D%7D%7D%20%3D%5Csqrt%7BKc_%7B2%7D%7D%20%3D8.8)