The equilibrium constant for the reaction:

2NO(g)+Br2(g) <-> 2NOBr(g) is Kc= 1.3x10^-2 at 1000 K.

1. At this temperature does the equilibrium favor NO and Br2, or does it favor NOBr?
2. Calculate Kc for 2NOBr(g) <-> 2NO(g)+Br2(g)
3. Calculate Kc for NOBr(g) <-> NO(g)+1/2 Br2(g)

Question

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dsdrajlin dsdrajlin · Answer 1 · 2019-12-15T15:34:29+01:00

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:

[tex]Kc_{1}=\frac{[NOBr]^{2}}{[NO]^{2}[Br_{2}]} =1.3 \times 10^{-2}[/tex]

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:

[tex]Kc_{2}=\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}} =\frac{1}{Kc_{1}} =77[/tex]

3. Calculate Kc for NOBr(g) ⇄ NO(g) + 1/2 Br₂(g)

The equilibrium constant is:

[tex]Kc_{3}=\frac{[NO][Br_{2}]^{1/2} }{[NOBr]} =\sqrt{\frac{[NO]^{2}[Br_{2}]}{[NOBr]^{2}}} =\sqrt{Kc_{2}} =8.8[/tex]