Question

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)

Answer 1

1. NO and Br₂

2. 77

3. 8.8

Step-by-step explanation:

Let's consider the following reaction.

2 NO(g) + Br₂(g) ⇄ 2 NOBr(g)

The equilibrium constant for this reaction is:

`Kc_(1)=([NOBr]^(2))/([NO]^(2)[Br_(2)]) =1.3 * 10^(-2)`

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)=([NO]^(2)[Br_(2)])/([NOBr]^(2)) =(1)/(Kc_(1)) =77`

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

The equilibrium constant is:

`Kc_(3)=([NO][Br_(2)]^(1/2) )/([NOBr]) =\sqrt{([NO]^(2)[Br_(2)])/([NOBr]^(2))} =\sqrt{Kc_(2)} =8.8`