2009 ada dat #65

[UCF FINatic]

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Consider the equilibrium reaction:
2NO + O2 <-> 2NO2 K=2.0

Which of the following is the correct value of K for the reaction:
NO2 <-> NO + 1/2 O2

I obviously got this one wrong and was wondering why.

The correct answer is (1/2.0)^1/2

I guess I was expecting it to be negative... and for it to be the reciprocal times half.

 

[AlbinoPolarBear]

Consider the equilibrium reaction:
2NO + O2 <-> 2NO2, K=2.0

Which of the following is the correct value of K for the reaction:
NO2 <-> NO + 1/2 O2

I obviously got this one wrong and was wondering why.

The correct answer is (1/2.0)^1/2

I guess I was expecting it to be negative... and for it to be the reciprocal times half.

Just take this question 1 step at a time. There are 2 transformations to get to the final answer.

(1) 2NO + O2 <-> 2NO2, K=2.0
(2) 2NO2 <-> 2NO + O2

K = [P]/[R]

So if we flip P and R, we are gonna get [R]/[P] = 1/K, therefore
(2 )2NO2 <-> 2NO + O2, K=1/2.0

Next, we half the reaction, and this what you have to memorize.

A <->B, K= K
2A <-> 2B, K= K^2
0.5A <-> 0.5B, K = K^0.5

So the final transformation, we half it from 2 to 3.

(2) 2NO2 <-> 2NO + O2, K=1/2.0
(3) NO2 <-> NO + 1/2 O2, K=(1/2.0)^0.5

Hope that helps!