Use this information to answer the following question.

Question

At a certain temperature, the equilibrium constant K for the following reaction is ×2.51x10¹²

N₂(g) + O₂(g)  2NO(g)

Use this information to answer the following question. 

Suppose a 44. L reaction vessel is filled with 1.2 mol of NO. What can you say about the composition of the mixture in the vessel at equilibrium?

A. There will be very little N2 and O2.
B.There will be very little NO.
C. Neither of the above is true.

Answer 1

N₂(g) + O₂(g)  2NO(g) .. K = 2.51x10¹²

Because the equilibrium constant (K) is so large (so much greater than 1), we know that the reaction will lie very far to the right. That is to say, we know that there will be very little of the N₂ and O₂ left and the amount of NO will be very large. So option A would be correct.

If you want to actually do the calculations, you can set up an ICE table as follows:

N₂(g) + O₂(g)  2NO(g)

0..........0...............1.2.........Initial

+x........+x.............-2x.........Change

x...........x..............1.2-2x....Equilibrium

The K for the reverse reaction is 1 / 2.51x10¹² = 3.98x10^-13

3.98x10^-13 = x² / (1.2-x)² and ignore the 2x since K is so small

3.98x10^-13 = x² / (1.2)²

x = 7.6x10^-7 moles of N2 and O2 formed

So @ equilibrium we can predict the following concentrations:

[N₂] = 7.6x10^-7 moles / 44 L = 1.73x10^-8 M

[O_2] = 7.6x10^-7 moles / 44 L = 1.73x10^-8 M

[NO] = ~1.2 moles / 44 L = ~0.027 M