Reducing reagents????

Question

The equilibrium constant for the reaction N2(g) + ½ O2 (g) ó N2O (g) is 7.1 x 10-19 at 25 oC. The equilibrium constant for the reaction N2(g) + O2 (g) ó 2NO (g) is 4.32 x 10-31 at 25 oC.

a) Given this data, what is the equilibrium constant for the reaction below at 25 oC? N2O (g) + ½ O2 (g) ó 2 NO (g)

b) What is the value of Kc at 25oC?

c) Assume this system is in equilibrium in a 1.000 L container, if the volume of the container is now reduced to 0.500 L, what happens to the composition of the container?

i understand how to do A, B, but i don't understand how to do C. Can anyone help me out with this problem?

Answer 1

The volume is decreased so the reactions will occur that would tend to reduce the volume.

That would be N2(g) + ½ O2 (g) -> N2O (g). There are 1½ moles of gas on the left side and only 1 mole on the right.

The moles of N2O(g) will increase. The moles of N2(g), O2 (g) and 2NO (g) will all decrease.