Question

An equilibrium mixture in a vessel of capacity
100 L
contains
1
mole of
N
2
,
2
moles of
O
2
, and
3
of
NO
. No.of moles of
O
2
to be added so that, at new equilibrium, the concentration of
NO
is found to be
0.04 mol/L
?

Answer 1

The first thing that you need to do here is to calculate the equilibrium constant of the reaction at this particular temperature.

You know that

O2(g)+N2(g)⇌2NO2(g)

For this reaction, the equilibrium constant, Kc, is equal to

Kc=[NO]2[O2]⋅[N2]

The equilibrium concentrations of the three chemical species will be

[O2]=2 moles100 L=0.02 M

[N2]=1 mole100 L=0.01 M

[NO]=3 moles100 L=0.03 M

This means that you have--I'll leave the expression of the equilibrium constant without added units!

Kc=(0.03)20.02⋅0.01

Kc=32⋅1100222⋅1100⋅1⋅1100

Kc=92

Now, you know that after some oxygen gas is added to the reaction vessel, the equilibrium concentration of nitrogen oxide is equal to

[NO]new=0.04 M

The balanced chemical equation tells you that in order for the reaction to produce 1mole of nitrogen oxide, it must consume 12 moles of oxygen gas and 12 moles of nitrogen gas.

This means that in order for the concentration of nitrogen oxide to increase by

0.04 M − 0.03 M=0.01 M

the concentrations of nitrogen gas and of oxygen gas must decrease by 12⋅0.01 M. Since you didn't add any nitrogen gas to the reaction vessel, you can say that the new equilibrium concentration of nitrogen gas will be

[N2]new=[N2]−12⋅0.01 M

[N2]new=0.01 M−12⋅0.01 M

[N2]new=0.005 M

Next, use the expression of the equilibrium constant to find the new equilibrium concentration of oxygen gas

[O2]new=[NO]2newKc⋅[N2]new

[O2]new=(0.04)292⋅0.005=16225

This means that, when the new equilibrium is established, you have

[O2]new=16225  M

Use the fact that in order for the new equilibrium to be established, the reaction consumed 12⋅0.01 M of oxygen gas to find the concentration of oxygen gas afteryou increased the number of moles of this reactant,

[O2]increased=16225  M+12⋅0.01 M

[O2]increased=1371800  M

This means that the concentration of oxygen gas increased by

Δ[O2]=1371800  M−0.02 M

Δ[O2]=1011800  M

Finally, to find the number of moles of oxygen gas added to the reaction vessel, use the volume of the vessel

100L⋅1011800  moles O21L=10118  moles O2