In the reaction: $$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$ Suppose the system has 3 moles of $\ce{N2}$, 1 mole of $\ce{H2}$ and 1 mole of $\ce{NH3}$ at equilibrium. $Q_\chi=K_\chi=8.33$ and $Q_p=K_p$.

Now I add 5 moles of $\ce{N2}$ in the container by keeping (i) pressure and (ii) volume constant (temperature constant in both).

According to Le Chatelier's principle:

  • If pressure is constant: equilibrium shifts to backward direction.
  • If volume is constant: equilibrium shifts to forward direction.

But if I look at the equilibrium in terms of mole fraction, $Q_\chi$ will be the same in both the cases and hence equilibrium should shift in the same direction i.e. in backward direction.

Why do they predict different equilibrium states?

[$K_\chi$ is equilibrium constant in terms of mole fraction, $K_p$ is equilibrium constant in terms of partial pressures].


Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature. It depends on total pressure as well as temperature. Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed. In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.


Le Chatelier's principle says that, for a system at equilibrium, a change in temperature, pressure or concentration of ONE of the components will tend to shift to counteract the change.

Consider first case ii, constant volume. If additional nitrogen is introduced into this volume, the concentration (pressure) of nitrogen will increase, while all other concentrations (partial pressures) remain the same. By Le Chatelier's principle, this single change tends to drive the reaction toward a higher concentration of NH3 (forward, not backward).

Keeping total pressure constant (case i) requires expanding the volume by a factor of two (3+1+1 moles increases to 3+1+1+5 moles). This changes ALL of the concentrations (pressures).

The concentration of N2 increases by a factor of 6. The effect of H2 is 8x in the backward direction (1/2 cubed) and the effect of decreased NH3 is 4x in the forward direction (1/2 squared). The overall effect will be 6 - 8 + 4 = 2; i.e., the addition of extra N2 and volume expansion still favors the forward direction.

However, keeping the total pressure constant violates Le Chatelier's principle by changing more than one variable (concentrations of reactants and product). The agreement of the predictions in the two cases is fortuitous. If more N2 were added (10 moles), the reaction would be predicted to favor the backward direction.

BTW, mole fraction is not an appropriate unit of concentration or pressure.


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