For the Haber-Bosch equilibrium $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ with $\Delta H = -92\,\text{kJ}$, what happens to the equilibrium position when the volume of the container is decreased (choose: shift to reactant side, shift to product side, or no change), and what would the resulting concentration vs. time graph look like?

What this change does to the system

Decreasing the container volume squeezes the same amount of gas into less space. That raises the total pressure (and the partial pressures or concentrations of all gases) instantly.

Using Le Chatelier’s principle (moles of gas)

The equilibrium is: $$N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$$ Count gas moles on each side:

• Reactant side: $1 + 3 = 4$ mol gas
• Product side: $2$ mol gas

Higher pressure favors the side with fewer gas moles, so the system shifts right to reduce pressure.

Effect: shift to product side (more $NH_3$).

How the concentration vs. time graph should look

At the instant the volume is decreased (time $t_0$):

• $[N_2]$, $[H_2]$, and $[NH_3]$ all jump up immediately (same moles, smaller volume).

Then, as the equilibrium shifts right:

• $[N_2]$ decreases from that jumped-up value to a new steady value
• $[H_2]$ decreases from that jumped-up value to a new steady value
• $[NH_3]$ increases further to a new steady value

A simple sketch (concentration vs time):

concentration
 ^
 | NH3 _________
 | / \
 | _________/ \______
 | | (instant jump)
 | N2 |\__________
 | | \____________
 | H2 |\__________
 | | \____________
 +-----+------------------------------> time
 t0 (volume decreases)

The key features are the immediate jump for all species at $t_0$, followed by opposite trends: reactants down, product up, until a new equilibrium is reached.