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 $[N_2(g)]$ is increased (shift to reactant side, shift to product side, or no change), and what would the resulting concentration vs time graph look like?

What the change is doing to the system

You are adding more $N_2(g)$, which is a reactant. That is a “concentration stress” on the equilibrium mixture.

Using Le Chatelier’s principle (and $Q$ vs $K$)

Right after adding $N_2$, the reaction quotient $$Q = \frac{[NH_3]^2}{[N_2][H_2]^3}$$ gets smaller because the denominator increases. So $Q < K$, and the system responds by moving forward (to the right) until $Q$ increases back to $K$. That forward shift consumes $N_2$ and $H_2$ and produces more $NH_3$.

What the concentration-time graph should show

Assume you plot concentration vs time and the disturbance happens at time $t_0$.

• $[N_2]$: instant vertical increase at $t_0$ (because you added it), then a gradual drop as it is consumed, ending above the original level but below the spike.
• $[NH_3]$: no instant jump at $t_0$ (you did not add $NH_3$), then a gradual increase to a higher equilibrium plateau.
• $[H_2]$: no instant jump at $t_0$, then a gradual decrease to a lower equilibrium plateau.

Simple sketch (qualitative)

concentration
^
| N2 |\
| | \
| | \_____
| NH3 | /''''''
| | /
| |___/__________
| H2 |\
| | \
| | \____
+-----------------+-----------------> time
 t0 (add N2)

(Shapes are qualitative: vertical jumps only for the species you directly add; others change smoothly as equilibrium re-establishes.)