For the equilibrium $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO…

Question

For the equilibrium $\mathrm{N_2O_4(g)} \rightleftharpoons 2\mathrm{NO_2(g)}$, the system is suddenly COMPRESSED (volume decreased) at constant temperature. According to Le Chatelier's principle, the equilibrium will

Accepted Answer

B. shift toward $\mathrm{N_2O_4}$ (less dissociation)

1. Le Chatelier's principle: when a system at equilibrium is disturbed, it shifts to PARTIALLY undo the disturbance.
2. Compression (volume decrease) raises the total pressure. The system shifts to REDUCE the number of moles of gas.
3. Forward reaction: $1$ mol $\mathrm{N_2O_4}$ → $2$ mol $\mathrm{NO_2}$ (mole count INCREASES).
4. Reverse reaction: $2$ mol $\mathrm{NO_2}$ → $1$ mol $\mathrm{N_2O_4}$ (mole count DECREASES).
5. To relieve the higher pressure, the system favours the reverse reaction → shifts toward $\mathrm{N_2O_4}$. So option B.
6. Option C is half-right ($K_c$ doesn't change with pressure), but the COMPOSITION still shifts. Option A is the opposite direction. Temperature isn't relevant here (option D).
_Source: NCERT Class 11 Chemistry, Ch 6, §6.8.1 (Effect of Change in Pressure / Le Chatelier on $\mathrm{N_2O_4} \rightleftharpoons 2\mathrm{NO_2}$), p. 19–20._

Answer

shift toward $2\mathrm{NO_2}$ (more dissociation)

Answer

remain unchanged because $K$ is constant at constant temperature

Answer

shift in a direction that depends on whether the system absorbs or releases heat