Two different equilibrium mixtures of the same reaction $\mathrm{A}(\mathrm{g})\rightleftharpoons 2\mathrm{B}(\mathrm{g})$ at the same temperature are prepared in two vessels of equal volume: mixture 1 has $[\mathrm{A}]=0.10$, $[\mathrm{B}]=0.20$ and mixture 2 has $[\mathrm{A}]=0.40$, $[\mathrm{B}]=0.40$ (all in $\mathrm{mol}{\mathrm{L}}^{-1}$). The vessels are connected and the gases mix, giving a total volume that is twice one vessel, at the same temperature. Immediately after mixing, the reaction

1. Aproceeds backward, because $Q_c>K_c$
2. Bis at equilibrium, because both parts were at equilibrium
3. Ccannot be decided without knowing the volume of each vessel
4. Dproceeds forward, because $Q_c<K_c$