For $\mathrm{N}{}_2(\mathrm{g})+3\mathrm{H}{}_2(\mathrm{g})\rightleftharpoons 2\mathrm{NH}{}_3(\mathrm{g})$ at equilibrium in a rigid vessel, $[\mathrm{N}{}_2]=0.50\mathrm{M}$, $[\mathrm{H}{}_2]=0.50\mathrm{M}$ and $[\mathrm{NH}{}_3]=0.25\mathrm{M}$. Then enough $\mathrm{N}{}_2$ is added to make $[\mathrm{N}{}_2]=1.00\mathrm{M}$ and at the same instant $\mathrm{NH}{}_3$ is removed to make $[\mathrm{NH}{}_3]=0.10\mathrm{M}$ ($[\mathrm{H}{}_2]$ unchanged). Which statement describes the situation immediately afterwards?

1. A$Q_c=0.08$, which is less than $K_c=1.00$, so the net reaction is backward
2. B$Q_c=0.08$, which is less than $K_c=1.00$, so the net reaction is forward
3. C$Q_c=12.5$, which is greater than $K_c=1.00$, so the net reaction is backward
4. D$Q_c=1.00=K_c$, so the system stays at equilibrium