JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Equilibrium

At equilibrium the constant for $\mathrm{A}\rightleftharpoons \mathrm{B}$ is the ratio of the amounts of product to reactant present. Counting the particles in the figure, there are $6$ squares representing $\mathrm{A}$ and $11$ circles representing $\mathrm{B}$, so $K_{eq}=\frac{[\mathrm{B}]}{[\mathrm{A}]}=\frac{11}{6}\approx 1.8$, which rounds to $2$.