For the chemical reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\leftrightharpoons 2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right)$, the correct option is:

For the reaction ${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2\mathrm{N}{\mathrm{H}}_3\left(\mathrm{g}\right)$,

Rate=$\frac{\mathrm{Change} \mathrm{in} \mathrm{concentration} \mathrm{of} \mathrm{reactants} \mathrm{or} \mathrm{products}}{\mathrm{time} \mathrm{taken} \mathrm{for} \mathrm{change}}$

Rate $=\frac{-\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{-1}{3}\frac{\mathrm{d}\left[{\mathrm{H}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{N}{\mathrm{H}}_3\right]}{\mathrm{d}\mathrm{t}}$

$\therefore \frac{-\mathrm{d}\left[{\mathrm{N}}_2\right]}{\mathrm{d}\mathrm{t}}=\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{N}{\mathrm{H}}_3\right]}{\mathrm{d}\mathrm{t}}$

A negative sign before a reaction rate symbol signifies a decrease in concentration of reactant with increase of time.