Decomposition of NH₃ on gold surface follows zero order kinetics. If rate constant K is $5\times {10}^{-4}\mathrm{M}-{\mathrm{s}}^{-1}\text{, rate of formation of }{\mathrm{N}}_2\text{ will be }$ 

     N₂​+3H₂​→2NH_3​ 

 K =5$\times$10^-4 M s−1

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

Rate of formation of N₂​ = $\frac{\mathrm{d}\left[{\mathrm{N}}_2\right] }{\mathrm{dt}}$=  K =5$\times$10^-4 M s⁻¹