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Q.
The decomposition of $NH_3$ on platinum surface is zero order reaction. The rates of production of $N_2$ and $H_2$ are respectively $(k = 2.5 \times 10^{-4} mol\, L^{-1} s^{-1})$
Given reaction is $2NH_3 \xrightarrow{pt} N_2 + 3 H_2 $
For zero order reaction, $ rate = k $
$\therefore -\frac{1}{2}\frac{d\left[NH\right]_{3}}{dt}= + \frac{d\left[N_{2}\right]}{dt} =+ \frac{1}{3}\frac{d\left[H_{2}\right]}{dt} = 2.5\times10^{-4} mol \,L^{-1} s^{-1 } $
Rate of production of $ N_{2} =\frac{d\left[N_{2}\right]}{dt} =2.5\times10^{-4} mol \,L^{-1}s^{-1} $
Rate of production of $H_{2} =\frac{d\left[H_{2}\right]}{dt }=3\times\left(2.5\times10^{-4}\right) $
$=7.5\times10^{-4} mol \,L^{-1} s^{-1 }$