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Chemistry Question on Chemical Kinetics

The decomposition of NH3NH_3 on platinum surface is zero order reaction. The rates of production of N2N_2 and H2H_2 are respectively (k=2.5×10−4mol L−1s−1)(k = 2.5 \times 10^{-4} mol\, L^{-1} s^{-1})

A

7.5×10−4mol L−1s−1;2.5×10−4mol L−1s−17.5 \times 10^{-4} mol \,L^{-1} s^{-1} ;2.5 \times 10^{-4} mol\,L^{-1}s^{-1}

B

2.5×10−4mol L−1s−1;7.5×10−4mol L−1s−12.5 \times 10^{-4} mol \,L^{-1} s^{-1} ;7.5 \times 10^{-4} mol\,L^{-1}s^{-1}

C

2.5×10−4mol L−1s−1;7.5×10−4mol L−1s−12.5 \times 10^{-4} mol \, L^{-1} s^{-1} ;7.5 \times 10^{-4} mol\,L^{-1}s^{-1}

D

2.5×10−4mol L−1s−1;5.0×10−4mol L−1s−12.5 \times 10^{-4} mol \, L^{-1} s^{-1} ;5.0 \times 10^{-4} mol\,L^{-1}s^{-1}

Answer

2.5×10−4mol L−1s−1;7.5×10−4mol L−1s−12.5 \times 10^{-4} mol \,L^{-1} s^{-1} ;7.5 \times 10^{-4} mol\,L^{-1}s^{-1}

Explanation

Solution

Given reaction is 2NH3−>[pt]N2+3H22NH_3 { ->[pt] } N_2 + 3 H_2 For zero order reaction, rate=krate = k ∴−12d[NH]3dt=+d[N2]dt=+13d[H2]dt=2.5×10−4mol L−1s−1\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 N2=d[N2]dt=2.5×10−4mol L−1s−1N_{2} =\frac{d\left[N_{2}\right]}{dt} =2.5\times10^{-4} mol \,L^{-1}s^{-1} Rate of production of H2=d[H2]dt=3×(2.5×10−4)H_{2} =\frac{d\left[H_{2}\right]}{dt }=3\times\left(2.5\times10^{-4}\right) =7.5×10−4mol L−1s−1=7.5\times10^{-4} mol \,L^{-1} s^{-1 }