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Chemistry Question on Rate of a Chemical Reaction

For the reaction N2+3H2→2NH3N_{2}+3H_{2} \rightarrow 2NH_{3} if Δ[NH3]Δt=2×10−4 mol l−1s−1,\frac{\Delta\left[NH_{3}\right]}{\Delta t}=2\times10^{-4}\,mol\,l^{-1}s^{-1}, then value of −Δ[H2]Δt\frac{-\Delta\left[H_{2}\right]}{\Delta t} would be

A

1�10−4 mol L−1s−11 � 10^{-4}\, mol\, L^{-1}s^{-1}

B

3�10−4 mol L−1s−13 � 10^{-4}\, mol\, L^{-1}s^{-1}

C

4�10−4 mol L−1s−14 � 10^{-4}\, mol\, L^{-1}s^{-1}

D

6�10−4 mol L−1s−16 � 10^{-4}\, mol\, L^{-1}s^{-1}

Answer

3�10−4 mol L−1s−13 � 10^{-4}\, mol\, L^{-1}s^{-1}

Explanation

Solution

N2+3H2→2NH3N_{2}+3H_{2} \rightarrow 2NH_{3}
−[N2]Δt=13Δ[H2]Δt=12Δ[NH3]Δt\frac{-\left[N_{2}\right]}{\Delta t}=\frac{1}{3} \frac{\Delta\left[H_{2}\right]}{\Delta t}=\frac{1}{2} \frac{\Delta\left[NH_{3}\right]}{\Delta t}
∴−Δ[H2]Δt=32×Δ[NH3]Δt=32×2×10−4\therefore \frac{-\Delta\left[H_{2}\right]}{\Delta t}=\frac{3}{2}\times\frac{\Delta\left[NH_{3}\right]}{\Delta t}=\frac{3}{2}\times2\times10^{-4}
=3×10−4 mol L−1S−1=3\times10^{-4}\,mol\,L^{-1}S^{-1}