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Chemistry Question on Chemical Kinetics

For the reaction, N2+3H2→2NH3N _{2}+3 H _{2} \rightarrow 2 NH _{3}, If d[NH3]dt=2×10−4 mol L−1s−1\frac{ d \left[ NH _{3}\right]}{ dt }=2 \times 10^{-4}\, mol\, L ^{-1} s ^{-1}, The value of −d[H2]dt\frac{- d \left[ H _{2}\right]}{ dt } would be -

A

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

B

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

C

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

D

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

Answer

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

Explanation

Solution

13−d[H2]dt=12[NH3]dt\frac{1}{3} \frac{- d \left[ H _{2}\right]}{ dt }= \frac{1}{2} \frac{\left[ NH _{3}\right]}{ dt }
−d[H2]dt=32d[NH3]dt\frac{- d \left[ H _{2}\right]}{ dt }=\frac{3}{2} \frac{ d \left[ NH _{3}\right]}{ dt }
−d[H2]dt=32×2×10−4\frac{- d \left[ H _{2}\right]}{ dt }=\frac{3}{2} \times 2 \times 10^{-4}
=3×10−4 mol L−1 s−1=3 \times 10^{-4}\, mol\, L ^{-1}\, s ^{-1}