For the reaction N_2 + 3H_2 → 2NH_3 , if d[NH_3]/dt = 2 × 10^-4 mol L^-1 s^-1 , the value of -d[H_2]/dt would be: - Chemistry MCQ Chemistry | ExamTest.live | ExamTest.live

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For the reaction N_{2} + 3 H_{2} \to 2 NH_{3}, if \frac{d [ NH _{3} ]}{d t} = 2 \times 1 0^{- 4} mol L^{- 1} s^{- 1}, the value of - \frac{d [ H _{2} ]}{d t} would be:

A $4 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$
B $6 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$
C $1 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$
D $3 \times 1 0^{- 4} mol L^{- 1} s^{- 1}$

Solution & Step-by-step Explanation

The rate of reaction is given by: Rate = - \frac{d [ N _{2} ]}{d t} = - \frac{1}{3} \frac{d [ H _{2} ]}{d t} = \frac{1}{2} \frac{d [ NH _{3} ]}{d t} We need to find - \frac{d [ H _{2} ]}{d t} : \frac{1}{3} (- \frac{d [ H _{2} ]}{d t}) = \frac{1}{2} \frac{d [ NH _{3} ]}{d t} - \frac{d [ H _{2} ]}{d t} = \frac{3}{2} \times (2 \times 1 0^{- 4}) - \frac{d [ H _{2} ]}{d t} = 3 \times 1 0^{- 4} mol L^{- 1} s^{- 1}

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For the reaction N_{2} + 3 H_{2} \to 2 NH_{3}, if \frac{d [ NH _{3} ]}{d t} = 2 \times 1 0^{- 4} mol L^{- 1} s^{- 1}, the value of - \frac{d [ H _{2} ]}{d t} would be:

A

4 \times 1 0^{- 4} mol L^{- 1} s^{- 1}

B

6 \times 1 0^{- 4} mol L^{- 1} s^{- 1}

C

1 \times 1 0^{- 4} mol L^{- 1} s^{- 1}

D

3 \times 1 0^{- 4} mol L^{- 1} s^{- 1}

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Chemical Kinetics Rate of Reaction

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