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For the reaction, N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3, if d[NH3]dt=2×10−4 mol L−1 s−1\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}, the value of −d[H2]dt-\frac{d[H_2]}{dt} would be:

A

3 × 10⁻⁴ mol L⁻¹ s⁻¹

B

4 × 10⁻⁴ mol L⁻¹ s⁻¹

C

6 × 10⁻⁴ mol L⁻¹ s⁻¹

D

1 × 10⁻⁴ mol L⁻¹ s⁻¹

Step-by-Step Solution

The rate of a chemical reaction can be expressed in terms of any of the reactants or products. For a reaction aA+bB→cC+dDaA + bB \rightarrow cC + dD, the rate is related to the stoichiometric coefficients as: Rate=−1ad[A]dt=−1bd[B]dt=+1cd[C]dt=+1dd[D]dt\text{Rate} = -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} = +\frac{1}{d}\frac{d[D]}{dt}

For the specific reaction N2+3H2→2NH3N_2 + 3H_2 \rightarrow 2NH_3: Rate=−d[N2]dt=−13d[H2]dt=+12d[NH3]dt\text{Rate} = -\frac{d[N_2]}{dt} = -\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt}

We are given the rate of appearance of ammonia, d[NH3]dt=2×10−4 mol L−1 s−1\frac{d[NH_3]}{dt} = 2 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}. We need to find the rate of disappearance of hydrogen, −d[H2]dt-\frac{d[H_2]}{dt}.

Using the equality: −13d[H2]dt=+12d[NH3]dt-\frac{1}{3}\frac{d[H_2]}{dt} = +\frac{1}{2}\frac{d[NH_3]}{dt} −d[H2]dt=32×d[NH3]dt-\frac{d[H_2]}{dt} = \frac{3}{2} \times \frac{d[NH_3]}{dt} −d[H2]dt=32×(2×10−4)-\frac{d[H_2]}{dt} = \frac{3}{2} \times (2 \times 10^{-4}) −d[H2]dt=3×10−4 mol L−1 s−1-\frac{d[H_2]}{dt} = 3 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}

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