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Write the general rate expression for the reaction.

Chemistry

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a) For the reaction N2(g)+3H2(g)→2NH3(g)\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \to 2\text{NH}_3(\text{g}), the rate of reaction can be expressed in terms of the change in concentration of reactants and products over time.

  1. Write the general rate expression for the reaction.

    The rate of reaction is related to the rate of change of concentration of each species divided by its stoichiometric coefficient. For reactants, a negative sign is used as their concentration decreases.

    Rate=−d[N2]dt=−13d[H2]dt=12d[NH3]dt\text{Rate} = -\frac{d[\text{N}_2]}{dt} = -\frac{1}{3}\frac{d[\text{H}_2]}{dt} = \frac{1}{2}\frac{d[\text{NH}_3]}{dt}

    To express the rate in terms of the formation of ammonia, we use the term involving NH3\text{NH}_3.

    Rate=12d[NH3]dt\text{Rate} = \boxed{\frac{1}{2}\frac{d[\text{NH}_3]}{dt}}

    b) For the reaction 3A+2B→C+2D3A + 2B \to C + 2D, the rate of disappearance of reactant A is given as 4.6×10−2 Ms−14.6 \times 10^{-2} \text{ Ms}^{-1}.
    The general rate expression for this reaction is:

    Rate=−13d[A]dt=−12d[B]dt=d[C]dt=12d[D]dt\text{Rate} = -\frac{1}{3}\frac{d[A]}{dt} = -\frac{1}{2}\frac{d[B]}{dt} = \frac{d[C]}{dt} = \frac{1}{2}\frac{d[D]}{dt}

    We are given −d[A]dt=4.6×10−2 Ms−1-\frac{d[A]}{dt} = 4.6 \times 10^{-2} \text{ Ms}^{-1}.

    xiii) What is the rate of disappearance of the reactant B?

  2. Relate the rate of disappearance of B to the rate of disappearance of A using the stoichiometric coefficients.

    −12d[B]dt=−13d[A]dt-\frac{1}{2}\frac{d[B]}{dt} = -\frac{1}{3}\frac{d[A]}{dt}

  3. Solve for the rate of disappearance of B, −d[B]dt-\frac{d[B]}{dt}.

    −d[B]dt=23×(−d[A]dt)-\frac{d[B]}{dt} = \frac{2}{3} \times \left(-\frac{d[A]}{dt}\right)

  4. Substitute the given value for −d[A]dt-\frac{d[A]}{dt}.

    −d[B]dt=23×(4.6×10−2 Ms−1)-\frac{d[B]}{dt} = \frac{2}{3} \times (4.6 \times 10^{-2} \text{ Ms}^{-1}) −d[B]dt=9.23×10−2 Ms−1-\frac{d[B]}{dt} = \frac{9.2}{3} \times 10^{-2} \text{ Ms}^{-1} −d[B]dt=3.066...×10−2 Ms−1-\frac{d[B]}{dt} = 3.066... \times 10^{-2} \text{ Ms}^{-1}

    Rounding to two significant figures:

    −d[B]dt=3.1×10−2 Ms−1-\frac{d[B]}{dt} = \boxed{3.1 \times 10^{-2} \text{ Ms}^{-1}}

    xiv) What is the rate of appearance for the product D?

  5. Relate the rate of appearance of D to the rate of disappearance of A using the stoichiometric coefficients.

    12d[D]dt=−13d[A]dt\frac{1}{2}\frac{d[D]}{dt} = -\frac{1}{3}\frac{d[A]}{dt}

  6. Solve for the rate of appearance of D, d[D]dt\frac{d[D]}{dt}.

    d[D]dt=23×(−d[A]dt)\frac{d[D]}{dt} = \frac{2}{3} \times \left(-\frac{d[A]}{dt}\right)

  7. Substitute the given value for −d[A]dt-\frac{d[A]}{dt}.

    d[D]dt=23×(4.6×10−2 Ms−1)\frac{d[D]}{dt} = \frac{2}{3} \times (4.6 \times 10^{-2} \text{ Ms}^{-1}) d[D]dt=9.23×10−2 Ms−1\frac{d[D]}{dt} = \frac{9.2}{3} \times 10^{-2} \text{ Ms}^{-1} d[D]dt=3.066...×10−2 Ms−1\frac{d[D]}{dt} = 3.066... \times 10^{-2} \text{ Ms}^{-1}

    Rounding to two significant figures:

    d[D]dt=3.1×10−2 Ms−1\frac{d[D]}{dt} = \boxed{3.1 \times 10^{-2} \text{ Ms}^{-1}}
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