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1 the rate of reaction 2 the half life

Chemistry

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Answer

2.0×10−5 mol dm−3 s−1\text{2.0} \times 10^{-5} \text{ mol dm}^{-3}\text{ s}^{-1}

346.5 s\text{346.5 s}

Step-by-step solution

3 steps

Here's how to solve this problem:

  1. Identify the given information.

    The reaction is first order.
    The concentration of the reactant, [N2O5]=0.01 mol/dm3[\text{N}_2\text{O}_5] = 0.01 \text{ mol/dm}^3.
    The specific rate constant, k=2.0×10−3 s−1k = 2.0 \times 10^{-3} \text{ s}^{-1}.

  2. Calculate the rate of reaction.

    For a first-order reaction, the rate law is:

    Rate=k[N2O5]\text{Rate} = k[\text{N}_2\text{O}_5]

    Substitute the given values:

    Rate=(2.0×10−3 s−1)×(0.01 mol/dm3)\text{Rate} = (2.0 \times 10^{-3} \text{ s}^{-1}) \times (0.01 \text{ mol/dm}^3) Rate=2.0×10−5 mol dm−3 s−1\text{Rate} = 2.0 \times 10^{-5} \text{ mol dm}^{-3}\text{ s}^{-1}

    The rate of reaction is 2.0×10−5 mol dm−3 s−1\boxed{\text{2.0} \times 10^{-5} \text{ mol dm}^{-3}\text{ s}^{-1}}.

  3. Calculate the half-life.

    For a first-order reaction, the half-life (t1/2t_{1/2}) is given by the formula:

    t1/2=0.693kt_{1/2} = \frac{0.693}{k}

    Substitute the value of kk:

    t1/2=0.6932.0×10−3 s−1t_{1/2} = \frac{0.693}{2.0 \times 10^{-3} \text{ s}^{-1}} t1/2=346.5 st_{1/2} = 346.5 \text{ s}

    The half-life is 346.5 s\boxed{\text{346.5 s}}.

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