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Question: A first order reaction has a half life period of 69.3 sec. At 0.10 mol lit$^{-1}$ reactant concentra...

A first order reaction has a half life period of 69.3 sec. At 0.10 mol lit−1^{-1} reactant concentration rate will be

A

10−4^{-4} M sec−1^{-1}

B

10−3^{-3} M sec−1^{-1}

Answer

10−3^{-3} M sec−1^{-1}

Explanation

Solution

  1. For a first-order reaction, the half-life (t1/2t_{1/2}) is related to the rate constant (kk) by the formula:

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

  2. Calculate the rate constant (kk) using the given half-life:

    k=0.69369.3 sec=0.01 sec−1=10−2 sec−1k = \frac{0.693}{69.3 \text{ sec}} = 0.01 \text{ sec}^{-1} = 10^{-2} \text{ sec}^{-1}

  3. The rate law for a first-order reaction is given by:

    Rate = k[A]k[A]

    where [A][A] is the concentration of the reactant.

  4. Calculate the rate using the calculated rate constant and the given reactant concentration (0.10 M):

    Rate = (10−2 sec−1)×(0.10 M)(10^{-2} \text{ sec}^{-1}) \times (0.10 \text{ M})

    Rate = (10−2)×(10−1) M sec−1(10^{-2}) \times (10^{-1}) \text{ M sec}^{-1}

    Rate = 10−3 M sec−110^{-3} \text{ M sec}^{-1}