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Question: The rate constant of a certain first order reaction is \( 3.12\times {{10}^{-3}}mi{{n}^{-1}}. \) How...

The rate constant of a certain first order reaction is 3.12×10−3min−1.3.12\times {{10}^{-3}}mi{{n}^{-1}}. How many minutes does it take for the reactant concentration to drop to 0.02M0.02M if the initial concentration of the reactant is 0.045M0.045M ?

Explanation

Solution

We know that the chemical kinetics is based on the rate of different chemical reactions. First order reaction is the reaction in which the rate of the reaction is based on only one reactant concentration. The rate is directly proportional to the concentration of the reactant. The formula that will be used is t=2.303klog[R]n[R].t=\dfrac{2.303}{k}log\dfrac{{{\left[ R \right]}^{n}}}{\left[ R \right]}. This formula will be used for the calculation of the total time taken in the performance of the first order reaction.

Complete answer:
Given that the concentration of reactant is halved. So the time required for the concentration of reactant to be halved tells us that we have to calculate the half-life of the reactant, first order reaction is the reaction that depends on the concentration of only one reactant. Other reactants can be present, but each will be zero-order. First order reaction is a unimolecular reaction.
Order of a reaction is defined as the number of reactants which determine rate of reaction or number of reactants whose molar mass concentration changes during the chemical reaction or also can be defined as it is the sum of exponents raised on active masses of reactants in a rate law equation. It is an experimental value, it may be zero, negative, or in fraction and order of a reaction depends upon the temperature, pressure and concentration.
Here, we know that t=2.303klog[R]n[R].t=\dfrac{2.303}{k}log\dfrac{{{\left[ R \right]}^{n}}}{\left[ R \right]}.
Thus after substituting the values in above equation we get;
t=2.3033.12×10−3log[0.2][0.045]t=\dfrac{2.303}{3.12\times {{10}^{-3}}}log\dfrac{\left[ 0.2 \right]}{\left[ 0.045 \right]}
Hence, since you know the time and the concentration we can calculate the R that can be used in the above equation. On further solving we get;
t=259.96 or 260min.t=259.96\text{ }or\text{ }260min.

Note:
Remember that during each half-life, half of the substance decay into atoms or molecules. Each radioactive nuclide has its own half-life. Half-lives can be as short as a fraction of a second or as long as billions of years. Half-life is the time required to reach steady state.