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Question: The half-life period and the initial concentration for a reaction are as follows. What is the order ...

The half-life period and the initial concentration for a reaction are as follows. What is the order of reaction?

Explanation

Solution

Half-life is the time required to consume the half amount of a substance in a reaction. The order of a reaction is an experimental value. This is basically the total number of atoms or molecules that take part in a reaction at the rate determining step. For elementary reaction the order is equal to the molecularity of the reaction.

Complete step by step answer:
Now to calculate the order of a reaction from the half-life, there is a general equation of relation between the half-life and the initial concentration of the reactant. Which is shown below.
t12(a)(1 - n){{\text{t}}_{\dfrac{{\text{1}}}{{\text{2}}}}}\infty {\left( {\text{a}} \right)^{\left( {{\text{1 - n}}} \right)}}, where a is the initial concentration of reactant and n is the order of the reaction.
Now from the given values,
For initial concentration 350 the value of half-life is 425,
425(350)(1n){\text{425}}\infty {\left( {350} \right)^{\left( {1 - n} \right)}}
For initial concentration 540 the value of half-life is 275,
275(540)(1n){\text{275}}\infty {\left( {540} \right)^{\left( {1 - n} \right)}}
Now take a ration of both reaction as follows,
425275=(350540)(1n)\dfrac{{{\text{425}}}}{{275}} = {\left( {\dfrac{{350}}{{540}}} \right)^{\left( {1 - n} \right)}}
Now calculate this equation and find out the value of n as follows,

425275 = (350540)(1 - n) log(425275) = (1 - n)log(350540) log1.54 = (1 - n)log0.6481 0.189 = - (1 - n)0.188 0.1890.188 = n - 1 1 + 1 = n n = 2  \dfrac{{{\text{425}}}}{{{\text{275}}}}{\text{ = }}{\left( {\dfrac{{{\text{350}}}}{{{\text{540}}}}} \right)^{\left( {{\text{1 - n}}} \right)}} \\\ {\text{log}}\left( {\dfrac{{{\text{425}}}}{{{\text{275}}}}} \right){\text{ = }}\left( {{\text{1 - n}}} \right){\text{log}}\left( {\dfrac{{{\text{350}}}}{{{\text{540}}}}} \right) \\\ {\text{log1}}{\text{.54 = }}\left( {{\text{1 - n}}} \right){\text{log0}}{\text{.6481}} \\\ {\text{0}}{\text{.189 = - }}\left( {{\text{1 - n}}} \right){\text{0}}{\text{.188}} \\\ \dfrac{{{\text{0}}{\text{.189}}}}{{{\text{0}}{\text{.188}}}}{\text{ = n - 1}} \\\ {\text{1 + 1 = n}} \\\ {\text{n = 2}} \\\

Therefore, the order of the reaction is 2.

Note:
The definition of rate of a reaction is the speed of a reaction by which the concentration of reactants decreases and the concentrations of products increases per unit time. The reaction A \to products. is a first order reaction. The rate equation of first order reaction is r = k[A]{\text{r = k}}\left[ {\text{A}} \right]. Where, rate is r, rate constant is k and [A]\left[ {\text{A}} \right] is concentration of reactant A at a time t. The unit of rate depends upon the concentration of reactant and rate constant.