Question

The gaseous reaction $ {A_2} \to 2A $ is first order in $ {A_2} $ . After $ 12.3 $ minutes $ 65\% $ of $ {A_2} $ remains undecomposed. how long will it take to decompose $ 90\% $ of $ {A_2} $ ? What is the half life of the reaction?

Answer 1

Hint: The first order of the reaction is the reaction that depends only on the concentration of one reactant. Further half life of a reaction is a time required for the reactant concentration to decrease to one half its initial value.

Complete step by step solution:
Firstly we are given that the $ {A_2} \to 2A $ is first order in $ {A_2} $ as we know that the first order of the reaction is the reaction that depends only on the concentration of one reactant. Now we need to need to calculate the half life in the above question so we will discuss that in detail. Half life of a reaction is a time required for the reactant concentration to decrease to one half its initial value.The half life of a first order reaction is basically a constant that is related to the rate constant for the reaction.
Now we are given the time when the $ 65\% $ of $ {A_2} $ remains undecomposed after $ 12.3 $ minutes.
Then as we will calculate the k when $ 65\% $ of $ {A_2} $ remains and this is given by the formula
 $ k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {{A_2}} \right]}_0}}}{{\left[ {{A_2}} \right]}} $
Now $ \dfrac{{{{\left[ {{A_2}} \right]}_0}}}{{\left[ {{A_2}} \right]}} $ is $ \dfrac{{100}}{{65}} $ and t is the time.
Now substituting the values we get:
 $ \dfrac{{2.303}}{{12.3\min }}{\log _{10}}\dfrac{{100}}{{65}} $
 $ k = 0.03503{\min ^{ - 1}} $
Then we should know here when then the $ 90\% $ gets decomposed then only $ 10\% $ remains then we will again apply the same formula and then we will substitute
 $ \dfrac{{2.303}}{{0.03503{{\min }^{ - 1}}}}{\log _{10}}\dfrac{{100}}{{10}} $ here we will calculate the time so therefore the value of calculated k need to be substituted.
 $ t = 65.7\min $ then further we will calculate the half life of the reaction with the formula:
 $ {t_{1/2}} = \dfrac{{0.69}}{{0.03503}} $
 $ {t_{1/2}} = 19.78\min $ so this is the required answer.

Note:
We should know that if the two reactions have the same order, the faster reaction will have the shorter half life,and slower reaction will have a longer half life. the half life of a first order reaction under a given set of conditions is a constant.