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Question: The gaseous reaction \( {A_2} \to 2A \) is first order in \( {A_2} \) . After \( 12.3 \) minutes \( ...

The gaseous reaction A22A{A_2} \to 2A is first order in A2{A_2} . After 12.312.3 minutes 65%65\% of A2{A_2} remains undecomposed. how long will it take to decompose 90%90\% of A2{A_2} ? What is the half life of the reaction?

Explanation

Solution

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 A22A{A_2} \to 2A is first order in A2{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%65\% of A2{A_2} remains undecomposed after 12.312.3 minutes.
Then as we will calculate the k when 65%65\% of A2{A_2} remains and this is given by the formula
k=2.303tlog10[A2]0[A2]k = \dfrac{{2.303}}{t}{\log _{10}}\dfrac{{{{\left[ {{A_2}} \right]}_0}}}{{\left[ {{A_2}} \right]}}
Now [A2]0[A2]\dfrac{{{{\left[ {{A_2}} \right]}_0}}}{{\left[ {{A_2}} \right]}} is 10065\dfrac{{100}}{{65}} and t is the time.
Now substituting the values we get:
2.30312.3minlog1010065\dfrac{{2.303}}{{12.3\min }}{\log _{10}}\dfrac{{100}}{{65}}
k=0.03503min1k = 0.03503{\min ^{ - 1}}
Then we should know here when then the 90%90\% gets decomposed then only 10%10\% remains then we will again apply the same formula and then we will substitute
2.3030.03503min1log1010010\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=65.7mint = 65.7\min then further we will calculate the half life of the reaction with the formula:
t1/2=0.690.03503{t_{1/2}} = \dfrac{{0.69}}{{0.03503}}
t1/2=19.78min{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.