Question

The time required for \[10\% \] completion of a first-order reaction at \[298K\] is equal to that required for its \[25\% \] completion at\[308K\]. If the value of A is \[4 \times {10^{10}}{s^{ - 1}}\]. Calculate k at \[318K\] and \[{E_{a}}\].

Answer 1

Hint: We are going to use the Arrhenius equation to solve this problem. As per the Arrhenius equation, Activation energy (\[{E_{a}}\]​) and rate constant (\[k1\; and \;k2\]) of a chemical reaction at two different temperatures of T1​ and T2 are related.
Formula used:
Arrhenius equation, Activation energy (\[{E_{a}}\]​) and rate constant (\[k1\; and \;k2\]) of a chemical reaction at two different temperatures of T1​and T2 are related \[ln\dfrac{{k2}}{{k1}} = - \dfrac{{Ea}}{R}(\dfrac{1}{{T2}} - \dfrac{1}{{T1}})\]

Complete step by step answer:
First we have to find the rate constant for the reaction.
As mentioned in the question, time (T1) required for 10% completion of the reaction at298K is equal to time (T2) required for 25% completion of the reaction at 308K
So, k for 10% reaction completion is equal to
\[k\left( {298} \right){\text{ }} = \;\dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}\]
similarly, k for 25% reaction completion is equal to
\[k\left( {308} \right){\text{ }} = \;\dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}\]
Therefore,
Taking log on numerator and denominator,
\[\dfrac{{k(308)}}{{k(298)}} = \dfrac{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{75}}}}{{\dfrac{{2.303}}{t}\log \dfrac{{100}}{{90}}}}\]
​​\[\dfrac{{k(308)}}{{k(298)}} = \dfrac{{\log \dfrac{{100}}{{75}}}}{{\log \dfrac{{100}}{{90}}}}\]


\[\dfrac{{k(308)}}{{k(298)}} = \dfrac{{\log 1.33}}{{\log 1.11}}\]
\[\dfrac{{k(308)}}{{k(298)}} = \dfrac{{0.12385}}{{0.0453}}\]
∴\[\;\dfrac{{k(308)}}{{k(298)}}\]=2.73
According to Arrhenius equation,
Converting into natural log,
\[ln\dfrac{{k(308)}}{{k(298)}} = - \dfrac{{Ea}}{{2.303 \times 8.314}}(\dfrac{1}{{T2}} - \dfrac{1}{{T1}})\]
Substituting the value of T1 and T2
\[{\text{ln 2}}{\text{.73 }}{\text{ = }}\dfrac{{{\text{Ea}}}}{{2.303 \times 8.314}}{\text{(}}\dfrac{{308 - 298}}{{308 \times 298}}{\text{)}}\]
Multiplying the given values,
\[{\text{Ea}} = \dfrac{{2.303 \times 8.314 \times 308 \times 298 \times \log (2.73)}}{{308 - 298}}\]
Simplifying the above equation we will get
\[{\text{Ea}} = \dfrac{{766508.142}}{{10}}\]
\[Ea = 76650.814\]
By solving above equation, we have found the value of $Ea$
\[Ea\; = {\text{ }}76650Joule/mol{\text{ }} = \;76.650kJ/mol\]
Now, we have to calculate the value of k at \[318{\text{ }}K\]
\[Log{\text{ }}k\]=\[logA - \dfrac{{Ea}}{{2.303RT}}\]
We know the values of A and we found the value of \[Ea\] .
Therefore we substituting the value in this equation
\[log{\text{ }}k = \]\[log{\text{ (}}4 \times {10^{10}}) - \dfrac{{76650}}{{2.303 \times 8.314 \times 318}}\]

\[log{\text{ }}k = \]\[(0.60205 + 10) - \dfrac{{76650}}{{6088.7911}}\]
\[log{\text{ }}k = \]\[10.60205 - 12.588\]
\[log{\text{ }}k{\text{ }} = {\text{ }} - 1.9823\]
Therefore, \[k{\text{ }} = {\text{ }}Antilog{\text{ }}\left( { - 1.9823} \right)\]
\[ = {\text{ }}1.042 \times {10^{ - 2}}/s\]
Therefore the value of K at \[318K\] is \[{\text{ }}1.042 \times {10^{ - 2}}/s\]
Hence, we have calculated the \[\;k\left( {318} \right){\text{ }} = {\text{ }}1.042 \times {10^{ - 2}}/s\]and \[Ea\; = {\text{ }}76623Joule/mol{\text{ }} = \;76.623kJ/mol\]


Note: We can use the Arrhenius equation to find the effect of a change of temperature on the rate constant, and consequently on the rate of the reaction. In the Arrhenius equation A is the frequency or pre-exponential factor. The term $k_1$ in the Arrhenius equation is used for the first-order rate constant for the reaction, and $k_2$ is used for the second-order rate constant for the reaction.