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

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.