# 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 \[{E_{a}}\].

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**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 T1and 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.