Question

A first order reaction is 50 percent completed in two minutes at ${{27}^{0}}C$. and 5 minutes at ${{47}^{0}}C$. The energy is activation of the reaction is:
A. 43.85 KJ/mol
B. 55.26KJ/mol
C. 11.97 KJ/mol
D. 6.65 KJ/mol

Answer 1

Hint: You need to remember that for any reaction, the value of $T\times K$ remains constant, where T is the time and K is the rate constant. After that, use the formula involved with the Arrhenius equation.

Complete step by step answer:
In order to calculate the activation energy of the reaction, we will start from the basic knowledge we know. Now, as we know that for any reaction, $T\times K$remains constant so we can write it as:
${{T}_{1}}\times {{K}_{1}}={{T}_{2}}\times {{K}_{2}}$, so
$20\times {{K}_{1}}=5\times {{K}_{2}}$
Which means that $\dfrac{{{K}_{2}}}{{{K}_{1}}}=4$……….( I )
- Let us convert the respective temperatures in celsius to K
So,
\[\begin{array}{*{35}{l}}
   {{\text{T}}_{1}}\text{ }=\text{ }{{27}^{0}}C\text{ }=\text{ }27+273\text{ }K\text{ }=\text{ }300\text{ }K \\
   {{\text{T}}_{2}}\text{ }=~{{47}^{0}}C\text{ }=\text{ }47+273\text{ }K\text{ }=\text{ }320\text{ }K \\
\end{array}\]
- Now, according to the Arrhenius equation we have
$\log (\dfrac{{{K}_{2}}}{{{K}_{1}}})=\dfrac{{{E}_{a}}}{2.303R}[\dfrac{1}{{{T}_{1}}}-\dfrac{1}{{{T}_{2}}}]$
So,
$\begin{align}
 & \log 4=\dfrac{{{E}_{a}}}{2.303\times 8.314}[\dfrac{1}{300}-\dfrac{1}{320}] \\
 & \log 4=\dfrac{{{E}_{a}}}{19.14}[\dfrac{320-300}{320\times 300}] \\
 & 2\log 2=\dfrac{{{E}_{a}}\times 20}{19.14\times 320\times 300}\,\,\,\,\,,\log 2=0.693 \\
 & {{E}_{a}}=\dfrac{2\times 0.693\times 19.14\times 320\times 300}{20} \\
\end{align}$
On calculating, activation energy comes out to be 55.26 KJ/mol. That means 55.26 KJ/mol energy is required so that the first order reaction is 50 percent completed in to minutes at ${{27}^{0}}C$ and 5 minutes at ${{47}^{0}}C$. The correct option is option “B” .

Note: Always remember to look into the question properly and check for the units. In this case, the units of temperature are not in the SI unit. So, it is needed to be converted into the SI unit i.e kelvin first. Generally, questions which have the term “activation energy” in them have something related to the Arrhenius equation. You need to be very careful regarding the Arrhenius equation calculations, as they get very complex and messy, due to the involvement of the logarithm function. Also, you do not need to find the answer till decimal places. Just check for the first one or two digits and match with the most appropriate option.