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Chemical Kinetics and Nuclear Chemistry question

2021 · 25 Feb · Shift 2 · Q22

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Chemical Kinetics and Nuclear Chemistry
2021 · 25 Feb · Shift 2 · Q22

Chemical Kinetics and Nuclear Chemistry question

2021 · 25 Feb · Shift 2 · Q22

JEE MainChemistryChemical Kinetics and Nuclear ChemistryNumerical+4 / −1

The rate constant of a reaction increases by five times on increase in temperature from 27

^\circ

C to 52

^\circ

C. The value of activation energy in kJ mol

-

1 is

\underline{\hspace{2cm}}

. (Rounded off to the nearest integer) [R = 8.314 J K

-

1mol

-

1]

Numerical answer

View written solutionFree

Correct answer: 52

Use the Arrhenius equation in two-temperature form

For a reaction,

\ln\left(\frac{k_2}{k_1}\right)=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)

Given data

$k_2 = 5k_1 \Rightarrow \dfrac{k_2}{k_1}=5$
$T_1 = 27^\circ\text{C} = 300\,\text{K}$
$T_2 = 52^\circ\text{C} = 325\,\text{K}$
$R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$

So,

\ln 5 = \frac{E_a}{8.314}\left(\frac{1}{300}-\frac{1}{325}\right)

Evaluate the temperature term

\frac{1}{300}-\frac{1}{325} = \frac{325-300}{300\times 325} = \frac{25}{97500} = \frac{1}{3900}

Hence,

\ln 5 = \frac{E_a}{8.314}\cdot \frac{1}{3900}

Solve for activation energy

E_a = 8.314 \times 3900 \times \ln 5

Using $\ln 5 \approx 1.609$ ,

E_a \approx 8.314 \times 3900 \times 1.609

First,

8.314 \times 3900 = 32424.6

Then,

E_a \approx 32424.6 \times 1.609 \approx 52171\,\text{J mol}^{-1}

E_a \approx 52.17\,\text{kJ mol}^{-1}

Rounded to nearest integer

E_a \approx 52\,\text{kJ mol}^{-1}

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