Practicing Success

Target Exam

CUET

Subject

Chemistry

Chapter

Physical: Chemical Kinetics

Question:

The rate constant of a reaction is 1.5 × 107 s–1 at 50°C and 4.5 × 107 s–1 at 100°C. What is the value of activation energy?

Options:

2.2 × 103 J mol–1

2300 J mol–1

2.2 × 104 J mol–1

220 J mol–1

Correct Answer:

2.2 × 104 J mol–1

Explanation:

The correct answer is option 3. \(2.2 \times 10^4\, \ kJ/mol\)

Given,

\(k_1 = 1.5 × 10^7 s^{−1}\)          \(T_1 =50ºC = 323 K\)   

\(k_2 = 4.5 × 10^7 s^{−1}\)          \(T_1 =100ºC = 373 K\)   

According to Arrhenius equation,

\(log\left(\frac{k_2}{k_1}\right) = \frac{E_a}{2.303 × R}\left[\frac{T_2 − T_1}{T_1T_2}\right]\)

\(⇒ log\left(\frac{4.5 × 10^7 s^{−1}}{1.5 × 10^7 s^{−1}}\right) = \frac{E_a}{2.303 × 8.314}\left[\frac{373 − 323}{323 × 373}\right]\)

\(⇒ log (3) = \frac{E_a}{19.147}\left[\frac{50}{323 × 373}\right]\)

\(⇒ E_a =\frac{log(3) × 19.147 × 323 × 373 }{50}\)

\(⇒ E_a =22012.74 \text{ J/mol}\)

\(or E_a =2.2 × 10^{4}\text{ J/mol}\)