Question

1 What is the activation energy for a reaction if its
rate doubles when the temperature is raised from
20°C to 35°C? (R = 8.314 J mol-1K-1) [NEET-2013]
(1) 269 kJ mol
(2) 34.7 kJ mol
(3) 15.1 kJ mol-1
(4) 342 kJ mot

Answer 1

Answer:

Explanation:

Since \log \frac{K_{2}}{K_{1}} = \frac{Ea}{2.303\:R}\left ( \frac{1}{T_{1}} - \frac{1}{T_{2}} \right )

K2  = 2K1 , T1 = 273 + 20 K = 293 K

T2 = 273 + 35 K = 308 K

R = 8.314 J mol-1 K-1

\therefore \log 2 = \frac{Ea}{2.303 \times 8.314}\left ( \frac{1}{293} - \frac{1}{308}\right )

\Rightarrow Ea = 34.7 \:KJ\:mol^{-1}

Option 1)

15.1 kJ mol-1

This answer is incorrect

Option 2)

342 kJ mol-1

This answer is incorrect

Option 3)

269 kJ mol-1

This answer is incorrect

Option 4)

34.7 kJ mol-1