Question

The values of the rate constants of the reaction C2H5I + OH–  C2H5OH + I– at 30o and 60oC temperatures are 0.325 and 6.735 litre mole–1 second–1 than what will be the value of activation energy? [CBSE PMT–2003] (a) 20260 cal (b) 20260 K cal (c) 361.44 cal (d) 84773 cal

Answer 1

Answer:

Option B = 20260 Kcal is the correct answer.

Explanation:

We know that the formula for calculating the activation energy whenever we have two rate constants given is as follows:

$log\frac{k_{2} }{k_{1} }$= $\frac{E_{a} }{2.303R}$$[\frac{1}{T_{1} }$$- \frac{1}{T_{2} } ]$

where T = temperature and $E_{a}$= activation energy k = rate constants

We have in this question the two rate constants as,

$k_{1}$ = 0.325 L$mol^{-1}$ $sec^{-1}$   $k_{2}$ = 6.735 L$mol^{-1} sec^{-1}$

log$\frac{6.735}{0.325}$ = $\frac{E_{a} }{2.303 R}$$[\frac{1}{303K }$$- \frac{1}{313K } ]$

log 20.72 = $\frac{E_{a} }{2.303R}$$[\frac{333K - 303K}{(333K)(303K)} ]$

log 20.72 = $\frac{E_{a} }{2.303(8.314)}$ ×  $\frac{30}{100899}$

1.31 = $E_{a}$ × 1.55

$E_{a}$ = 84360.833 J/ mol = 84.360833 K J / mol = 84.360833/ 4.184 = 20162.7231 K cal = 20260 K cal approx.

Therefore the answer is option b = 20260 K cal (approx) .

Answer 2

Given:   Temperature of the reaction$30$°$C and 60$°$C$ and $K_{2}=6.735,K_{1}=0.325$

We have to find the value of activation energy.

For this, we are using$\frac{\log K_{2}}{K_{1}}=\frac{E_{a}}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)$

We have,

$k_{1}=0.325 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{sec}^{-1} \quad k_{2}=6.735 \mathrm{~L} \mathrm{~mol}^{-1} \mathrm{sec}^{-1}$

Then,

$=>\frac{\log K_{2}}{K_{1}}=\frac{E_{a}}{2.303 R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right)\\\\=>\therefore \frac{\log 6.735}{0.325}=\frac{E_{a}}{2.303} \times 1.987\left(\frac{1}{303}-\frac{1}{333}\right)\\\\=>\therefore E_{a}=20260 \text { cal }$

Hence, the value of activation energy$E_{a}=20260 \text { cal }$.