### Neet Solved Paper 2015 Question 15

##### Question: The activation energy of a reaction can be determined from the slope of which of the following graphs?

#### Options:

A) In $ K,vsT $

B) $ \frac{lnK}{T}vsT $

C) In $ Kvs\frac{l}{t} $

D) $ \frac{T}{\ln ,k}vs\frac{l}{T} $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

By Arrhenius equation $ K=A{e^{-E _{a}/RT}} $

where, $ E _{a} $ = energy of activation Applying log on both the side, $ \ln ,k=\ln A-\frac{E _{a}}{RT}\ $ …(i)

or $ \log ,k=-\frac{E _{a}}{2.303RT}+\log A, $ …(ii)

This equation is of the form of $ y=mx+c $ i.e. the equation of a straight line.

Thus, if a plote of $ \log k,vs\frac{1}{T} $ is a straight line, the validity of the equation is confirmed.

Slope of the line $ =-\frac{E _{a}}{2.303R} $

Thus, measuring the slope of the line, the value of $ E _{a} $ can be calculated.