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} $

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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.