📖 generic · 12th TN - English Medium · CHEMISTRY-VOLUME 1 · Page 229question

of temperature on reaction rate · Part 2

Chapter 8: 7 · CHEMISTRY-VOLUME 1

temperature (in K) The frequency factor (A) is related to the frequency of collisions (number of collisions per second) between the reactant molecules. The factor A does not vary significantly with temperature and hence it may be taken as a constant. E a is the activation energy of the reaction, which Arrhenius considered as the minimum energy that a molecule must have to posses to react. Taking logarithm on both side of the equation ( ) ln ln ln k - e E RT ln ln k E RT ln e = −  ∴ ln ln k E R T −  ....( ) y = c + m x The above equation is of the form of a straight line y = mx + c.

A plot of ln k Vs T gives a straight line with a negative slope − E R a . If the rate constant for a reaction at two different temperatures is known, we can calculate the activation energy as follows. At temperature T = T ; the rate constant k = k ln ln k A- E RT ....( ) XII U7 kinetics - Jerald XII U7 kinetics - Jerald - - - - At temperature T = T ; the rate constant k = k ln ln k A- E RT ....( ) ( ) –( ) ln ln k - k - E RT E RT =   +  ln k k E R T - T  = log k k E R T T TT  = log k k E R T T TT  = ln ln k k E RT E RT = −   +  This equation can be used to calculate E a from rate constants k and k at temperatures T and T . Example The rate constant of a reaction at and 200K are .

and . s - respectively. Calculate the value of activation energy. Solution According to Arrhenius equation log k

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