Q. 26

(i) Graphically e

Answer :


• Rate constant of a reaction depends on temperature and it is said that rate constant increases almost doubly for evey increase of 10 of temperatureand this dependency can be expressed quantitatively by Arrhenius equation :

k = A e Ea/RT

Where , K= rate constant, A= frequency factor or Arrhenius factor R= universal gas constant and T= temperature (Kelvin) , Ea = energy of activation for the reaction.

• To represent this relationship graphically we have to reform the equation as :

By taking ln on the both sides we get,

lnK = - Ea/RT + lnA

and by plotting ln k values of a certain reaction against the (1/T) values we get the following straight line graph with a negative slope = - Ea/R and an intercept of ln A



• From Arrhenius equation k = A e Ea/RT .

• R= N0K , where R = ideal gas constant , k= boltzman constant and N0 = avogadro’s number.

Hence K =R/ N0 .

From the given equation , e Ea/kT = e-28000N0/RT = e Ea/RT

Hence ,


Or, Ea = 28000 X 6.023 X 1023

=1.744 X 1028 joule/molecule .



• From the given plot the order of the reaction should be of zero order.

Hence, Q + R Products is a zero order reaction .

• Because, the concentration of R ,

[R] varies with time which is shown by the plot. And that means the rate of the reaction is actually proportional to the concentration of reactant R (zero power on [R] ).

i.e. rate = d[R]/dt =k R0 = K (R0 = 1] which upon integration results in,

[R] = -kt + [R]0( where , [R] is the final concentration and [R]0 is the initial concentration

or, k = [R] – [R]0


• The dimension of the rate constant for the reaction is concentration/time hence, unit = M/s.

• The rate expression for the given zero order reaction is ;

rate = d[R]/dt =kR0 = K

hence, rate=k


• For the given 1st order reaction K=2.54 x 10-3 s-1

• For a 1st order reaction t= 2.303 log a

k a-x

where, a = initial amount of the reactant and a-x denotes final amount until the reaction is taken into consideration.

• Now , for 3/4 th life a – x =a – (3/4) a= a/4

• t3/4 = 2.303 log a

2.54x10-3 a/4

= 2.303 log 4


=2.303 X 0.6020 [ log 4 =0.6020]


= 545.82 second.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses

Note: In the follChemistry - Exemplar

Which of the follChemistry - Exemplar