Answer :

(i)

• 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) , E_{a} = 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 = - E_{a}/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 = - E_{a}/R and an intercept of ln A

.

(ii)

• From Arrhenius equation k = A e ^{–} ^{Ea/RT} .

• R= N_{0}K , where R = ideal gas constant , k= boltzman constant and N_{0} = avogadro’s number.

Hence K =R/ N_{0} .

From the given equation , e ^{–} ^{Ea/kT} = e^{-}^{28000N}_{0}^{/RT =} e ^{–} ^{Ea/RT}

Hence ,

=

Or, E_{a} = 28000 X 6.023 X 10^{23}

=1.744 X 10^{28} joule/molecule .

OR

(i)

• 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 R^{0} = K (R^{0 = 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}__

t

• 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 =kR^{0} = K

hence, rate=k

(ii)

• For the given 1^{st} 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

• t_{3/4} = __2.303__ log __a__

2.54x10^{-3} a/4

= __2.303__ log 4

2.54x10^{-3}

=__2.303__ X 0.6020 [ log 4 =0.6020]

2.54x10^{-3}

= 545.82 second.

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