Chemical Kinetics MCQs for NEET — Chemistry Questions with Answers

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The rate constant of a second order reaction is 10-2 mol-1 litre s-1 The rate constant expressed in cc molecule-1min-1 is:

 

 

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Explanation

(a) K = 10-2mol-1litresec-1

         = 10-2x1000x60/6.02x1023 cc molecule-1min-1

         = 9.9618 x 10-22 cc molecule-1min-1

The half-life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log 2 = 0.301):

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Explanation

(b) t= (2.303x6.93/0.693)log(100/1)(N0 = 100; N = 100-99 =1)

     t = 46.06 minute

A zero order reaction is one:

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Explanation

(c) Zero order reaction occur with constant rate.

For A+B C+D, H = -20 kJmol-1 the activation energy of the forward reaction is 85 kJ mol-1. The activation energy for backward reaction is...... kJ mol-1

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Explanation

(A) For a reaction Ea for forward reaction = Ea for backward reaction + H,

85 = A-20

A = 105KJmol-1

Given that K is the rate constant for some order of any reaction at temp. T then the value of limt logK = (where A is the Arrhenius constant):

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Explanation

(d) loge K = loge A -Ea/RT; (Arrhenius eq.)

     if   T, then  loge K = loge A

For the elementary reaction M  N, the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M. The order of the reaction with respect to M is:

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Explanation

(b) Consider, rate (r)=K[M]n where n is order of reaction

r1/r2=1/8=[M]n/[2M]nn=3

How much faster would a reaction proceed at 25°C than at 0°C if the activation energy is 65 kJ?

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Explanation

(c) 2.303logK2/k1 = Ea/Er[T2-T1/T1T2]

     2.303logK2/k1 = 65x103/8.314[25/298x273]

     K2/k1 = 11.05

K for a zero order reaction is 2 x10-2 mol L-1 sec-1. If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration must have been:

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Explanation

(d) For zero order [A]t = [A]0 - kt

     0.5 = [A]0 - 2x10-2x25

     [A]0 = 1.0M

The rate constant for a second order reaction is 8x10-5 M-1 min-1 . How long will it take a 1M solution to be reduced to 0.5M?

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Explanation

(c) For II order, t = (1/Ka) x/(a-x)

      t = 1/8x10-5x1(0.5/0.5)

      = 1.25x104 minute 

The activation energy for a reaction is 9.0 kcal/mol. The increase in the rate constant when its temperature is increased from 298K to 308K is:

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Explanation

(d) 2.303 log(K2/K1) = Ea/R{[T2-T1]/T1T2};

 2.303 log(K2/K1) = 9/2x10-3[10/298x308]

K2/K1 = 1.63; i.e ,63% increase 

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