Chemistry MCQs for NEET — Practice Questions with Answers

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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 

In the following first order competing reactions:

A + Reagent  Product

B + Reagent  Product

The ratio of K1/K2 if only 50% of B will have been reacted when 94% of A has been reacted in same time is:

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Explanation

(a) For 50% B reacted, K2 = (2.303/T2)log100/50

For 94% A reacted, K1 = (2.303/T1)log100/6

K2/K1 = t2/t1x0.3010/1.2218

Since t2 = t1, because 50% B has reacted when 94% A has reacted.

K2/K1 = 0.3010/1.2218 = 0.246 and K1/K2 = 4.06

 

For a reversible reaction A k2k1 B, Ist order in both the directions, the rate of reaction is given by:

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Explanation

(d) Rate = K1[A] - K2[B] for a reversible reaction of I order opposed by I order.

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