## A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to  10 kJ mole–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is

Q:  A catalyst lowers the activation energy of a reaction from 20 kJ mole–1 to  10 kJ mole–1. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at 27°C is

(A) –123°C

(B) 327°C

(C) 327°C

(D) + 23°C

Solution: $\large \frac{E_a’}{T_1} = \frac{E_a}{T_2}$

$\large \frac{10}{300} = \frac{20}{T_2}$

T2 = 600 K = 327° C

Ans: (B)

## The rate of a reaction increases 4-fold when concentration of reactant is increased 16 times. If the rate of reaction is 4 × 10^–6 mole L–1 S–1 mole L–1 when concentration of the reactant is 4 × 10^–4, the rate constant of the reaction will be

Q:  The rate of a reaction increases 4-fold when concentration of reactant is increased 16 times. If the rate of reaction is 4 × 10–6 mole L–1 S–1 mole L–1 when concentration of the reactant is 4 × 10–4, the rate constant of the reaction will be

(A) 2 × 10–4 mole1/2 L–1/2 S–1

(B) 1 × 10–2 S–1

(C) 2 × 10–4 mole–1/2, L1/2 S–1

(D) 25 mole–1 L min–1

Solution: $Rate \propto \sqrt{Concen}$

$Rate = k\sqrt{Concen}$

$k = \frac{Rate}{(Concen)^{1/2}} = \frac{4 \times 10^{-6}}{(4 \times 10^{-4})^{1/2}}$

= 2 × 10–4 mole1/2 L-1/2 S–1

Ans: (A)

## Consider a gaseous reaction, the rate of which is given by K[A] [B], the volume of the reaction vessel containing these gases is suddenly reduced to 1/4th of the initial volume. The rate of reaction relative to the original rate would be

Q: Consider a gaseous reaction, the rate of which is given by K[A] [B] , the volume of the reaction vessel containing these gases is suddenly reduced to 1/4th of the initial volume. The rate of reaction relative to the original rate would be

(A) 16/1

(B) 1/16

(C) 8/1

(D) 1/8

Solution: By reducing volume to 1/4th the concentration will become 4 times hence rate 16 times.

Ans: (A)

## For the reaction A + B → C, it is found that doubling the concentration of A, increases the rate 4 times and doubling the concentration of B doubles the reaction rate. What is the overall order of the reaction ?

Q:  For the reaction A + B → C, it is found that doubling the concentration of A, increases the rate 4 times and doubling the concentration of B doubles the reaction rate. What is the overall order of the reaction ?

(A) 3/2

(B) 4

(C) 1

(D) 3

Solution: Order with respect to A = – 2 , order with respect to B = 1 overall order = 3

Ans: (D)

## Fill in the blank : 92U235 + 0n1 → ? + 36Kr92 + 3 0n1

Q: Fill in the blank

92U235 + 0n1 → ? + 36Kr92 + 3 0n1

(A) 56Ba141

(B) 56Ba139

(C) 54Ba139

(D) 54Ba141

Solution: 92 + 0 = Z + 36 + 0  ⇒ Z = 56

235 + 1 → A + 92 + 3

⇒ A = 141

Missing nucleide is 56Ba141

Ans: (A)

## In the nuclear reaction 92U235 → 82Pb207 , the number of  α and β particles lost would be

Q: In the nuclear reaction 92U23582Pb207 , the number of  α and β particles lost would be

(A) 8 , 4

(B) 6 , 2

(C) 7 , 4

(D) 4 , 3

Solution: 92U23582Pb207 + x (2α4) + y (-1β0)

235 = 207 + 4x  ⇒ x = 7

92 = 82 + 2x – y or y = 4

Ans: (C)

## In uranium mineral, the atomic ratio $\frac{N^{U-238}}{N^{Pb-206}}$ is nearly equal to one…

Problem : In uranium mineral, the atomic ratio $\frac{N^{U-238}}{N^{Pb-206}}$ is nearly equal to one. The age (in years) of the mineral is nearly (given that half-life of U238 is 4.5 × 109 years).

(A) 3.0 × 108

(B) 4.5 × 108

(C) 3.0 × 109

(D) 4.5 × 109

Solution: Equal no. of atoms of U238 and Pb206 means half of U has distintegrated

time taken = t1/2

Ans: (D)