1

### JEE Advanced 2014 Paper 2 Offline

X and Y are two volatile liquids with molar weights of 10 g mol$$-$$1 and 40 g mol$$-$$1, respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is due to

A
larger mean free path for X as compared to that of Y.
B
larger mean free path for Y as compared to that of X.
C
increased collision frequency of Y with the inert gas as compared to that of X with the inert gas.
D
increased collision frequency of X with the inert gas as compared to that of Y with the inert gas.

## Explanation

Collision frequency of X is greater than Y.

2

### JEE Advanced 2014 Paper 2 Offline

X and Y are two volatile liquids with molar weights of 10 g mol$$-$$1 and 40 g mol$$-$$1, respectively. Two cotton plugs, one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm, as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of 300 K. Vapours of X and Y react to form a product which is first observed at a distance d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour for the inert gas and the two vapours. The value of d in cm (shown in the figure), as estimated from Graham's law, is
A
8
B
12
C
16
D
20

## Explanation

Using Graham's law of diffusion, we have that for identical conditions, rate of diffusion varies inversely with square root of mass of the gases. So,

$${x \over {24 - x}} = {\left( {{{40} \over {10}}} \right)^{1/2}} \Rightarrow x = 16$$

3

### JEE Advanced 2014 Paper 2 Offline

For the elementary reaction M $$\to$$ 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
A
4
B
3
C
2
D
1
4

### JEE Advanced 2013 Paper 1 Offline

In the reaction, P + Q $$\to$$ R + S, the time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q varies with reaction time as shown in the figure. The overall order of the reaction is A
2
B
3
C
0
D
1

## Explanation

This is a first-order reaction because t75% = 2 $$\times$$ t50%. The graph shows that the order with respect to Q is 0, so we can write the rate expression as

Rate = k[P]1[Q]0

#### Questions Asked from Chemical Kinetics and Nuclear Chemistry

On those following papers in MCQ (Single Correct Answer)
Number in Brackets after Paper Indicates No. of Questions
JEE Advanced 2014 Paper 2 Offline (3)
JEE Advanced 2013 Paper 1 Offline (1)

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