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
Collision frequency of X is greater than Y.
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.
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,
$${x \over {24 - x}} = {\left( {{{40} \over {10}}} \right)^{1/2}} \Rightarrow x = 16$$
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
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
