Question

$ x $ mole of $ {N_2}{O_4} $ is taken at $ {P_1} $ atm in a closed vessel and heated when $ 75\% $ $ {N_2}{O_4} $ dissociated at equilibrium, total pressure is found to be $ {P_2} $ atm. The relation between $ {P_1} $ and $ {P_2} $ are:
(A) $ {P_1}:{P_2} = 7:4 $
(B) $ {P_1}:{P_2} = 7:2 $
(C) $ {P_1}:{P_2} = 4:7 $
(D) $ {P_1}:{P_2} = 3:4 $

Answer 1

Hint: Here, the question says that the $ x $ mole of $ {N_2}{O_4} $ is taken at $ {P_1} $ atm in a closed vessel and heated when $ 75\% $ $ {N_2}{O_4} $ dissociated at equilibrium, total pressure is found to be $ {P_2} $ atm. Now, we have to understand the relation between moles and pressure that pressure is directly proportional to the number of moles of a given substance. And we have to find out the ratio between two pressures, one when the $ x $ mole is taken and one after the dissociation at equilibrium.

Complete answer:
Here, we have to represent the equilibrium of the reaction as below:
 $ {N_2}{O_4} \rightleftharpoons 2N{O_2} $
Now the number of moles of this reaction at time $ t = 0 $ and after attaining equilibrium at time $ t = eq. $ is given in the following table as:

time	No. of moles of reactant	No. of moles of product
$ t = 0 $	$ x $	$ 0 $
$ t = eq. $	$ x(1 - \alpha ) $	$ 2x\alpha $

Here, in the above table the equilibrium is attained after the $ 75\% $ of dissociation of $ {N_2}{O_4} $ and it is represented by $ \alpha = 75\% = \dfrac{{75}}{{100}} = 0.75 $
Thus, after equilibrium is attained the reactant moles is equal to:
 $ x(1 - \alpha ) = x(1 - 0.75) = 0.25x $
And for product moles is equal to:
 $ 2x\alpha = 2 \times 0.75 \times x = 1.5x $
Thus, the total moles are calculated as:
 $ 0.25x + 1.5x = 1.75x $
Now, we know that the pressure is directly proportional to the number of moles of the compounds in the reactions.
Therefore, $ {\text{P }}\alpha {\text{ no}}{\text{. of moles}} $
So, we have
 $ {P_1} = x $ and $ {P_2} = 1.75x $
Now, here we are asked to find the ratio between these two pressures as below:
 $ \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{x}{{1.75x}} $
 $ \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{1}{{1.75}} $
 $ \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{100}}{{175}} $
 $ \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{4}{7} $
Thus, the ratio between the pressures is $ {P_1}:{P_2} = 4:7 $
The correct answer is option C.

Note:
Here, the ratio of the pressures is obtained since the pressure is directly proportional to the number of moles of the reactions. We have observed that there is change in moles when the reaction attains the equilibrium after dissociation as we have shown in the table above. On calculating the values we obtained the answer.