Question

In the given reaction \[{{A(s)}} \rightleftharpoons {{B(g) + C(g)}}\]. Total pressure at time of equilibrium is 40 atmospheres. If all the contents of this reactor have been shifted to another reactor of double capacity, then the total equilibrium pressure in the new reactor will be:(in equilibrium)
A) 20
B) 40
C) 400
D) 1600

Answer 1

Hint: Here the reaction mixture is in equilibrium. As the reaction mixture experiences change in volume, the change in pressure can be obtained using the relation between pressure and volume.

Complete step by step solution:
We have the equation of the equilibrium reaction as follows;
\[{{A(s)}} \rightleftharpoons {{B(g) + C(g)}}\]

Let the pressure at equilibrium be 1 in the reactor of volume ‘V’ litre. Suppose ‘x’ mole is dissociated at equilibrium, then;

\[{{A(s)}} \rightleftharpoons {{B(g) + C(g)}}\]

$1\;\;\;\;\;\;\;\;\;\;\;\; 0 \;\;\;\;\;\;\;\;\;\;\;\;\; 0$ - Initial concentration

$ (1-x) \;\;\;\;x\;\;\;\;\;\;\;\; x$ - At equilibrium

Then the total pressure at the time of equilibrium is given as \[{{40 atm}}\].
Therefore \[{{2x = 40}}\] and \[{{x = 20}}\]
Then we are shifting the entire reaction mixture to a reactor having the double capacity of the first one. So
The volume of the new reactor= \[{{2V}}\]
Here we are using Boyle's Law. Boyle’s law is one of the gas laws that constitute the ideal gas equation. Boyle’s law states that for a given mass of gas, the volume is inversely proportional to the pressure, provided the temperature remains constant.
That is,
\[{{PV = K}}\]

\[{{{P}}_{{1}}}{{{V}}_{{1}}}{{ = }}{{{P}}_{{2}}}{{{V}}_{{2}}}\]

We are applying this gas law in this question and we have

\[{{{P}}_{{1}}}{{ = 40}}\] (pressure in the first reactor)

\[{{{V}}_{{1}}}{{ = V}}\] (volume of the first reactor)

\[{{{P}}_{{2}}}{{ = ?}}\] (pressure of the second reactor)

\[{{{V}}_{{2}}}{{ = 2V}}\] (volume of the second reactor which is double the volume of the first)

Substituting these values in \[{{{P}}_{{1}}}{{{V}}_{{1}}}{{ = }}{{{P}}_{{2}}}{{{V}}_{{2}}}\]we have

\[{{40 \times V = P \times 2V}}\]

\[{{P = }}\dfrac{{{{40 \times V}}}}{{{{2V}}}}\]

\[{{P = 20 atm}}\]

Therefore, the equilibrium pressure in the new reactor having double the volume of the first reactor will be \[{{P = 20 atm}}\].

So, the correct answer is Option A.

Note: Equilibrium state can only be achieved if a reversible reaction is carried out in a closed vessel. There the rate of forward reaction will be equal to the rate of the backward reaction.

The reversible reactions in which the reactants and products are in the same phase are called homogeneous reactions. The reversible reactions in which the reactants and products are in different phases are called heterogeneous reactions.