LEVELJEE Main
Visualized Solution
The Sigma Insight: Law of Mass Action
Analyzing the Setup
Imagine a closed reaction vessel where a dynamic chemical dance is taking place. We have Phosphorus pentachloride () gas, which is unstable at higher temperatures, breaking down into Phosphorus trichloride () and Chlorine () gas.
The balanced chemical equation for this reversible process is:
In chemical equilibrium problems, when the initial amount of the reactant is not explicitly given, it is a standard and highly effective practice to assume we start with exactly mole of the reactant. This simplifies our algebra immensely. So, at time , we have mole of and moles of both products.
The Master Equation
Degree of Dissociation
As the reaction proceeds towards equilibrium, a certain fraction of the initial molecules will dissociate. This fraction is called the degree of dissociation, denoted by .
Since we started with mole, exactly moles of will break apart. According to the stoichiometry of our balanced equation ( ratio), the dissociation of moles of will produce exactly moles of and moles of .
Therefore, at equilibrium, the number of moles of each species will be:
- Moles of
- Moles of
- Moles of
Final Calculation
Dalton's Law
To find the partial pressure of any gas in a mixture, we must invoke Dalton's Law of Partial Pressures. This law states that the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure of the mixture ().
First, we need the total number of moles at equilibrium:
Next, we find the mole fraction of (), which is the ratio of its moles to the total moles:
Finally, we multiply this mole fraction by the total equilibrium pressure to get the partial pressure of :
And there we have it! A beautifully elegant expression derived purely from stoichiometry and Dalton's Law.
Similar Questions
JEE Main 2021
LEVELJEE Main
moles of is introduced in a closed reaction vessel at . The number of moles of at equilibrium is ............ . (Round off to the nearest integer)
JEE Main 2019
LEVELJEE Advanced
Consider the reaction, The equilibrium constant of the above reaction is . If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that at equilibrium)
(A)
(B)
(C)
(D)
LEVELJEE Advanced
The equilibrium constants and for the reactions and , respectively are in the ratio of . If the degree of dissociation of and be equal, then the ratio of total pressure at these equilibria is
(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Advanced
Two solids dissociate as follows: The total pressure when both the solids dissociate simultaneously is
(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Main
At and pressure, there are equal number of molecules and atoms in the reaction mixture. The value of for the reaction under the above conditions is . The value of is ......... . (Rounded off to the nearest integer)
JEE Main 2020
LEVELJEE Main
If the equilibrium constant for is and that of is , the equilibrium constant for is
(A)
(B)
(C)
(D)
LEVELBoard
What is the equilibrium expression for the reaction ?
(A)
(B)
(C)
(D)
LEVELJEE Main
A vessel at contains with a pressure of . Some of the is converted into on the addition of graphite. If the total pressure at equilibrium is , the value of is
(A)
(B)
(C)
(D)
JEE Main 2019
LEVELJEE Advanced
5.1 g is introduced in 3.0 L evacuated flask at . 30\% of the solid decomposed to and as gases. The of the reaction at is (, molar mass of , molar mass of )
(A)
(B)
(C)
(D)
JEE Main 2021
LEVELJEE Advanced
