Predict expression from α in terms of \(K_{eq}\) and concentration C :
\(A_2 B_3(aq) \leftrightharpoons 2{A_3} (aq)+3B_{{2-}}(aq)\)
Predict expression from α in terms of \(K_{eq}\) and concentration C :
\(A_2 B_3(aq) \leftrightharpoons 2{A_3} (aq)+3B_{{2-}}(aq)\)
\(\left(\frac{{K_{eq}}}{{108C^4}}\right)^{\frac{1}{5}}\)
\((\frac{K_{eq}}{5C^4})^\frac{1}{5}\)
\((\frac{4K_{eq}}{5C^4})^\frac{1}{5}\)
\((\frac{9K_{eq}}{108C^4})^\frac{1}{5}\)
The Correct Option is A
Solution and Explanation
The expression in terms of \(α\) (degree of dissociation), \(K_{eq}\) (equilibrium constant), and concentration \(C\) for the given reaction:
\[ A_2B_3(aq) \leftrightharpoons 2A_3(aq) + 3B_{2-}(aq) \]
can be derived as follows:
Assume the initial concentration of \(A_2B_3\) is \(C\). At equilibrium:
- The concentration of \(A_2B_3\) will be \((1 - α)C\).
- The concentration of \(A_3\) will be \(2αC\).
- The concentration of \(B_{2-}\) will be \(3αC\).
The equilibrium constant (\(K_{eq}\)) for the reaction is given by:
\[ K_{eq} = \frac{[A_3]^2 [B_{2-}]^3}{[A_2B_3]} \]
Substituting the equilibrium concentrations:
\[ K_{eq} = \frac{[(2αC)^2 \times (3αC)^3]}{[(1 - α)C]} \]
Expand and simplify:
\[ K_{eq} = \frac{(4α^2C^2) \times (27α^3C^3)}{(1 - α)C} \]
\[ K_{eq} = \frac{108α^5C^5}{(1 - α)C} \]
Simplify further:
\[ K_{eq} = \frac{108α^5C^4}{(1 - α)} \]
To express in terms of \(α\), rearrange the equation:
\[ \frac{K_{eq}}{108C^4} = \frac{α^5}{(1 - α)} \]
Taking the fifth root:
\[ \left(\frac{K_{eq}}{108C^4}\right)^{\frac{1}{5}} = α \]
Final expression:
\[ \left(\frac{K_{eq}}{108C^4}\right)^{\frac{1}{5}} \]
Hence, the correct answer is option (A):
\[ \left(\frac{K_{eq}}{108C^4}\right)^{\frac{1}{5}} \]
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Concepts Used:
Equilibrium
An equilibrium represents a state in a process when the observable properties such as color, temperature, pressure, concentration etc do not show any change.
The word equilibrium means ‘balance’ which indicates that a chemical reaction represents a balance between the reactants and products taking part in the reaction. The equilibrium state is also noticed in certain physical processes such as the melting point of ice at 0℃, both ice and water are present at equilibrium.
In the case of physical processes such as the melting of solid, dissolution of salt in water etc., the equilibrium is called physical equilibrium while the equilibrium associated with chemical reaction is known as chemical equilibrium.
Equilibrium in Chemical changes
The chemical equilibrium in a reversible reaction is the state at which both forward and backward reactions occur at the same speed.
The stage of the reversible reaction at which the concentration of the reactants and products do not change with time is called the equilibrium state.
Read More: Calculating Equilibrium Concentration
Types of Chemical Equilibrium
There are two types of chemical equilibrium:
- Homogeneous Equilibrium
- Heterogeneous Equilibrium
Homogenous Chemical Equilibrium
In this type, the reactants and the products of chemical equilibrium are all in the same phase. Homogenous equilibrium can be further divided into two types: Reactions in which the number of molecules of the products is equal to the number of molecules of the reactants. For example,
- H2 (g) + I2 (g) ⇌ 2HI (g)
- N2 (g) + O2 (g) ⇌ 2NO (g)
Reactions in which the number of molecules of the products is not equal to the total number of reactant molecules. For example,
- 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)
- COCl2 (g) ⇌ CO (g) + Cl2 (g)
Heterogeneous Chemical Equilibrium
In this type, the reactants and the products of chemical equilibrium are present in different phases. A few examples of heterogeneous equilibrium are listed below.
- CO2 (g) + C (s) ⇌ 2CO (g)
- CaCO3 (s) ⇌ CaO (s) + CO2 (g)
Thus, the different types of chemical equilibrium are based on the phase of the reactants and products.
Check Out: Equilibrium Important Questions