Answer :
Certainly! Let's go through the details step-by-step for the given reaction during lightning storms:
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
### (i) Equilibrium Constant Expression
The equilibrium constant expression for a reaction is derived from the balanced chemical equation. For the reaction:
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
The equilibrium constant expression, [tex]\( K_{eq} \)[/tex], is given by the ratio of the product of the concentrations of the products, raised to their respective stoichiometric coefficients, to the product of the concentrations of the reactants, raised to their respective stoichiometric coefficients.
Thus, for this reaction:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
### (ii) Determine the Units of Equilibrium Constant
To determine the units of the equilibrium constant, we need to consider the units of concentration, which is typically moles per liter (M). The equilibrium constant expression involves the concentrations with their respective powers:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
In terms of their units:
[tex]\[ K_{eq} = \frac{(\text{M})^2}{(\text{M})^3} = \frac{\text{M}^2}{\text{M}^3} = \frac{1}{\text{M}} \][/tex]
So, the units of the equilibrium constant, [tex]\( K_{eq} \)[/tex], for this reaction are:
[tex]\[ \text{Units of } K_{eq} = \frac{1}{\text{M}} \][/tex]
### (iii) Forward and Reverse Reactions
For the given reaction:
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
The forward reaction is the reaction proceeding from left to right, where oxygen gas ([tex]\(\text{O}_2\)[/tex]) converts into ozone ([tex]\(\text{O}_3\)[/tex]):
[tex]\[ \text{Forward Reaction: } 3 \text{O}_2 (g) \rightarrow 2 \text{O}_3 (g) \][/tex]
The reverse reaction is the reaction proceeding from right to left, where ozone ([tex]\(\text{O}_3\)[/tex]) decomposes back into oxygen gas ([tex]\(\text{O}_2\)[/tex]):
[tex]\[ \text{Reverse Reaction: } 2 \text{O}_3 (g) \rightarrow 3 \text{O}_2 (g) \][/tex]
To summarize, here are the answers to the given parts of the question:
(i) The equilibrium constant expression:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
(ii) The units of the equilibrium constant:
[tex]\[ \text{Units of } K_{eq} = \frac{1}{\text{M}} \][/tex]
(iii) The forward and reverse reactions:
- Forward Reaction: [tex]\( 3 \text{O}_2 (g) \rightarrow 2 \text{O}_3 (g) \)[/tex]
- Reverse Reaction: [tex]\( 2 \text{O}_3 (g) \rightarrow 3 \text{O}_2 (g) \)[/tex]
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
### (i) Equilibrium Constant Expression
The equilibrium constant expression for a reaction is derived from the balanced chemical equation. For the reaction:
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
The equilibrium constant expression, [tex]\( K_{eq} \)[/tex], is given by the ratio of the product of the concentrations of the products, raised to their respective stoichiometric coefficients, to the product of the concentrations of the reactants, raised to their respective stoichiometric coefficients.
Thus, for this reaction:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
### (ii) Determine the Units of Equilibrium Constant
To determine the units of the equilibrium constant, we need to consider the units of concentration, which is typically moles per liter (M). The equilibrium constant expression involves the concentrations with their respective powers:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
In terms of their units:
[tex]\[ K_{eq} = \frac{(\text{M})^2}{(\text{M})^3} = \frac{\text{M}^2}{\text{M}^3} = \frac{1}{\text{M}} \][/tex]
So, the units of the equilibrium constant, [tex]\( K_{eq} \)[/tex], for this reaction are:
[tex]\[ \text{Units of } K_{eq} = \frac{1}{\text{M}} \][/tex]
### (iii) Forward and Reverse Reactions
For the given reaction:
[tex]\[ 3 \text{O}_2 (g) \rightleftharpoons 2 \text{O}_3 (g) \][/tex]
The forward reaction is the reaction proceeding from left to right, where oxygen gas ([tex]\(\text{O}_2\)[/tex]) converts into ozone ([tex]\(\text{O}_3\)[/tex]):
[tex]\[ \text{Forward Reaction: } 3 \text{O}_2 (g) \rightarrow 2 \text{O}_3 (g) \][/tex]
The reverse reaction is the reaction proceeding from right to left, where ozone ([tex]\(\text{O}_3\)[/tex]) decomposes back into oxygen gas ([tex]\(\text{O}_2\)[/tex]):
[tex]\[ \text{Reverse Reaction: } 2 \text{O}_3 (g) \rightarrow 3 \text{O}_2 (g) \][/tex]
To summarize, here are the answers to the given parts of the question:
(i) The equilibrium constant expression:
[tex]\[ K_{eq} = \frac{[\text{O}_3]^2}{[\text{O}_2]^3} \][/tex]
(ii) The units of the equilibrium constant:
[tex]\[ \text{Units of } K_{eq} = \frac{1}{\text{M}} \][/tex]
(iii) The forward and reverse reactions:
- Forward Reaction: [tex]\( 3 \text{O}_2 (g) \rightarrow 2 \text{O}_3 (g) \)[/tex]
- Reverse Reaction: [tex]\( 2 \text{O}_3 (g) \rightarrow 3 \text{O}_2 (g) \)[/tex]
