Answer :
Let's consider the reaction:
[tex]\[ H_2 (g) + I_2 (g) \rightarrow 2 HI (g) \][/tex]
Initially, we have 1.0 mole each of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] in a volume of 600 L. At equilibrium, we are given that the mixture contains 2.0 moles of [tex]\( HI \)[/tex].
We need to determine the concentrations of [tex]\( H_2 \)[/tex], [tex]\( I_2 \)[/tex], and [tex]\( HI \)[/tex] at equilibrium and also calculate the equilibrium constant [tex]\( K_{eq} \)[/tex].
### Step-by-Step Solution:
1. Initial Moles and Volume:
[tex]\[ \text{Initial moles of } H_2 = 1.0 \text{ mole}, \quad \text{Initial moles of } I_2 = 1.0 \text{ mole} \][/tex]
[tex]\[ \text{Volume of the mixture} = 600 \text{ L} \][/tex]
2. Change in Moles:
Since at equilibrium there are 2.0 moles of [tex]\( HI \)[/tex], let’s consider [tex]\( x \)[/tex] moles of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] reacted to form [tex]\( HI \)[/tex]. For each mole of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] reacted, 2 moles of [tex]\( HI \)[/tex] are produced.
Hence, [tex]\( 2HI \)[/tex] means [tex]\( x = \frac{2.0 \text{ moles}}{2} = 1.0 \text{ mole} \)[/tex]
3. Equilibrium Moles:
Therefore:
[tex]\[ \text{Moles of } H_2 \text{ at equilibrium} = 1.0 - x = 1.0 - 1.0 = 0 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } I_2 \text{ at equilibrium} = 1.0 - x = 1.0 - 1.0 = 0 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } HI \text{ at equilibrium} = 2.0 \text{ moles} \][/tex]
4. Concentration Calculations:
The concentration is given by the number of moles divided by the volume of the mixture.
[tex]\[ \text{Concentration of } H_2 \text{ at equilibrium} = \frac{\text{moles of } H_2}{\text{volume}} = \frac{0 \text{ moles}}{600 \text{ L}} = 0 \text{ M} \][/tex]
[tex]\[ \text{Concentration of } I_2 \text{ at equilibrium} = \frac{\text{moles of } I_2}{\text{volume}} = \frac{0 \text{ moles}}{600 \text{ L}} = 0 \text{ M} \][/tex]
[tex]\[ \text{Concentration of } HI \text{ at equilibrium} = \frac{\text{moles of } HI}{\text{volume}} = \frac{2.0 \text{ moles}}{600 \text{ L}} = \frac{2.0}{600} \text{ M} = \frac{1.0}{300} \text{ M} \approx 0.00333 \text{ M} \][/tex]
5. Equilibrium Constant Calculation:
The equilibrium constant expression for the reaction is:
[tex]\[ K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \][/tex]
Since the concentrations of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] are both [tex]\( 0 \text{ M} \)[/tex]:
[tex]\[ K_{eq} = \frac{(0.00333)^2}{0 \times 0} = \text{undefined since both concentrations are zero,} \][/tex]
this scenario suggests that in our idealized setup there are no reactants left to sustain any dynamic equilibrium; therefore it is typically inconsistent with finite [tex]\( K_{eq} \)[/tex] concept.
### Conclusion:
- The concentrations at equilibrium are:
- Concentration of [tex]\( H_2 \)[/tex] at equilibrium: [tex]\( 0 \text{ M} \)[/tex]
- Concentration of [tex]\( I_2 \)[/tex] at equilibrium: [tex]\( 0 \text{ M} \)[/tex]
- Concentration of [tex]\( HI \)[/tex] at equilibrium: [tex]\( \approx 0.00333 \text{ M} \)[/tex]
- This result highlights conceptual limitations, as true equilibrium requires non-zero reactants to accurately define conventional [tex]\( K_{eq} \)[/tex]
[tex]\[ H_2 (g) + I_2 (g) \rightarrow 2 HI (g) \][/tex]
Initially, we have 1.0 mole each of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] in a volume of 600 L. At equilibrium, we are given that the mixture contains 2.0 moles of [tex]\( HI \)[/tex].
We need to determine the concentrations of [tex]\( H_2 \)[/tex], [tex]\( I_2 \)[/tex], and [tex]\( HI \)[/tex] at equilibrium and also calculate the equilibrium constant [tex]\( K_{eq} \)[/tex].
### Step-by-Step Solution:
1. Initial Moles and Volume:
[tex]\[ \text{Initial moles of } H_2 = 1.0 \text{ mole}, \quad \text{Initial moles of } I_2 = 1.0 \text{ mole} \][/tex]
[tex]\[ \text{Volume of the mixture} = 600 \text{ L} \][/tex]
2. Change in Moles:
Since at equilibrium there are 2.0 moles of [tex]\( HI \)[/tex], let’s consider [tex]\( x \)[/tex] moles of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] reacted to form [tex]\( HI \)[/tex]. For each mole of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] reacted, 2 moles of [tex]\( HI \)[/tex] are produced.
Hence, [tex]\( 2HI \)[/tex] means [tex]\( x = \frac{2.0 \text{ moles}}{2} = 1.0 \text{ mole} \)[/tex]
3. Equilibrium Moles:
Therefore:
[tex]\[ \text{Moles of } H_2 \text{ at equilibrium} = 1.0 - x = 1.0 - 1.0 = 0 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } I_2 \text{ at equilibrium} = 1.0 - x = 1.0 - 1.0 = 0 \text{ moles} \][/tex]
[tex]\[ \text{Moles of } HI \text{ at equilibrium} = 2.0 \text{ moles} \][/tex]
4. Concentration Calculations:
The concentration is given by the number of moles divided by the volume of the mixture.
[tex]\[ \text{Concentration of } H_2 \text{ at equilibrium} = \frac{\text{moles of } H_2}{\text{volume}} = \frac{0 \text{ moles}}{600 \text{ L}} = 0 \text{ M} \][/tex]
[tex]\[ \text{Concentration of } I_2 \text{ at equilibrium} = \frac{\text{moles of } I_2}{\text{volume}} = \frac{0 \text{ moles}}{600 \text{ L}} = 0 \text{ M} \][/tex]
[tex]\[ \text{Concentration of } HI \text{ at equilibrium} = \frac{\text{moles of } HI}{\text{volume}} = \frac{2.0 \text{ moles}}{600 \text{ L}} = \frac{2.0}{600} \text{ M} = \frac{1.0}{300} \text{ M} \approx 0.00333 \text{ M} \][/tex]
5. Equilibrium Constant Calculation:
The equilibrium constant expression for the reaction is:
[tex]\[ K_{eq} = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \][/tex]
Since the concentrations of [tex]\( H_2 \)[/tex] and [tex]\( I_2 \)[/tex] are both [tex]\( 0 \text{ M} \)[/tex]:
[tex]\[ K_{eq} = \frac{(0.00333)^2}{0 \times 0} = \text{undefined since both concentrations are zero,} \][/tex]
this scenario suggests that in our idealized setup there are no reactants left to sustain any dynamic equilibrium; therefore it is typically inconsistent with finite [tex]\( K_{eq} \)[/tex] concept.
### Conclusion:
- The concentrations at equilibrium are:
- Concentration of [tex]\( H_2 \)[/tex] at equilibrium: [tex]\( 0 \text{ M} \)[/tex]
- Concentration of [tex]\( I_2 \)[/tex] at equilibrium: [tex]\( 0 \text{ M} \)[/tex]
- Concentration of [tex]\( HI \)[/tex] at equilibrium: [tex]\( \approx 0.00333 \text{ M} \)[/tex]
- This result highlights conceptual limitations, as true equilibrium requires non-zero reactants to accurately define conventional [tex]\( K_{eq} \)[/tex]
