Westonci.ca is your go-to source for answers, with a community ready to provide accurate and timely information. Discover comprehensive solutions to your questions from a wide network of experts on our user-friendly platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
