Answer :
To solve this problem, let's follow the steps systematically.
### Initial Setup:
- Volume of reaction vessel (V): 5000 L
- Initial moles of H₂ (initial H₂): 1.00 mole
- Initial moles of I₂ (initial I₂): 1.00 mole
- Moles of HI at equilibrium (equilibrium HI): 1.58 mole
### Chemical Reaction:
[tex]\( \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \)[/tex]
Since we have an equilibrium reaction, we must determine the moles of each reactant and product at equilibrium. First, define the change in moles for each substance during the reaction.
### Equilibrium Changes:
For every mole of H₂ and I₂ that reacts, 2 moles of HI are formed:
- Let [tex]\( x \)[/tex] be the moles of H₂ (and I₂) that react to form HI.
- Given that 2x moles of HI are formed whenever x moles of H2 and I2 react.
From the data provided:
- At equilibrium, we have 1.58 moles of HI present.
- Therefore, [tex]\( 2x = 1.58 \)[/tex]
- Solving for [tex]\( x \)[/tex]: [tex]\( x = \frac{1.58}{2} = 0.79 \)[/tex]
### Calculating Remaining Moles:
- Remaining H₂: original moles of H₂ - reacted moles of H₂
[tex]\(= 1.00 \text{ mole} - 0.79 \text{ mole} = 0.21 \text{ mole} \)[/tex]
- Remaining I₂: original moles of I₂ - reacted moles of I₂
[tex]\(= 1.00 \text{ mole} - 0.79 \text{ mole} = 0.21 \text{ mole} \)[/tex]
### Concentrations at Equilibrium:
Concentrations are given by the moles divided by the volume of the vessel (in liters).
- Concentration of H₂ ([tex]\( [\text{H}_2] \)[/tex]):
[tex]\[ [\text{H}_2] = \frac{\text{remaining moles of H}_2}{\text{V}} = \frac{0.21 \text{ mole}}{5000 \text{ L}} = 4.2 \times 10^{-5} \text{ M} \][/tex]
- Concentration of I₂ ([tex]\( [\text{I}_2] \)[/tex]):
[tex]\[ [\text{I}_2] = \frac{\text{remaining moles of I}_2}{\text{V}} = \frac{0.21 \text{ mole}}{5000 \text{ L}} = 4.2 \times 10^{-5} \text{ M} \][/tex]
- Concentration of HI ([tex]\( [\text{HI}] \)[/tex]):
[tex]\[ [\text{HI}] = \frac{\text{moles of HI at equilibrium}}{\text{V}} = \frac{1.58 \text{ mole}}{5000 \text{ L}} = 3.16 \times 10^{-4} \text{ M} \][/tex]
### Calculating the Equilibrium Constant ([tex]\( K_c \)[/tex]):
The equilibrium constant for the reaction [tex]\( \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \)[/tex] is given by:
[tex]\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \][/tex]
- Substitute the values:
[tex]\[ K_c = \frac{(3.16 \times 10^{-4})^2}{(4.2 \times 10^{-5})(4.2 \times 10^{-5})} \][/tex]
- Perform the calculation:
[tex]\[ K_c = \frac{9.9856 \times 10^{-8}}{1.764 \times 10^{-9}} = 56.6 \][/tex]
To summarize:
- Concentration of H₂ at equilibrium: [tex]\(4.2 \times 10^{-5} \text{ M}\)[/tex]
- Concentration of I₂ at equilibrium: [tex]\(4.2 \times 10^{-5} \text{ M}\)[/tex]
- Concentration of HI at equilibrium: [tex]\(3.16 \times 10^{-4} \text{ M}\)[/tex]
- Equilibrium constant [tex]\( K_c \)[/tex]: [tex]\( 56.6 \)[/tex]
Therefore, the concentrations of [tex]\( H_2 \)[/tex], [tex]\( I_2 \)[/tex], and [tex]\( HI \)[/tex] at equilibrium are [tex]\( 4.2 \times 10^{-5} \text{ M} \)[/tex], [tex]\( 4.2 \times 10^{-5} \text{ M} \)[/tex], and [tex]\( 3.16 \times 10^{-4} \text{ M} \)[/tex] respectively. The equilibrium constant [tex]\( K_c \)[/tex] is [tex]\( 56.6 \)[/tex].
### Initial Setup:
- Volume of reaction vessel (V): 5000 L
- Initial moles of H₂ (initial H₂): 1.00 mole
- Initial moles of I₂ (initial I₂): 1.00 mole
- Moles of HI at equilibrium (equilibrium HI): 1.58 mole
### Chemical Reaction:
[tex]\( \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \)[/tex]
Since we have an equilibrium reaction, we must determine the moles of each reactant and product at equilibrium. First, define the change in moles for each substance during the reaction.
### Equilibrium Changes:
For every mole of H₂ and I₂ that reacts, 2 moles of HI are formed:
- Let [tex]\( x \)[/tex] be the moles of H₂ (and I₂) that react to form HI.
- Given that 2x moles of HI are formed whenever x moles of H2 and I2 react.
From the data provided:
- At equilibrium, we have 1.58 moles of HI present.
- Therefore, [tex]\( 2x = 1.58 \)[/tex]
- Solving for [tex]\( x \)[/tex]: [tex]\( x = \frac{1.58}{2} = 0.79 \)[/tex]
### Calculating Remaining Moles:
- Remaining H₂: original moles of H₂ - reacted moles of H₂
[tex]\(= 1.00 \text{ mole} - 0.79 \text{ mole} = 0.21 \text{ mole} \)[/tex]
- Remaining I₂: original moles of I₂ - reacted moles of I₂
[tex]\(= 1.00 \text{ mole} - 0.79 \text{ mole} = 0.21 \text{ mole} \)[/tex]
### Concentrations at Equilibrium:
Concentrations are given by the moles divided by the volume of the vessel (in liters).
- Concentration of H₂ ([tex]\( [\text{H}_2] \)[/tex]):
[tex]\[ [\text{H}_2] = \frac{\text{remaining moles of H}_2}{\text{V}} = \frac{0.21 \text{ mole}}{5000 \text{ L}} = 4.2 \times 10^{-5} \text{ M} \][/tex]
- Concentration of I₂ ([tex]\( [\text{I}_2] \)[/tex]):
[tex]\[ [\text{I}_2] = \frac{\text{remaining moles of I}_2}{\text{V}} = \frac{0.21 \text{ mole}}{5000 \text{ L}} = 4.2 \times 10^{-5} \text{ M} \][/tex]
- Concentration of HI ([tex]\( [\text{HI}] \)[/tex]):
[tex]\[ [\text{HI}] = \frac{\text{moles of HI at equilibrium}}{\text{V}} = \frac{1.58 \text{ mole}}{5000 \text{ L}} = 3.16 \times 10^{-4} \text{ M} \][/tex]
### Calculating the Equilibrium Constant ([tex]\( K_c \)[/tex]):
The equilibrium constant for the reaction [tex]\( \text{H}_2 + \text{I}_2 \rightleftharpoons 2 \text{HI} \)[/tex] is given by:
[tex]\[ K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} \][/tex]
- Substitute the values:
[tex]\[ K_c = \frac{(3.16 \times 10^{-4})^2}{(4.2 \times 10^{-5})(4.2 \times 10^{-5})} \][/tex]
- Perform the calculation:
[tex]\[ K_c = \frac{9.9856 \times 10^{-8}}{1.764 \times 10^{-9}} = 56.6 \][/tex]
To summarize:
- Concentration of H₂ at equilibrium: [tex]\(4.2 \times 10^{-5} \text{ M}\)[/tex]
- Concentration of I₂ at equilibrium: [tex]\(4.2 \times 10^{-5} \text{ M}\)[/tex]
- Concentration of HI at equilibrium: [tex]\(3.16 \times 10^{-4} \text{ M}\)[/tex]
- Equilibrium constant [tex]\( K_c \)[/tex]: [tex]\( 56.6 \)[/tex]
Therefore, the concentrations of [tex]\( H_2 \)[/tex], [tex]\( I_2 \)[/tex], and [tex]\( HI \)[/tex] at equilibrium are [tex]\( 4.2 \times 10^{-5} \text{ M} \)[/tex], [tex]\( 4.2 \times 10^{-5} \text{ M} \)[/tex], and [tex]\( 3.16 \times 10^{-4} \text{ M} \)[/tex] respectively. The equilibrium constant [tex]\( K_c \)[/tex] is [tex]\( 56.6 \)[/tex].
