IDNLearn.com offers a unique blend of expert answers and community-driven insights. Find the solutions you need quickly and accurately with help from our knowledgeable community.
Sagot :
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].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your reliable source for accurate answers. Thank you for visiting, and we hope to assist you again.
