Answer :
The reaction in question is:
[tex]\[ H_2 (g) + I_2 (g) \rightleftharpoons 2 HI (g) \][/tex]
We need to determine the equilibrium concentrations when 1.00 mole of [tex]\( I_2 \)[/tex] and 1.00 mole of [tex]\( H_2 \)[/tex] are placed in a 5-liter container. The equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is 64.0.
### Step-by-Step Solution:
1. Calculate Initial Concentrations:
Given:
- Initial moles of [tex]\( I_2 \)[/tex] = 1.00 mole
- Initial moles of [tex]\( H_2 \)[/tex] = 1.00 mole
- Volume of container = 5.0 liters
The initial concentrations (Molarity) are:
[tex]\[ [I_2]_{\text{initial}} = \frac{1.00 \, \text{mole}}{5.0 \, \text{L}} = 0.20 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{initial}} = \frac{1.00 \, \text{mole}}{5.0 \, \text{L}} = 0.20 \, \text{M} \][/tex]
2. Define Change in Concentration:
Let [tex]\( x \)[/tex] be the amount of [tex]\( I_2 \)[/tex] and [tex]\( H_2 \)[/tex] reacted to form [tex]\( HI \)[/tex]. Since the stoichiometry of the balanced equation is 1:1:2, the changes in concentrations are:
[tex]\[ [I_2] = 0.20 - x \][/tex]
[tex]\[ [H_2] = 0.20 - x \][/tex]
[tex]\[ [HI] = 2x \][/tex]
3. Establish the Equilibrium Expression:
[tex]\[ K_{\text{eq}} = \frac{[HI]^2}{[H_2][I_2]} \][/tex]
Substituting the equilibrium concentrations,
[tex]\[ 64.0 = \frac{(2x)^2}{(0.20 - x)(0.20 - x)} \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 64.0 = \frac{4x^2}{(0.20 - x)^2} \][/tex]
[tex]\[ 64.0 = \frac{4x^2}{0.04 - 0.4x + x^2} \][/tex]
[tex]\[ 64(0.04 - 0.4x + x^2) = 4x^2 \][/tex]
[tex]\[ 2.56 - 25.6x + 64x^2 = 4x^2 \][/tex]
[tex]\[ 2.56 - 25.6x + 60x^2 = 0 \][/tex]
4. Solve the Quadratic Equation:
The quadratic equation is:
[tex]\[ 60x^2 - 25.6x + 2.56 = 0 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 60 \)[/tex], [tex]\( b = -25.6 \)[/tex], and [tex]\( c = 2.56 \)[/tex].
[tex]\[ x = \frac{25.6 \pm \sqrt{(-25.6)^2 - 4 \cdot 60 \cdot 2.56}}{2 \cdot 60} \][/tex]
[tex]\[ x = \frac{25.6 \pm \sqrt{655.36 - 614.40}}{120} \][/tex]
[tex]\[ x = \frac{25.6 \pm \sqrt{40.96}}{120} \][/tex]
[tex]\[ x = \frac{25.6 \pm 6.4}{120} \][/tex]
The two solutions are:
[tex]\[ x_1 = \frac{32}{120} = \frac{8}{30} = \frac{4}{15} \approx 0.267 \, \text{M} \][/tex]
[tex]\[ x_2 = \frac{19.2}{120} = \frac{16}{100} = \frac{4}{25} = 0.16 \, \text{M} \][/tex]
Only the value [tex]\( x = 0.16 \, \text{M} \)[/tex] is physically feasible because it keeps the concentrations of [tex]\( I_2 \)[/tex] and [tex]\( H_2 \)[/tex] positive.
5. Calculate the Equilibrium Concentrations:
Using [tex]\( x = 0.16 \)[/tex]:
[tex]\[ [I_2]_{\text{eq}} = 0.20 - 0.16 = 0.04 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{eq}} = 0.20 - 0.16 = 0.04 \, \text{M} \][/tex]
[tex]\[ [HI]_{\text{eq}} = 2 \times 0.16 = 0.32 \, \text{M} \][/tex]
### Conclusion
The equilibrium concentrations are:
[tex]\[ [I_2]_{\text{eq}} = 0.04 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{eq}} = 0.04 \, \text{M} \][/tex]
[tex]\[ [HI]_{\text{eq}} = 0.32 \, \text{M} \][/tex]
[tex]\[ H_2 (g) + I_2 (g) \rightleftharpoons 2 HI (g) \][/tex]
We need to determine the equilibrium concentrations when 1.00 mole of [tex]\( I_2 \)[/tex] and 1.00 mole of [tex]\( H_2 \)[/tex] are placed in a 5-liter container. The equilibrium constant [tex]\( K_{\text{eq}} \)[/tex] is 64.0.
### Step-by-Step Solution:
1. Calculate Initial Concentrations:
Given:
- Initial moles of [tex]\( I_2 \)[/tex] = 1.00 mole
- Initial moles of [tex]\( H_2 \)[/tex] = 1.00 mole
- Volume of container = 5.0 liters
The initial concentrations (Molarity) are:
[tex]\[ [I_2]_{\text{initial}} = \frac{1.00 \, \text{mole}}{5.0 \, \text{L}} = 0.20 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{initial}} = \frac{1.00 \, \text{mole}}{5.0 \, \text{L}} = 0.20 \, \text{M} \][/tex]
2. Define Change in Concentration:
Let [tex]\( x \)[/tex] be the amount of [tex]\( I_2 \)[/tex] and [tex]\( H_2 \)[/tex] reacted to form [tex]\( HI \)[/tex]. Since the stoichiometry of the balanced equation is 1:1:2, the changes in concentrations are:
[tex]\[ [I_2] = 0.20 - x \][/tex]
[tex]\[ [H_2] = 0.20 - x \][/tex]
[tex]\[ [HI] = 2x \][/tex]
3. Establish the Equilibrium Expression:
[tex]\[ K_{\text{eq}} = \frac{[HI]^2}{[H_2][I_2]} \][/tex]
Substituting the equilibrium concentrations,
[tex]\[ 64.0 = \frac{(2x)^2}{(0.20 - x)(0.20 - x)} \][/tex]
Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 64.0 = \frac{4x^2}{(0.20 - x)^2} \][/tex]
[tex]\[ 64.0 = \frac{4x^2}{0.04 - 0.4x + x^2} \][/tex]
[tex]\[ 64(0.04 - 0.4x + x^2) = 4x^2 \][/tex]
[tex]\[ 2.56 - 25.6x + 64x^2 = 4x^2 \][/tex]
[tex]\[ 2.56 - 25.6x + 60x^2 = 0 \][/tex]
4. Solve the Quadratic Equation:
The quadratic equation is:
[tex]\[ 60x^2 - 25.6x + 2.56 = 0 \][/tex]
Solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
Here, [tex]\( a = 60 \)[/tex], [tex]\( b = -25.6 \)[/tex], and [tex]\( c = 2.56 \)[/tex].
[tex]\[ x = \frac{25.6 \pm \sqrt{(-25.6)^2 - 4 \cdot 60 \cdot 2.56}}{2 \cdot 60} \][/tex]
[tex]\[ x = \frac{25.6 \pm \sqrt{655.36 - 614.40}}{120} \][/tex]
[tex]\[ x = \frac{25.6 \pm \sqrt{40.96}}{120} \][/tex]
[tex]\[ x = \frac{25.6 \pm 6.4}{120} \][/tex]
The two solutions are:
[tex]\[ x_1 = \frac{32}{120} = \frac{8}{30} = \frac{4}{15} \approx 0.267 \, \text{M} \][/tex]
[tex]\[ x_2 = \frac{19.2}{120} = \frac{16}{100} = \frac{4}{25} = 0.16 \, \text{M} \][/tex]
Only the value [tex]\( x = 0.16 \, \text{M} \)[/tex] is physically feasible because it keeps the concentrations of [tex]\( I_2 \)[/tex] and [tex]\( H_2 \)[/tex] positive.
5. Calculate the Equilibrium Concentrations:
Using [tex]\( x = 0.16 \)[/tex]:
[tex]\[ [I_2]_{\text{eq}} = 0.20 - 0.16 = 0.04 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{eq}} = 0.20 - 0.16 = 0.04 \, \text{M} \][/tex]
[tex]\[ [HI]_{\text{eq}} = 2 \times 0.16 = 0.32 \, \text{M} \][/tex]
### Conclusion
The equilibrium concentrations are:
[tex]\[ [I_2]_{\text{eq}} = 0.04 \, \text{M} \][/tex]
[tex]\[ [H_2]_{\text{eq}} = 0.04 \, \text{M} \][/tex]
[tex]\[ [HI]_{\text{eq}} = 0.32 \, \text{M} \][/tex]
