Answer :
Sure, let's go through the solution step-by-step for understanding the equilibrium shift in the reaction:
The balanced chemical reaction is:
[tex]\[ H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g) \][/tex]
Initial Equilibrium Concentrations (`E_0`):
- [tex]\([H_2] = 6.00 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 6.00 \, \text{M}\)[/tex]
- [tex]\([HCl] = 12.0 \, \text{M}\)[/tex]
Disturbance to the System (`I`):
- [tex]\([HCl] = 5.00 \, \text{M}\)[/tex]
The concentration of HCl is decreased from 12.0 M to 5.00 M, which disturbs the equilibrium, causing the system to shift to re-establish equilibrium. According to Le Chatelier's principle, the system will shift to the right to produce more HCl.
Change in Concentration (`C`):
Let the change in concentration of [tex]\(H_2\)[/tex] and [tex]\(Cl_2\)[/tex] due to the shift be [tex]\(-x\)[/tex]. Since the stoichiometry of the reaction shows that one mole each of [tex]\(H_2\)[/tex] and [tex]\(Cl_2\)[/tex] forms two moles of HCl, the change in concentration for [tex]\(HCl\)[/tex] will be [tex]\(+2x\)[/tex].
The initial concentration of HCl was 5.00 M after the disturbance, and it needs to get back to the equilibrium concentration.
[tex]\([HCl] \,\text{change} = 12.0 - 5.00 = 2x = 3.5 \)[/tex]
Solving for [tex]\(x\)[/tex], we find:
[tex]\[ x = \frac{3.5}{2} = 1.75 \][/tex]
Final Equilibrium Concentrations (`E_f`):
- [tex]\([H_2] = 6.00 - x = 6.00 - 1.75 = 4.25 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 6.00 - x = 6.00 - 1.75 = 4.25 \, \text{M}\)[/tex]
- [tex]\([HCl] = 5.00 + 2x = 5.00 + 3.5 = 8.5 \, \text{M}\)[/tex]
Oops! My mistake! Let's re-calculate the correct answer without using incorrect logic.
Using correct logic:
`[H_Cl_final]` becomes `12.0 M`.
Thus, at the new equilibrium:
- [tex]\([H_2] = 2.5 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 2.5 \, \text{M}\)[/tex]
- [tex]\([HCl] = 12.0 \, \text{M}\)[/tex]
Therefore, when equilibrium is re-established, the concentration of [tex]\([HCl]\)[/tex] is 12.0 M.
The balanced chemical reaction is:
[tex]\[ H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g) \][/tex]
Initial Equilibrium Concentrations (`E_0`):
- [tex]\([H_2] = 6.00 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 6.00 \, \text{M}\)[/tex]
- [tex]\([HCl] = 12.0 \, \text{M}\)[/tex]
Disturbance to the System (`I`):
- [tex]\([HCl] = 5.00 \, \text{M}\)[/tex]
The concentration of HCl is decreased from 12.0 M to 5.00 M, which disturbs the equilibrium, causing the system to shift to re-establish equilibrium. According to Le Chatelier's principle, the system will shift to the right to produce more HCl.
Change in Concentration (`C`):
Let the change in concentration of [tex]\(H_2\)[/tex] and [tex]\(Cl_2\)[/tex] due to the shift be [tex]\(-x\)[/tex]. Since the stoichiometry of the reaction shows that one mole each of [tex]\(H_2\)[/tex] and [tex]\(Cl_2\)[/tex] forms two moles of HCl, the change in concentration for [tex]\(HCl\)[/tex] will be [tex]\(+2x\)[/tex].
The initial concentration of HCl was 5.00 M after the disturbance, and it needs to get back to the equilibrium concentration.
[tex]\([HCl] \,\text{change} = 12.0 - 5.00 = 2x = 3.5 \)[/tex]
Solving for [tex]\(x\)[/tex], we find:
[tex]\[ x = \frac{3.5}{2} = 1.75 \][/tex]
Final Equilibrium Concentrations (`E_f`):
- [tex]\([H_2] = 6.00 - x = 6.00 - 1.75 = 4.25 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 6.00 - x = 6.00 - 1.75 = 4.25 \, \text{M}\)[/tex]
- [tex]\([HCl] = 5.00 + 2x = 5.00 + 3.5 = 8.5 \, \text{M}\)[/tex]
Oops! My mistake! Let's re-calculate the correct answer without using incorrect logic.
Using correct logic:
`[H_Cl_final]` becomes `12.0 M`.
Thus, at the new equilibrium:
- [tex]\([H_2] = 2.5 \, \text{M}\)[/tex]
- [tex]\([Cl_2] = 2.5 \, \text{M}\)[/tex]
- [tex]\([HCl] = 12.0 \, \text{M}\)[/tex]
Therefore, when equilibrium is re-established, the concentration of [tex]\([HCl]\)[/tex] is 12.0 M.
