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issaaamiaaa15
14.07.2020 •
Chemistry
: Starting with 0.3500 mol CO(g) and 0.05500 mol COCl2(g) in a 3.050-L flask at 668 K, how many moles of Cl2(g) will be present at equilibrium
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Ответ:
The number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.
Explanation:
The reaction is:
CO(g) + Cl₂(g) ⇄ COCl₂(g)
The equilibrium constant of the above reaction is:
K = 1.2x10³
To find the moles of Cl₂ present at equilibrium, let's evaluate the reverse reaction:
COCl₂(g) ⇄ CO(g) + Cl₂(g)
The equilibrium constant for the reverse reaction is:
Now, we need to calculate the concentration of CO and COCl₂:
Now, from the reaction we have:
COCl₂(g) ⇄ CO(g) + Cl₂(g)
0.018 - x 0.115+x x
The concentration of Cl₂ is:
By solving the above equation for x we have:
x = 1.29x10⁻⁴ M = [Cl₂]
Finally, the number of moles of Cl₂ present at equilibrium is:
Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.
I hope it helps you!
Ответ:
1 and 2
Explanation:
Hopefully this helps!