Answer:
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:
[tex] K_{r} = \frac{1}{1.2 \cdot 10^{3}} = 8.3 \cdot 10^{-4} [/tex]
Now, we need to calculate the concentration of CO and COCl₂:
[tex] C_{CO} = \frac{\eta_{CO}}{V} = \frac{0.3500 moles}{3.050 L} = 0.115 M [/tex]
[tex] C_{COCl_{2}} = \frac{\eta_{COCl_{2}}}{V} = \frac{0.05500 moles}{3.050 L} = 0.018 M [/tex]
Now, from the reaction we have:
COCl₂(g) ⇄ CO(g) + Cl₂(g)
0.018 - x 0.115+x x
The concentration of Cl₂ is:
[tex] K_{r} = \frac{[CO][Cl_{2}]}{[COCl_{2}]} [/tex]
[tex] 8.3 \cdot 10^{-4} = \frac{(0.115 + x)(x)}{0.018 - x} [/tex]
[tex] 8.3 \cdot 10^{-4}*(0.018 - x) - (0.115 + x)(x) = 0 [/tex]
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:
[tex] \eta_{Cl_{2}} = C_{Cl_{2}}*V = 1.29 \cdot 10^{-4} mol/L*3.050 L = 3.94 \cdot 10^{-4} moles [/tex]
Therefore, the number of moles of Cl₂ present at equilibrium is 3.94x10⁻⁴ moles.
I hope it helps you!
