Respuesta :
Answer:
see explanation below
Explanation:
The question is incomplete. Here's the complete question:
Consider the following reaction:
SO2Cl2 -----> SO2(g) + Cl2(g)
Kc= 2.99 x 10^-7 at 227 degrees celcius
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
This is a problem of equilibrium, therefore, we need to solve this using the expression of equilibrium constant. To do that, we need to wirte an ICE chart and solve from there:
SOCl2 ---------> SO2 + Cl2 Kc = 2.99x10⁻⁷
i) 0.168 0 0
c) -x +x +x
e) 0.168-x x x
Writting the Kc expression:
Kc = [SO2] [Cl2] / [SOCl2]
Replacing the values from the chart:
2.99x10⁻⁷ = x² / 0.168 - x
However, Kc is a very very small value, therefore, we can assume that the value of "x" would be very small too, and we can neglect the 0.168-x and just round it to 0.168:
2.99x10⁻⁷ = x²/0.168
2.99x10⁻⁷ * 0.168 = x²
√5.02x10⁻⁸ = x
x = 2.24x10⁻⁴ M
This means then, that the concentration of Cl2 in equilibrium would be:
[Cl₂] = 2.24x10⁻⁴ M
The equilibrium concentration of [Cl₂] = 2.24x10⁻⁴ M.
Calculation of an equilibrium;
Here we need to write an ICE chart and solve from there:
Since
SOCl2 ---------> SO2 + Cl2 Kc = 2.99x10⁻⁷
So,
i) 0.168 0 0
c) -x +x +x
e) 0.168-x x x
Now Writing the Kc expression:
Kc = [SO2] [Cl2] / [SOCl2]
Now
2.99x10⁻⁷ = x² / 0.168 - x
2.99x10⁻⁷ = x²/0.168
2.99x10⁻⁷ * 0.168 = x²
√5.02x10⁻⁸ = x
x = 2.24x10⁻⁴ M
This question is incomplete. Please find the full question below.
Consider the following reaction:
SO2Cl2 -----> SO2(g) + Cl2(g)
Kc= 2.99 x 10^-7 at 227 degrees celcius
If a reaction mixture initially contains 0.168 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
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