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Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kc=2.99×10−7 at 227∘C Part A If a reaction mixture initially contains...

Consider the following reaction:
SO2Cl2(g)⇌SO2(g)+Cl2(g)
Kc=2.99×10−7 at 227∘C

Part A

If a reaction mixture initially contains 0.195 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?
[Cl2] =

Solutions

Expert Solution

SO2Cl2 <---> SO2 + Cl2
I      0.195            0    0
C       -x               x    x
Eq   0.195-x x    x

Kc = [SO2]*[Cl2]/[SO2Cl2]
2.99*10^-7 = x2/(0.195 - x)
Lets ignore the value of -x, because this value of Kc it's too small
2.99*10^-7 = x2/(0.195)
x = 0.24*10^-3
This value confirmats that we could ignore the -x, it's too small in comparison to the initial concentration.

In equilibrium:
[SO2Cl2] = 0.195 - 0.24*10^-3 = 0.19476 mol/l
[SO2] = 2.41x10^-4 mol/l
[Cl2] = 2.41x10^-4 mol/l

queen_honey_blossom answered 2 years ago
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