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Chemistry question regarding equilibrium and ICE table?
A mixture of 0.1532 M Cl2 (g) and 0.1532 M I2(g) was allowed to react and come to equilibrium at some temperature. If Kc = 1.492e5, what is the equilibrium molar concentration of ICl for the reaction?
Cl2(g) + I2(g) <-> 2ICl(g)
I made an ICE table and when I plugged all the values into the Kc equation and then did the quadratic formula, I got a negative number underneath the square root. I'm not sure what I'm doing wrong, please help!
1 Answer
- 9 years agoFavorite Answer
Hello,
R: Cl2(g) + I2(g) <-> 2ICl(g)
I: .1532M .1532M 0
C: -x -x +2x
eq: .1532-x .1532-x 2x
Kc = 1.492E5 = [ICI]^(2) / [Cl2][I2]
1.492E5 = [2x]^(2) / [.1532-x][.1532-x]
**The trick here is, since the denominator of [.1532-x][.1532-x] is the same as [.1532-x]^(2) & the numerator is squared as well, we can square root both sides (thus the "quadratic equation" is not necessary).
1.492E5 = [2x]^(2) / [.1532-x]^(2)
sqrt (1.492E5) = sqrt ([2x]^(2)) / sqrt([.1532-x]^(2))
386.3 = [2x] / [.1532-x]
59.2 - 386.3x = 2x
59.2 = 388.3x
.152 M = x
Since the concentration of ICl(g) at equilibrium is equal to 2x....
[ICl]equilibrium = 2x = 2 (.152M) = .304M
Source(s): U of H Brain
