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3 years ago

initially, only a and b are present, each at 2.00 m. what is the final concentration of a once equilibrium is reached?

1 answer:

nataly862011 [7]3 years ago

4 0

The "REVERSIBLE" chemical reaction-
A + B <--> C + D
Kc = [C][D] / [A][B] = 6.8
Concentration at the start and after equilibrium-
[A] = 2.00 M --> (2.00 - X) M
[B] = 2.00 M --> (2.00 - X) M
[C] = 0.00 M --> X M
[D] = 0.00 M --> X M
6.8 = X^2 / (2.00 - X)^2
take the square root of both sides-
2.6077 = X / (2.00 - X)
5.2154 - 2.6077 X = X
5.2154 = 3.6077 X
X = 1.4456
at equilibrium-
[A] = 0.55 M
[B] = 0.55 M
[C] = 1.45 M
[D] = 1.45 M

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