Question

following equilibrium system with a Kc of 1.23E-03:

C3H6O (aq) + 2C2H6O (aq)--> C7H16O2 (aq) + 2H2O (l)
1.What is the equilibrium expression for this system?

2.If the molar concentrations of C3H6O2 and C2H6O are both 0.255 at equilibrium, what is the equilibrium concentration of C7H16O2?

Answer 1

So write the equilibrium expression and substitute the numbers from the problem to evaluate Keq.

Answer 2

can you check to see I did it right.
1.[C3H6O][C2H6O]2/[C7H16O2]
2.2.04E-06

Answer 3

The Keq is not right. It's products/reactants with coefficients becoming exponents.
2 is also wrong because 1 is wrong.

Answer 4

Plese check: thank u.
1.[C3H6O][C2H6O]2/[C7H16O2][H2O]2
2.8.00E-05

Answer 5

You wrote it the same way. Its products/reactants. Products are on the right. Reactants are on the left.
Keq = 1.23E-3 = [C7H16O2]/[C3H6O][C2H6O]^2
and I omitted the water. Pure solids and pure liquids are not included.
8E-5 is correct.

Answer 6

The expression I wrote is correct.
The answer of #2 of 8E-5 is not right. Neither you nor I squared the the term.
1.23E-3 = x/(0.255)(0.255)^2 = and x = 2.04E-5