Homework Help: chemistry

Posted by cateye on Sunday, April 22, 2012 at 11:45pm.

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?

chemistry - DrBob222, Monday, April 23, 2012 at 12:09am
So write the equilibrium expression and substitute the numbers from the problem to evaluate Keq.
chemistry - cateye, Monday, April 23, 2012 at 12:26am
can you check to see I did it right.
1.[C3H6O][C2H6O]2/[C7H16O2]
2.2.04E-06
chemistry - DrBob222, Monday, April 23, 2012 at 12:40am
The Keq is not right. It's products/reactants with coefficients becoming exponents.
2 is also wrong because 1 is wrong.
chemistry - cateye, Monday, April 23, 2012 at 12:45am
Plese check: thank u.
1.[C3H6O][C2H6O]2/[C7H16O2][H2O]2
2.8.00E-05
chemistry - DrBob222, Monday, April 23, 2012 at 1:05am
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.
oops--chemistry - DrBob222, Monday, April 23, 2012 at 1:53am
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

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