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.
- 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.
- 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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