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?
6 answers
So write the equilibrium expression and substitute the numbers from the problem to evaluate Keq.
can you check to see I did it right.
1.[C3H6O][C2H6O]2/[C7H16O2]
2.2.04E-06
1.[C3H6O][C2H6O]2/[C7H16O2]
2.2.04E-06
The Keq is not right. It's products/reactants with coefficients becoming exponents.
2 is also wrong because 1 is wrong.
2 is also wrong because 1 is wrong.
Plese check: thank u.
1.[C3H6O][C2H6O]2/[C7H16O2][H2O]2
2.8.00E-05
1.[C3H6O][C2H6O]2/[C7H16O2][H2O]2
2.8.00E-05
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.
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.
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
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
1 answer