Just made it a new post didn't what it to get overcrowning.I hope this will help us figure out now

For questions #1 through #10, refer to the 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?
[C7H16O2]/ [C3H6O][C2H6O]2

2.If the molar concentrations of C3H6O2 and C2H6O are both 0.255 at equilibrium, what is the equilibrium concentration of C7H16O2?
2.04E-06

3.What is true about the concentrations of the reactants and products at equilibrium?

The reactant concentrations are increasing, while the products are decreasing

4.What is true about the rate of the forward and reverse reactions at equilibrium?
They are identical

5.Is this an endothermic- or exothermic reaction, as written?
Exothermic

6.How would the equilibrium shift if the pressure of the system were doubled?
To the left

7.How would the equilibrium shift if C7H16O2 were added?
To the right

8.How would the equilibrium shift if the system were cooled?
No change

9.What would the Kc be for the reverse equilibrium? That is, for:

C7H16O2 (aq) + 2H2O (l)---> C3H6O (aq) + 2C2H6O (aq)

813

10.For the reverse equilibrium, as depicted in #9, is this an endo- or exothermic process?
endothermic

I answered all of this (that I could) at the original post site below.

To find the answers to the questions, you can use the given equilibrium expression and apply the principles of Le Chatelier’s principle and the relationship between Kc values for forward and reverse reactions.

1. The equilibrium expression for this system is [C7H16O2]/([C3H6O][C2H6O]2).

2. To find the equilibrium concentration of C7H16O2, you can use the equilibrium expression:

[C7H16O2]/([C3H6O][C2H6O]2) = 1.23E-03

Let the equilibrium concentration of C7H16O2 be x. Then the equilibrium molar concentrations of C3H6O and C2H6O will be 0.255. Substitute these values into the equilibrium expression:

x / (0.255 * 0.255^2) = 1.23E-03

Solve for x to find the equilibrium concentration of C7H16O2, which is 2.04E-06.

3. At equilibrium, the concentrations of the reactants and products are constant, and the rate of the forward and reverse reactions are equal. Therefore, the reactant concentrations are not increasing, and the product concentrations are not decreasing.

4. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. This means that the forward and reverse reactions are occurring at the same rate.

5. From the equation and the information given, you can determine the reaction is exothermic because heat is released when the product is formed.

6. According to Le Chatelier’s principle, if the pressure of the system is doubled, the equilibrium will shift to the side with fewer gas molecules to relieve the pressure. In this case, the left side has fewer gas molecules (1) compared to the right side (3), so the equilibrium will shift to the left.

7. If C7H16O2 is added, the equilibrium will shift to counteract the change. According to Le Chatelier's principle, if the concentration of a reactant is increased, the equilibrium will shift to the right to consume the added substance. Therefore, adding C7H16O2 will cause the equilibrium to shift to the right.

8. Cooling the system does not affect the equilibrium position. Temperature changes affect the value of the equilibrium constants (Kc), not the equilibrium position itself. Therefore, the equilibrium will not shift.

9. To find the Kc for the reverse equilibrium, you can use the relationship between Kc values for forward and reverse reactions. If the forward reaction has a Kc value of 1.23E-03, then the reverse reaction will have the reciprocal Kc value:

Kc(reverse) = 1 / Kc(forward)
Kc(reverse) = 1 / 1.23E-03
Kc(reverse) = 813

Thus, the Kc for the reverse equilibrium is 813.

10. Since the reverse reaction is the reaction written in the opposite direction, and it is the reverse of an exothermic reaction, it will be endothermic.