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

C3H6O (aq) + 2C2H6O (aq)--> C7H16O2 (aq) + 2H2O (l)
pleas check to see I answer these correclty..

1.Is this an endothermic- or exothermic reaction, as written?answer: Exothermic

2.How would the equilibrium shift if the pressure of the system were doubled?answer: To the left

3.How would the equilibrium shift if C7H16O2 were added?answer: To the right

4.How would the equilibrium shift if the system were cooled? answer: No change

5.What would the Kc be for the reverse equilibrium? That is, for:C7H16O2 (aq) + 2H2O (l)---> C3H6O (aq) + 2C2H6O (aq). answer:813

6.For the reverse equilibrium, as depicted in #24, is this an endo- or exothermic process?answer:endothermic

Answer the following for me.

What's C3H6O? C2H6O? C7H16O2?
How do you know it's exothermic?

I have to use the equation above:following equilibrium system with a Kc of 1.23E-03:

C3H6O (aq) + 2C2H6O (aq)--> C7H16O2 (aq) + 2H2O (l)
to complete the problems..this is what my teacher gave me to go on..So can u please help me out.. Tell me if I am go in the right direction

There isn't enough information to answer 1.

2, 3, are not right.
I need the information for 1 to answer 4 and 6.
5 is 1/1.23E-3 = 1/0.00123 = 813

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

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

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
ok.

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
OK

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

The reactant concentrations are increasing, while the products are decreasing
With a K of 0.00123 (less than 1), the reaction at equilibrium is far to the left which means that the concentrations of the reactants are much greater than the concentration of the products at equilibrium. Note that this doesn't say anything about the RATE of the forward or the reverse reaction.

4.What is true about the rate of the forward and reverse reactions at equilibrium?
They are identical
Correct, but note that this answer contradicts the incorrect answer you gave for the previous question.

5.Is this an endothermic- or exothermic reaction, as written?
Exothermic
There isn't enough information given to answer.

6.How would the equilibrium shift if the pressure of the system were doubled?
To the left
No change

7.How would the equilibrium shift if C7H16O2 were added?
To the right
When we add a reagent to a system in equilibrium, the system tries to undo what we've done. So if we add the product, the system will try to undo the addition, which means the system will try to use up the added material. It can do that by reacting to the left. If it shifts to the right you add MORE, not less, of the product.

8.How would the equilibrium shift if the system were cooled?
No change
Not enough information to answer

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

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

813
OK

1.For the reverse equilibrium, as depicted in #9, is this an endo- or exothermic process?
endothermic
Not enough information to answer.

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1. To determine whether the reaction is endothermic or exothermic, you would need to examine the reaction's enthalpy change (∆H). If ∆H is negative, the reaction is exothermic; if ∆H is positive, the reaction is endothermic. Unfortunately, the enthalpy change (∆H) is not provided in the given information, so we cannot definitively determine if the reaction is endothermic or exothermic.

2. If the pressure of the system is doubled, it would have no effect on the equilibrium position. Changes in pressure only affect systems involving gases, where a change in pressure alters the concentration of the gaseous components. In this reaction, all the compounds involved are in liquid or aqueous states, so changing the pressure will not cause any shift in equilibrium.

3. If C7H16O2 were added to the system, the equilibrium would shift towards the right. This is because adding a reactant increases its concentration, which according to Le Chatelier's principle, causes the equilibrium to shift in the direction that consumes the excess reactant, in this case, toward the products.

4. Cooling the system does not affect the equilibrium position. Changes in temperature only impact equilibrium reactions if they are exothermic or endothermic. Since we don't know the enthalpy change (∆H) of the reaction, we cannot determine if it is exothermic or endothermic. Without this knowledge, we cannot confidently state if cooling the system would have any effect on the equilibrium.

5. To calculate the equilibrium constant (Kc) for the reverse reaction, you need to calculate the reciprocal of the original Kc. So, the Kc for the reverse equilibrium reaction would be 1/Kc (1/1.23E-03), which is approximately 813.

6. The reverse reaction will be endothermic if the original reaction is exothermic, and vice versa. Since we don't know the original reaction's enthalpy change (∆H), we cannot definitively determine if the reverse reaction is endothermic or exothermic.