Equilibrium Problem: Relationship Between Q, K, and ∆G

In summary, The correct answer to the question is D, as the reaction is exothermic and at 1200˚K, more reactants will be produced, causing Kp to be smaller. ∆H˚rxn=-210 kj/mol.
  • #1
patrickbotros
34
1
I just took a test today, and I want to see if I got the right answer. I sort of guessed. Here's the question:
Which statement is correct?
  • a) When Q<K, then ∆G=1
  • b) When Q<K, then ∆G=-∆S
  • c) When Q=K, then ∆G=0
  • d) When Q>K, then ∆G=1
  • e) When Q>K, then ∆G=-RT
I put C, because when Q=K, the system is at equilibrium. I don't really know what ∆G is though other than that it's Gibb's Free Energy. Please explain your answer. Thanks so much! :smile::biggrin:
 
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  • #2
Its correct . change in gibbs free energy becomes zero when equilibrium is attained
 
  • #3
proton007007 said:
Its correct . change in gibbs free energy becomes zero when equilibrium is attained
Thanks! While I've got you, would you mind answering this one:
The reaction 2H2O2(g) ⇔ 2H2O(g)+O2(g) is exothermic: ∆H˚rxn=-210 kj/mol. Which one of the following is correct:
a) Kp at 800˚K is smaller than Kp at 1,200˚K
b) Temperature does not affect Kp
c) Kp depends only on pressure
d) Kp at 1,200˚K is smaller than Kp at 800˚K
e) Kp depends on total pressure as well as temperature.
 
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  • #4
But its better to know the reason as luck and logical reasoning may not always help you . Google up Gibbs Free Energy :)
 
  • #5
proton007007 said:
But its better to know the reason as luck and logical reasoning may not always help you . Google up Gibbs Free Energy :)
While I've got you, would you mind answering this one:
The reaction 2H2O2(g) ⇔ 2H2O(g)+O2(g) is exothermic: ∆H˚rxn=-210 kj/mol. Which one of the following is correct:
a) Kp at 800˚K is smaller than Kp at 1,200˚K
b) Temperature does not affect Kp
c) Kp depends only on pressure
d) Kp at 1,200˚K is smaller than Kp at 800˚K
e) Kp depends on total pressure as well as temperature.
 
  • #6
Well , giving out the answer directly won't do you good and it would (maybe) be against the forum rules . Think about Le Chateliers principle and the temperature variation of equilibrium constant . You must have surely learned this .
 
  • #7
proton007007 said:
Well , giving out the answer directly won't do you good and it would (maybe) be against the forum rules . Think about Le Chateliers principle and the temperature variation of equilibrium constant . You must have surely learned this .
I attempted that and I'm just seeing if I was right. I said that D was the answer, because at 1200˚K, more reactants will be produced (exothermic), so Kc would be smaller.
 
  • #8
I don't see you mentioning your answer and your attempt above . Anyway , your answer is correct .
 

FAQ: Equilibrium Problem: Relationship Between Q, K, and ∆G

1. What is the definition of equilibrium in a chemical reaction?

Equilibrium in a chemical reaction refers to the state in which the rate of the forward reaction is equal to the rate of the reverse reaction, resulting in no net change in the concentrations of the reactants and products.

2. How is equilibrium represented in a chemical equation?

Equilibrium is represented by a double arrow (⇌) in a chemical equation, with the reactants on the left side and the products on the right side.

3. What is the relationship between Q and K in an equilibrium problem?

Q (reaction quotient) and K (equilibrium constant) are both measures of the relative concentrations of reactants and products at equilibrium. Q is calculated using the concentrations of reactants and products at any point in the reaction, while K is a constant value that represents the equilibrium concentrations of reactants and products at a specific temperature.

4. How is the value of ∆G related to the direction of a chemical reaction at equilibrium?

The value of ∆G (change in Gibbs free energy) is a measure of the spontaneity of a chemical reaction. At equilibrium, ∆G is equal to 0, indicating that the reaction is neither spontaneous nor non-spontaneous. If ∆G is negative, the reaction is spontaneous in the forward direction, while a positive ∆G indicates a non-spontaneous reaction in the forward direction.

5. Can the equilibrium constant K change with temperature?

Yes, the equilibrium constant K is temperature-dependent. As the temperature of a system changes, the relative concentrations of reactants and products at equilibrium also change, leading to a different value for K. However, the value of K remains constant at a specific temperature, as long as the reaction conditions (such as pressure and concentration) remain constant.

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