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

  • #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:
 

Answers and Replies

  • #2
Its correct . change in gibbs free energy becomes zero when equilibrium is attained
 
  • #3
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.
 
Last edited:
  • #4
But its better to know the reason as luck and logical reasoning may not always help you . Google up Gibbs Free Energy :)
 
  • #5
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 wont 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 learnt this .
 
  • #7
Well , giving out the answer directly wont 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 learnt 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 dont see you mentioning your answer and your attempt above . Anyway , your answer is correct .
 

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