Van't Hoff Temperature Dependence
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Van't Hoff Temperature Dependence
Can someone explain to me why we are taking the natural log of Temperature in Van't Hoff's equation? I'm confused as to what the equation also is trying to convey also because I was confused about the in class example. Thank you!
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Re: Van't Hoff Temperature Dependence
He mentioned in lecture that you apply this equation to calculate K at different temperatures if deltaH is given.
He broke down the equation to the ln(K2/K1) = -deltaH/R (1/T2 - 1/T1) with the assumption that deltaH is constant.
He broke down the equation to the ln(K2/K1) = -deltaH/R (1/T2 - 1/T1) with the assumption that deltaH is constant.
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Re: Van't Hoff Temperature Dependence
Andrew F 2L wrote:Can someone explain to me why we are taking the natural log of Temperature in Van't Hoff's equation? I'm confused as to what the equation also is trying to convey also because I was confused about the in class example. Thank you!
The reason you take ln(K) is because you are assuming that the reaction is at equilibrium, meaning delta G is equal to zero.
You have the two equations ∆G= ∆Gnot +RTln(Q) and ∆Gnot=∆Hnot-T∆S
At equilibrium, Q=K and ∆G=0
You can then solve for ∆Gnot and set the two equations equal to each other. This is what gives you the ln(k) in the equation. I hope that this helps!
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Re: Van't Hoff Temperature Dependence
You have the natural log in the equation already given ∆G=∆G°+RTlnQ, and then ∆G°=-RTlnK at equilibrium(∆G=0 at equilibrium)
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Re: Van't Hoff Temperature Dependence
What equation do you use if you are not sure if the equation has a
equal to 0?
Is there a way to check that
is 0 if it is not explicitly stated in the problem?
What happens if you use the equation for a reaction that is not at equilibrium?
Is there a way to check that
What happens if you use the equation for a reaction that is not at equilibrium?
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