Can this question be solved using Le Chatlier's principle

[phattestlewt]

This is reaction 2

C is the correct answer.


Here's my reasoning.

R --> P
Because the enthalpy of the products is less than that of the reactants, the reaction is exothermic as written!
So, it releases heat energy.
R --> P + E

Looking at the question now, increasing the temperature, would shift the equilibrium to the left.

I understand how the equation G = H -TS can be used to figure out whether G increases/decreases, but how do you use le chatlier's principle to figure out the K/Q ratio?

Isn't there a pneumonic that goes when K < Q, the equilibrium shifts to the < (left)
and when K > Q, the equilibrium shifts to the > (right)?
Why doesn't the K/Q ratio decrease...Aghhh im confusing myself, someone please help!

Also whats the diff between G and G with the degree sign!

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[sazerac]

Don't confuse enthalpy (H) with energy (G). Reactions are driven forward by negative delta G.

Since they told you that dS is positive, you know that increasing the temperature will make dG even more negative, driving the reaction further to the right. The enthalpy graph is just a huge distractor for this question.

Since a rising temperature will create more products, the K will increase, so the K/Q ratio will increase too.

 

[phattestlewt]

AHH, good explanation!

So this question simply required using G = H - TS

 

[Tassatul]

Also using dG = -RTlnK. So since we have already established that dG is decreasing as T increases (due to dG = dH - TdS), we know that a decreasing dG leads to an increase in K because of the negative sign in dG = -RTlnK.

 

[phattestlewt]

What's the difference between dG and dG with a degree sign?

 

[Tassatul]

dG refers to any delta Gibbs Free Energy, at any temperature and pressure, while dG with a degrees sign refers to dG at either STP or 298K and 1 atm I can't remember which. I think the latter but youd better look it up

 

[phattestlewt]

Yes its 298K, thanks man!

 

[shinsoochoo]

To remember the pnemonic, just know that K comes before Q in the alphabet. Every time you do a problem, immediately write K __ Q. That way you can say that when K < Q, the equilibrium shifts to the < (left) and when K > Q, the equilibrium shifts to the > (right).

 

[Blueprint MCAT Tutor]

The comments above have done a great job explaining the problem, but the wording in the answer choices is kind of funky.

Using the reasoning above (if ΔS is positive then when it increases, then ΔG decreases AND according to the equation ΔG = -RTlnKeq then when ΔG decreases Keq increases) you can make a statement about ΔG and Keq.

However, the question speaks in terms of the Keq-to-Q ratio and we never really know what the Q is here. Since Q is just the proportions of reactants and products at any given time, we can't know what that number is unless we're given some info about it. So when you're looking through the choices, just focus on the part you understand.

As for the bit that's confusing: "Hey but it's exothermic! Shouldn't increasing T push the reaction back towards reactants, thus decreasing Keq?!?"

Well, yes, sort of. Remember that whether the reaction moves forward or backward depends centrally on ΔG, and ΔG accounts for BOTH ΔH and ΔS.

A lot of times we can afford pretty sloppy reasoning and just focus on ΔH (whether it's exo or endo thermic) - the MCAT will let us get away with that a huge proportion of the time, simply because the values associated with ΔH are so much more than ΔS. But once the problem *itself* brings ΔS into the picture, then you can NOT just fall back on the reasoning that "any time it's exothermic, an increase in T will push the reaction backwards".