# Change in Entropy and Temperature?

Discussion in 'MCAT Study Question Q&A' started by manohman, 09.27.14.

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1. ### manohman 2+ Year Member

Joined:
07.27.14
Messages:
109
Can someone explain how the change in entropy of a reaction decreases with Increasing temperature?

dS = q/T

I understand how entropy itself increases with increasing temperature but how does the change in entropy decrease with increasing temperature?

If we consider a chemical reaction A --> B for which entropy increaes from reactants to products, increasing temperature would increase the temperature for both sides, but does it increase the entropy of the lower entropy side more than it does for the higher entropy side?

Also shouldnt there be a dependence on whether the reaction is endothermic or exothermic? If its exothermic, then the increase in temperature would push the reaction to the reactants. Woudlnt' this affect the change in entropy?

Thanks guys relaly confused here exam krackers is killing me

3. ### BobRSanchez

Joined:
09.23.14
Messages:
6
Status:
Pre-Medical
Your equation is actually incomplete. This equation, dS=dq/T describes the second law of thermodynamics. It basically states that for a reversible ( key word) reaction, the change in entropy is the exchange of heat divided by the final equilibrium temperature. Raising T doesn't necessarily decrease dS because dq is also temperature dependent.

With that said, I don't think you need worry yourself too much about the logic behind the equation, just know that the change in entropy of a reversible (and irreversible) reaction can be found with that equation.

4. ### manohman 2+ Year Member

Joined:
07.27.14
Messages:
109
okay thank you!

Could you clarify what you mean by reversible and irreversible reactions with regard to entropy? Iv seen it around but it still confuses me a bit.