A2 Chemistry Entropy Question Help

Helooo,

I'm not sure what's going on with the question, it is true that for a spontaneous reaction
deltaS total > 0

but next to it as deltaH of the reaction is negative why would you need
deltaSsystem to be >0? Surely total entropy is positive anyway and it doesnt matter

entropy.jpg201.7KB

Reply 1

Kevin De Bruyne

I suspect that it's because of the inequality sign.

The line after 'For a feasible reaction...' if you divide both sides by the negative value deltaS then the inequality sign will flip round, and then you multiply by T to give T< -2..../3.31. If you specify that it's positive then you won't have this problem.

I hope that's right.

A simpler example would be that 5>3. If you divide by -1 you wouldn't get -5>-3 but -5<-3.

(edited 9 years ago)

Reply 2

madmenace

Original post by SeanFM

Hi, thanks alot for the reply.. I understand completely that when you divide an inequality by a negative number the sign flips but I'm still unsure about this question and why deltaSsystem > deltaH/T if because as the enthalpy change is negative it will come out positive so will not lower the entropy and instead increase it... I have reread your explanation but don't yet seem to understand it :-(:-(

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Reply 3

Kevin De Bruyne

Original post by madmenace

I'm with you up until the bold bit but not quite sure what you mean after.

I'm out of my depth here so would appreciate other input too :colondollar:

Reply 4

madmenace

Original post by SeanFM

I'm with you up until the bold bit but not quite sure what you mean after.

I'm out of my depth here so would appreciate other input too :colondollar:

the inequaltiy part is fine, for a reaction to be feasible deltaS total has to be greater than 0. deltaS total = deltsSsystem +deltaSsurroundings.

delta Ssystem is positive (the bit of working out at the top)...

now, deltaSsurroundings is given be -(deltaH)/T... so if delta H is +ve, the ASSurroundings will be negative and so ASSyestem will need to be greater and outweigh the negative one(which is what the question is getting at)

but in the case of the question deltaH is -ve already so the -(deltaH) term will be +ve... meaning so why would you need ASSyestem to be grater than surroundings if they are both positive anyway?

thanks in advance hope ive explained it better

Reply 5

Kevin De Bruyne

Original post by madmenace

Thanks, I understand what you're asking now. I'm not too sure. As deltaSsurroundings isn't mentioned at all (though you can see it from total and system) I suppose they just stated deltaStotal to be >0 because it can't be negative anyway?

I saw a question similar to this in A2 AQA, but all they needed us to state was about the activation energy, and never ever mentioned deltaSsurroundings. Let's hope someone with more knowledge posts or maybe see a teacher about it. Sorry that I wasn't of much help :s-smilie:

Reply 6

madmenace

Original post by SeanFM

thanks alot, I tried asking my teacher and he just waffled... he still hasnt given me a proper answer.... haha hope so.... been stuck on this for like 2 weeks :angry:

Reply 7

charco

Study Forum Helper

Original post by madmenace

thanks alot, I tried asking my teacher and he just waffled... he still hasnt given me a proper answer.... haha hope so.... been stuck on this for like 2 weeks :angry:

The entropy of the Universe must increase if the reaction is to be spontaneous, i.e. ΔS is positive.

ΔS_universe = ΔS_system + ΔS_surroundings

and as any increase in surroundings entropy is caused by a release of energy into the surroundings:

ΔS_surroundings = -ΔH_system/T

This makes sense as an exothermic reaction gives rise to an increase in thermal energy and hence an increase in entropy of the surroundings.

From the above equations:

ΔS_universe = ΔS_system - ΔH_system/T

So if the ΔS_universe is to be positive then:

ΔS_system > ΔH_system/T

(edited 9 years ago)

Reply 8

madmenace

Original post by charco

wahaaay so thats what it meant.... :eek:

.... to me that only makes sense for when deltaH is positive as -deltaH/T will be negative...... but when the deltaH is negative (as in the example)
ΔS_system - (-ΔH_system)/T would this not be positive???

so you dont need

ΔS_system > ΔH_system/T

(edited 9 years ago)

Reply 9

langlitz

Just like to add something to what's been said:

dS_{universe}=dS_{system}+dS_{surroundings}

dS_{sys}=dq_{sys}/T

and

dS_{surr}=dq_{surr}/T

For a spontaneous change:

|dq_{surr}|<|dq_{system}|

dS_{universe}=dS_{sys}+dS_{surr}=\frac{dq_{sys}+dq_{surr}}{T} \geq 0

\Delta S_{universe}\geq 0

=> Second law: For a spontaneous process the entropy of the universe increases.

dS_{universe}=dS_{system}-\frac{dq}{T}

and

\frac{dq}{T}=\frac{\Delta H}{T}

\Delta S_{universe}=\Delta S_{system}-\frac{\Delta H_{system}}{T}

T\Delta S_{universe}=T\Delta S_{system}-\Delta H_{system}

-T\Delta S_{universe}=\Delta H_{system}-T\Delta S_{system}

This

-T \Delta S_{universe}=\Delta G

, which is defined as the Gibb's Free Energy

(edited 9 years ago)

Reply 10

charco

Study Forum Helper

Original post by madmenace

wahaaay so thats what it meant.... :eek:

You are confusion the negative sign of ΔH.

The fact that it is exothermic (i.e. negative) means that it increases the entropy of the surroundings and gives the equation:

ΔS_universe = ΔS_system - ΔH_system/T