Question

A positive ∆H and positive ∆S __________.
Answer

will never produce a spontaneous reaction

will always produce a spontaneous reaction

will only produce a spontaneous reaction if the increase in entropy times temperature is greater than the increase in enthalpy

will only produce a spontaneous reaction if the increase in enthalpy is greater than - T times the increase in entropy

Answer 1

\[\Large \Delta G = \Delta H - T\Delta S \\ ~ \\
\large \Delta G >0:\text{non-spontaneous}\\
\large \Delta G<0: \text{spontaneous}\]
if \(\Delta H>0\) and \(\Delta S>0\) , then the sign of \(\Delta G\) depends on \(T\)
To see this, just imagine an easy situation and say \(\Delta H = 2\), \(\Delta S=1\), then

1) If \(T=2, \Delta G = 2-2(1)=0\implies \text{equilibrium}\)
2) If \(T = 3, \Delta G = 2-3(1)=-1 \implies \text{spontaneous}\)
3) If \(T=1, \Delta G = 2-1(1)=1\implies \text{non-spontaneous}\)

Also, you can analyze the question in this way:
\[ \Large \Delta G = \Delta H - T\Delta S \] If \(\Delta H = T\Delta S\), then \(\Delta G = 0\)
Now what happens if \(\Delta H > T \Delta S\)?
If \(\Delta H < T\Delta S\)?

Answer 2

@aaronq

Answer 3

Do you understand the first part of my analysis where I have the example with \(ΔH=2,ΔS=1\) and thus having both of these quantities positive means that you can get either an equilibrium, spontaneous or non-spontaneous reaction depending on the value of \(T \)?

Answer 4

i think i follow

Answer 5

The other part is checking to see if you will get a negative or positive value for \(\Delta G \) depending on whether \(\Delta H\) is larger or smaller than \(T \Delta S\). Maybe it's easier if we just pretend that we have the equation:
\[ G = x - y\] where \(\Delta G = G\), \(x = \Delta H\) and \(y = T \Delta S\),maybe the analysis will look easier.

If you have \(x = y\), then \(G = x-y \) is the same as \(G = x - x = 0\)

Now if \(x > y \), then you can think of say \(x = 3, y =2\). So \(G = x - y = 3 - 2 = 1 > 0\)... non-spontaneous.

Now if \(x < y\), then you can think of say \(x = 2, y = 3\). So \(G = x - y = 2 - 3 = -1 < 0 \).. spontaneous.

Answer 6

so what would be the answer to this since we dont have the numerical values to worj into the problem

Answer 7

I used numerical values just to give examples to help understand what is going on. But based on my first analysis, you should see whether the 1st two statements
(will never produce a spontaneous reaction
will always produce a spontaneous reaction)
are true or not very easily.

Answer 8

The last two statements look whether \(\Delta H \) is larger or smaller than \(T \Delta S\) and how that affects the spontaneity of the reaction.

Answer 9

ok but i still dont to how to find all that out

Answer 10

well it's an analysis on determining the sign of the free energy. Which part do you not understand more specifically?

Answer 11

i understand how you are analyzing this but it seems a lot to assume when the given question is seemingly blank. how do i reach the answer and what might it be?

Answer 12

Well you just use whatever you can to solve a question, even if it looks "blank". I mean the first step would be to imagine what does a positive quantity of H and S do in the equation? It looks like it would depend on the value of T, because there is no guarantee of getting a positive of negative G but just having a positive T or S. And I used an example to support that answer

Now, Given how the last two statements are given in your question, it looks like they are interested in comparing H and TS. And the statements say like "the reaction is spontaneous if TS is larger than H" , so that's why I analyzed H and TS in that way

Answer 13

but if there is no guarantee of getting a positive or negative G than the first two answers would not be correct right? leaving the bottom two?

Answer 14

exactly.

Answer 15

so now how do we decide between witch of them is the right one?

Answer 16

Try translating the choices in "symbols". The 3rd statement is basically saying:
The reaction will be \(\Delta G < 0\) (will be spontaneous) if \(T \Delta S > \Delta H \) (if the increase in entropy times temperature is greater than the increase in enthalpy)

Answer 17

so the forth answer is correct?

Answer 18

euh no

Answer 19

but the third option seems to not work out right? your saying its the third option?

Answer 20

The 3rd statement is equivalent to when I gave the example: Now if \(x<y\), then you can think of say \(x=2,y=3\). So \(G=x−y=2−3=−1<0\).. spontaneous.
Since \(x = \Delta H \), \(y = T\Delta S\)

Answer 21

oh wow now i get it, i feel dumb now.

Answer 22

:) don't worry.

Answer 23

so the equation listed above will only produce a spontaneous reaction if the increase in entropy times temperature is greater than the increase in enthalpy

Answer 24

yes