# Spontaneous reactions

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#1
What is the difference between the 2:

For working out whether a reaction will be spontaneous or not?

Thanks!
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6 years ago
#2
Using Gibbs free energy allows you to determine spontaneity by looking only at your reaction; you don't have to worry about the surroundings. You should get the same result by either method, but not having to calculate entropy changes for the rest of the universe saves a bit of time.
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6 years ago
#3
(Original post by Zenarthra)
What is the difference between the 2:

For working out whether a reaction will be spontaneous or not?

Thanks!
The first is the equation for the entropy change of a process.
The second in the equation for the change of gibbs free energy.

A reaction is spontaneous (this is different from actually going of it's own accord. There may be a very high activation barrier to the process so it would require heating to react, but being spontaneous means that energy is overall released as G becomes more negative in a spontaneous process) if dGrxn is negative.
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#4
(Original post by JMaydom)
The first is the equation for the entropy change of a process.
The second in the equation for the change of gibbs free energy.

A reaction is spontaneous (this is different from actually going of it's own accord. There may be a very high activation barrier to the process so it would require heating to react, but being spontaneous means that energy is overall released as G becomes more negative in a spontaneous process) if dGrxn is negative.
So which 1 is best to use and why?
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#5
(Original post by BJack)
Using Gibbs free energy allows you to determine spontaneity by looking only at your reaction; you don't have to worry about the surroundings. You should get the same result by either method, but not having to calculate entropy changes for the rest of the universe saves a bit of time.
Ahh ok, thanks!
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6 years ago
#6
(Original post by Zenarthra)
So which 1 is best to use and why?
They're totally dependant upon what data you have. You need to know dG to determine if it's spontaneous or not, but you cannot do this without knowing dS
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#7
(Original post by JMaydom)
They're totally dependant upon what data you have. You need to know dG to determine if it's spontaneous or not, but you cannot do this without knowing dS
Okkk, is it correct to say that if a reaction has negative enthalpy change and positive entropy change that the activation energy is higher?
And will be spontaneous at higher temperatures?
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6 years ago
#8
(Original post by Zenarthra)
Okkk, is it correct to say that if a reaction has negative enthalpy change and positive entropy change that the activation energy is higher?
And will be spontaneous at higher temperatures?
No, dG isn't the activation energy.

There is a correlation, but it isn't a direct cause and is a 3rd year undergrad course topic, so as far as you are concerned they are unrelated.

dG is the change in gibbs free energy. I would use delta but it isn't on the keyboard and dG is actually just as applicable (it has a different but very similar meaning)
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#9
(Original post by JMaydom)
No, dG isn't the activation energy.

There is a correlation, but it isn't a direct cause and is a 3rd year undergrad course topic, so as far as you are concerned they are unrelated.

dG is the change in gibbs free energy. I would use delta but it isn't on the keyboard and dG is actually just as applicable (it has a different but very similar meaning)
Ok thanks, i drew alittle image to show what i mean.
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6 years ago
#10
(Original post by Zenarthra)
Ok thanks, i drew alittle image to show what i mean.
OK this isn't very clear to me but i can't spot any gaping holes....

You have to understand the limitations of thermodynamics(TD) (this is thermodynamics btw!)

TD is excellent for telling us the overall result of a reaction, energy released, everything really but NOT the kinetics of the reaction.... i.e the rate of the reaction. The rate is (with a few complicating factors) scales inversely with activation energy. Look up the arrhenius equation for the exact relation. The complicating factors affect the constant A in the equation.

As an example, graphite is quite a lot more stable than diamond, and TD would predict a spontaneous conversion of diamond into graphite, which is does! But it happens over the most ridiculously long timescale as the activation energy for the reaction is so high.

With your plot, you've plotted a non-spontaneous reaction unless by E you mean H and there is a massive entropy gain.
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#11
(Original post by JMaydom)
OK this isn't very clear to me but i can't spot any gaping holes....

You have to understand the limitations of thermodynamics(TD) (this is thermodynamics btw!)

TD is excellent for telling us the overall result of a reaction, energy released, everything really but NOT the kinetics of the reaction.... i.e the rate of the reaction. The rate is (with a few complicating factors) scales inversely with activation energy. Look up the arrhenius equation for the exact relation. The complicating factors affect the constant A in the equation.

As an example, graphite is quite a lot more stable than diamond, and TD would predict a spontaneous conversion of diamond into graphite, which is does! But it happens over the most ridiculously long timescale as the activation energy for the reaction is so high.

With your plot, you've plotted a non-spontaneous reaction unless by E you mean H and there is a massive entropy gain.
Yes sorry, i meant E as in enthalpy lool.
Ok but lets say dH > dS(system) and dS(system)>0
is it then true that at higher temperatures the reaction is likely to be spontaneous?
0
6 years ago
#12
(Original post by Zenarthra)
Yes sorry, i meant E as in enthalpy lool.
Ok but lets say dH > dS(system) and dS(system)>0
is it then true that at higher temperatures the reaction is likely to be spontaneous?
Well think about it, dH is almost always gonna be greater than dS given we use KJ for H and J for S.

Think about the equation, if dG has to be negative to be spontaneous, how is T going to affect dG if dS is +ve?

I think you can answer that yourself......
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#13
(Original post by JMaydom)
Well think about it, dH is almost always gonna be greater than dS given we use KJ for H and J for S.

Think about the equation, if dG has to be negative to be spontaneous, how is T going to affect dG if dS is +ve?

I think you can answer that yourself......
why is it more likely to be spontaneous at higher temp?
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6 years ago
#14
as the TdS component changes with T
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