Well I got what the formula for gibbs free energy is. but what's the nature of this energy and why is it called 'free'? It does free work is what textbooks say but didn't get the intuitive feel.

The word "free" is not a very good one! In fact, IUPAC recommend calling it Gibbs energy or the Gibbs function, although most chemists still refer to it as Gibbs free energy. Gibbs originally called it _available energy_ and that is a good term because it is the energy associated with a chemical reaction that is available (or you could say _free_) to do work, assuming constant T and P.

Is there a difference between the notation ΔG and the notation ΔG˚, and if so, what is it?

STP is not standard conditions. Standard conditions are 1.0 M solutions and gases at 1.0 atm. Standard conditions does not actually specify a temperature but almost all thermodynamic data is given at 25C (298K) so many people assume this temperature.

In the subject heading, 'When is ΔG is negative?', is it a typo that it says 'When the process is endothermic, ΔHsystem > 0, and the entropy of the system decreases, ΔSsystem>0, the sign of ΔG is positive at all temperatures. Thus, the process is never spontaneous' shouldn't the entropy be < 0? if there is a decrease in entropy?

I think you are correct. Change in entropy must be smaller than zero, for the entropy to decrease. It is a typo. You can cross-check from the figure.

The Entropy change is given by Enthalpy change divided by the Temperature. Then how can the entropy change for a reaction be positive if the enthalpy change is negative?

Great question! Figuring out the answer has helped me learn this material. One way to define entropy is `Q/T` (where `Q` is the heat associated with a *reversible* process {also note that `∆H` is only equal to `Q` when `P` is constant}). For spontaneous and thus *irreversible* reactions, the `∆S` is the same as for a *reversible* reaction (because `S` is a state variable it doesn't depend on how we got from one condition to another). In contrast, `∆H` is not equal to `Q` (because this is not a *reversible* reaction). A later video helps explain this: https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/more-rigorous-gibbs-free-energy-spontaneity-relationship

Hey I´m stuck: The ∆G in a reaction is negative but the ∆H was positive and it is assumed that a change temperature doesn´ t significantly affect entropy and entalpy. What does this do to 1) spontanity 2) spontanity at high temp 3) value or sign of ∆S i read it 3 times now but i´m still insecure - :(

This looks like a homework question, so I'll give you some hints to get you on the riht path rather than answering directly. 3) We know that ∆G = ∆H - T∆S. Solving for ∆S, we have: ∆S=(∆H-∆G)/T. We know (from the question) that ∆G is negative and that ∆H is positive. Temperature is always positive (in Kelvin). From these values, we can know for certain whether ∆S is positive or negative (hint: remember that we are subtracting ∆G!). 1) Knowing the sign of ∆G is enough to say whether the reaction is spontaneous or not under these conditions. If ∆G is negative (from the question), is the reaction spontaneous or non-spontaneous? 2) Let's use ∆G = ∆H - T∆S again. Since ∆H and ∆S don't change significantly with temperature (given in the question), we can assume that they keep the same signs and values: i.e. ∆H is still positive and ∆S is still whatever sign you figured out above. As T increases, the T∆S component gets bigger. T is always positive, so if ∆S is positive then a bigger T∆S will make ∆G more negative (since we subtract T∆S). If ∆S is negative, then the negative signs (from the subtraction and the sign of ∆S) will cancel out, and so as T∆S gets bigger, ∆G will get more positive.

how do i see the sign of entropy when both reactant and product have the same phase

We have to look up the ΔS for the whole reaction in a table (or test the reaction ourselves... I'd rather look it up!). The value will be either positive or negative. If the reaction can result in a phase change then we might be lucky enough to find a list that has the reaction with reactant and products in the phases we need. Otherwise we could calculate the change in energy and the use the specific heat equations to see if the phase would change. The example above with melting ice looks a little different because the reaction was a phase change (ice to water) instead of the usual combining or splitting of molecules.

can an exothermic reaction be a not spontaneous reaction ?

Sure. Paper doesn't light itself on fire, right?

Is the reaction H2O(l) to H20(s) spontaneous or non spontaneous? The entropy, S, is positive when something goes from a solid to liquid, or liquid to gas, which is increasing in disorder. However, in this equation, water is going from a liquid to solid, so S is negative, and in the Gibbs free reaction equation, S must be positive for a reaction to be spontaneous. How do we determine, without any calculations, the spontaneity of the equation?

Using that grid from above, if it's an exothermic reaction (water is releasing heat into its surroundings in order to turn into ice), we know it's on the left column. The entropy of liquid water is higher than ice (water as a solid state)so therefore it is not always going to be spontaneous. Putting into the equation, ΔH<0 because it's exothermic, and ΔS<0 because entropy is decreased. Therefore, the reaction is only spontaneous at low temperatures (TΔS). If you think about its real-world application, it makes sense. Liquid water will turn into ice at low enough temperatures.

Main content

Course: Chemistry library > Unit 13

Lesson 3: Gibbs free energy

Gibbs free energy and spontaneity

Q: In the subject heading, 'When is ΔG is negative?', is it a typo that it says 'When the process is endothermic, ΔHsystem > 0, and the entropy of the system decreases, ΔSsystem>0, the sign of ΔG is positive at all temperatures. Thus, the process is never spontaneous' shouldn't the entropy be < 0? if there is a decrease in entropy?

I think you are correct. Change in entropy must be smaller than zero, for the entropy to decrease. It is a typo. You can cross-check from the figure.

Q: The Entropy change is given by Enthalpy change divided by the Temperature. Then how can the entropy change for a reaction be positive if the enthalpy change is negative?

Great question! Figuring out the answer has helped me learn this material. One way to define entropy is `Q/T` (where `Q` is the heat associated with a *reversible* process {also note that `∆H` is only equal to `Q` when `P` is constant}). For spontaneous and thus *irreversible* reactions, the `∆S` is the same as for a *reversible* reaction (because `S` is a state variable it doesn't depend on how we got from one condition to another). In contrast, `∆H` is not equal to `Q` (because this is not a *reversible* reaction). A later video helps explain this: https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/more-rigorous-gibbs-free-energy-spontaneity-relationship

Q: Hi all, Sal sir said we would prefer the reaction to proceed in a particular direction (the direction that makes our product!), but it's hard to argue with a positive ΔG! ( located before summary at other applications of del G) .can anybody please explain?

If ΔG is positive, then the only possible option is to vary the temperature but whether that would work depends on whether the reaction is exo- or endothermic and what the entropy change is.

Q: how do i see the sign of entropy when both reactant and product have the same phase

We have to look up the ΔS for the whole reaction in a table (or test the reaction ourselves... I'd rather look it up!). The value will be either positive or negative. If the reaction can result in a phase change then we might be lucky enough to find a list that has the reaction with reactant and products in the phases we need. Otherwise we could calculate the change in energy and the use the specific heat equations to see if the phase would change. The example above with melting ice looks a little different because the reaction was a phase change (ice to water) instead of the usual combining or splitting of molecules.

Q: can an exothermic reaction be a not spontaneous reaction ?

Sure. Paper doesn't light itself on fire, right?

Q: Is the reaction H2O(l) to H20(s) spontaneous or non spontaneous? The entropy, S, is positive when something goes from a solid to liquid, or liquid to gas, which is increasing in disorder. However, in this equation, water is going from a liquid to solid, so S is negative, and in the Gibbs free reaction equation, S must be positive for a reaction to be spontaneous. How do we determine, without any calculations, the spontaneity of the equation?

Using that grid from above, if it's an exothermic reaction (water is releasing heat into its surroundings in order to turn into ice), we know it's on the left column. The entropy of liquid water is higher than ice (water as a solid state)so therefore it is not always going to be spontaneous. Putting into the equation, ΔH<0 because it's exothermic, and ΔS<0 because entropy is decreased. Therefore, the reaction is only spontaneous at low temperatures (TΔS). If you think about its real-world application, it makes sense. Liquid water will turn into ice at low enough temperatures.

Google Classroom

How the second law of thermodynamics helps us determine whether a process will be spontaneous, and using changes in Gibbs free energy to predict whether a reaction will be spontaneous in the forward or reverse direction (or whether it is at equilibrium!).

Key points

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: $Δ S_{universe} = Δ S_{system} + Δ S_{surroundings} > 0$ ‍
At constant temperature and pressure, the change in Gibbs free energy is defined as $Δ G = Δ H - T Δ S$ ‍.
When $Δ G$ ‍ is negative, a process will proceed spontaneously and is referred to as exergonic.
The spontaneity of a process can depend on the temperature.

Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:

$C (s, diamond) \to C (s, graphite)$ ‍

This reaction takes so long that it is not detectable on the timescale of (ordinary) humans, hence the saying, "diamonds are forever." If we could wait long enough, we should be able to see carbon in the diamond form turn into the more stable but less shiny, graphite form.

Another thing to remember is that spontaneous processes can be exothermic or endothermic. That is another way of saying that spontaneity is not necessarily related to the enthalpy change of a process,

Δ H

How do we know if a process will occur spontaneously? The short but slightly complicated answer is that we can use the second law of thermodynamics. According to the second law of thermodynamics, any spontaneous process must increase the entropy in the universe. This can be expressed mathematically as follows:

$Δ S_{universe} = Δ S_{system} + Δ S_{surroundings} > 0 For a spontaneous process$ ‍

Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be a chemical reaction in a beaker. Do we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?)

Luckily, chemists can get around having to determine the entropy change of the universe by defining and using a new thermodynamic quantity called Gibbs free energy.

Gibbs free energy and spontaneity

When a process occurs at constant temperature

T

and pressure

P

, we can rearrange the second law of thermodynamics and define a new quantity known as Gibbs free energy:

$Gibbs free energy = G = H - TS$ ‍

where

H

is enthalpy,

T

is temperature (in kelvin,

K

), and

S

is the entropy. Gibbs free energy is represented using the symbol

G

and typically has units of

\frac{kJ}{mol-rxn}

A "mole of reaction," which is abbreviated as

mol-rxn

mol-reaction

, is defined as occurring when the number of moles given by the coefficients in your balanced equation react. That definition can sound rather confusing, but the idea is hopefully more clear in the context of an example. If we had the following balanced reaction:

$2 Al (s) + 3 {Cl}_{2} (g) \to 2 {AlCl}_{3} (s)$ ‍

we would say that

1

mole of reaction is when

2

moles of

Al

react with

3

moles of

{Cl}_{2}

to produce

2

moles of

{AlCl}_{3}

. We can write these relationships mathematically as well:

$1 mol-rxn = 2 mol Al = 3 {mol Cl}_{2} = 2 {mol AlCl}_{3}$ ‍

When using Gibbs free energy to determine the spontaneity of a process, we are only concerned with changes in

G

, rather than its absolute value. The change in Gibbs free energy for a process is thus written as

Δ G

, which is the difference between

G_{final}

, the Gibbs free energy of the products, and

G_{initial}

, the Gibbs free energy of the reactants.

$Δ G = G_{final} - G_{initial}$ ‍

For a process at constant

T

and constant

P

, we can rewrite the equation for Gibbs free energy in terms of changes in the enthalpy (

Δ H_{system}

) and entropy (

Δ S_{system}

) for our system:

$Δ G_{system} = Δ H_{system} - T Δ S_{system}$ ‍

You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for

Δ H

and

Δ S

are for the system of interest. This equation is exciting because it allows us to determine the change in Gibbs free energy using the enthalpy change,

Δ H

, and the entropy change ,

Δ S

, of the system. We can use the sign of

Δ G

to figure out whether a reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium.

When $Δ G < 0$ ‍, the process is exergonic and will proceed spontaneously in the forward direction to form more products.
When $Δ G > 0$ ‍, the process is endergonic and not spontaneous in the forward direction. Instead, it will proceed spontaneously in the reverse direction to make more starting materials.
When $Δ G = 0$ ‍, the system is in equilibrium and the concentrations of the products and reactants will remain constant.

Calculating change in Gibbs free energy

Although

Δ G

is temperature dependent, it's generally okay to assume that the

Δ H

and

Δ S

values are independent of temperature as long as the reaction does not involve a phase change. That means that if we know

Δ H

and

Δ S

, we can use those values to calculate

Δ G

at any temperature. We won't be talking in detail about how to calculate

Δ H

and

Δ S

in this article, but there are many methods to calculate those values including:

Estimating $Δ H_{reaction}$ ‍ using bond enthalpies
Calculating $Δ H$ ‍ using standard heats of formation, $Δ_{f} H^{\circ}$ ‍
Calculating $Δ H$ ‍ and $Δ S$ ‍ using tables of standard values

When the process occurs under standard conditions (all gases at

1 bar

pressure, all concentrations are

1 M

, and

T = 25^{\circ} C

), we can also calculate

Δ G

using the standard free energy of formation,

Δ_{f} G^{\circ}

Problem-solving tip: It is important to pay extra close attention to units when calculating

Δ G

from

Δ H

and

Δ S

! Although

Δ H

is usually given in

\frac{kJ}{mol-reaction}

Δ S

is most often reported in

\frac{J}{mol-reaction \cdot K}

. The difference is a factor of

1000

When is $Δ G$ ‍ negative?

If we look at our equation in greater detail, we see that

Δ G_{system}

depends on

3

values:

Δ G_{system} = Δ H_{system} - T Δ S_{system}

the change in enthalpy $Δ H_{system}$ ‍
the temperature $T$ ‍
the change in entropy $Δ S_{system}$ ‍

Temperature in this equation always positive (or zero) because it has units of

K

. Therefore, the second term in our equation,

T Δ S_{system}

, will always have the same sign as

Δ S_{system}

. We can make the following conclusions about when processes will have a negative

Δ G_{system}

When the process is exothermic ( $Δ H_{system} < 0$ ‍), and the entropy of the system increases ( $Δ S_{system} > 0$ ‍), the sign of $Δ G_{system}$ ‍ is negative at all temperatures. Thus, the process is always spontaneous.
When the process is endothermic, $Δ H_{system} > 0$ ‍, and the entropy of the system decreases, $Δ S_{system} < 0$ ‍, the sign of $Δ G$ ‍ is positive at all temperatures. Thus, the process is never spontaneous.

For other combinations of

Δ H_{system}

and

Δ S_{system}

, the spontaneity of a process depends on the temperature.

Exothermic reactions ( $Δ H_{system} < 0$ ‍) that decrease the entropy of the system ( $Δ S_{system} < 0$ ‍) are spontaneous at low temperatures.
Endothermic reactions ( $Δ H_{system} > 0$ ‍) that increase the entropy of the system ( $Δ S_{system} > 0$ ‍) are spontaneous at high temperatures.

Can you think of any reactions in your day-to-day life that are spontaneous at certain temperatures but not at others?

Example $1$ ‍: Calculating $Δ G$ ‍ for melting ice

Let's consider an example that looks at the effect of temperature on the spontaneity of a process. The enthalpy of fusion and entropy of fusion for water have the following values:

$Δ_{fus} H = 6.01 \frac{kJ}{mol-rxn}$ ‍

$Δ_{fus} S = 22.0 \frac{J}{mol-rxn \cdot K}$ ‍

What is $Δ G$ ‍ for the melting of ice at $20^{\circ} C$ ‍?

The process we are considering is water changing phase from solid to liquid:

$H_{2} O (s) \to H_{2} O (l)$ ‍

For this problem, we can use the following equation to calculate

Δ G_{rxn}

$Δ G = Δ H - T Δ S$ ‍

Luckily, we already know

Δ H

and

Δ S

for this process! We just need to check our units, which means making sure that entropy and enthalpy have the same energy units, and converting the temperature to Kelvin:

$T = 20^{\circ} C + 273 = 293 K$ ‍

If we plug the values for

Δ H

T

, and

Δ S

into our equation, we get:

\begin{aligned} Δ G & = Δ H - T Δ S \\ = 6.01 \frac{kJ}{mol-rxn} - (293 K) (0.022 \frac{kJ}{mol-rxn \cdot K)} \\ = 6.01 \frac{kJ}{mol-rxn} - 6.45 \frac{kJ}{mol-rxn} \\ = - 0.44 \frac{kJ}{mol-rxn} \end{aligned}

Since

Δ G

is negative, we would predict that ice spontaneously melts at

20^{\circ} C

. If you aren't convinced that result makes sense, you should go test it out!

Concept check: What is $Δ G$ ‍ for the melting of ice at $- 10^{\circ} C$ ‍?

First we need to calculate our temperature to kelvins:

$T = - 10^{\circ} C + 273 = 263 K$ ‍:

If we plug the values into our equation to calculate

Δ G

, we get:

\begin{aligned} Δ G & = Δ H - T Δ S \\ = 6.01 \frac{kJ}{mol-rxn} - (263 K) (0.022 \frac{kJ}{mol-rxn \cdot K)} \\ = 6.01 \frac{kJ}{mol-rxn} - 5.79 \frac{kJ}{mol-rxn} \\ = 0.22 \frac{kJ}{mol-rxn} \end{aligned}

Thus, we see that at

- 10^{\circ} C

the Gibbs free energy change

Δ G

is positive for the melting of water. Therefore, we would predict that the reaction is not spontaneous at

- 10^{\circ} C

. In fact, we would predict that the reverse reaction should be spontaneous: we would expect to see a puddle of water turn into ice.

We could test these results by observing what happens to an ice cube outside on a chilly winter day (or perhaps in your freezer, though it is harder to see what is happening when the door is closed). If the temperature outside is

- 10^{\circ} C

(or

14^{\circ} F

), we would expect the ice cube to stay solid.

Other applications for $Δ G$ ‍: A sneak preview

Being able to calculate

Δ G

can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction a reaction will proceed at a particular temperature, especially if we are trying to make a particular product. Chances are we would strongly prefer the reaction to proceed in a particular direction (the direction that makes our product!), but it's hard to argue with a positive

Δ G

Thermodynamics is also connected to concepts in other areas of chemistry. For example:

In chemical equilibrium, we can relate $Δ G$ ‍ with the equilibrium constant, $K$ ‍.
In electrochemistry, $Δ G$ ‍ is related to the cell voltage, $E_{cell}$ ‍.

Summary

The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: $Δ S_{universe} = Δ S_{system} + Δ S_{surroundings} > 0$ ‍
At constant temperature and pressure, the change in Gibbs free energy is defined as $Δ G = Δ H - T Δ S$ ‍.
When $Δ G$ ‍ is negative, a process will proceed spontaneously and is referred to as exergonic.
Depending on the signs of $Δ H$ ‍ and $Δ S$ ‍, the spontaneity of a process can change at different temperatures.

Try it!

For the following reaction,

Δ H_{rxn} = - 120 \frac{kJ}{mol-rxn}

and

Δ S_{rxn} = - 150 \frac{J}{mol-rxn \cdot K}

$2 NO (g) + O_{2} (g) \to 2 {NO}_{2} (g)$ ‍

At what temperatures will this reaction be spontaneous?

Note: Remember that we can assume that the

Δ H

and

Δ S

values are approximately independent of temperature.

To determine when the reaction will be spontaneous, we need to know when

Δ G

is negative. We can begin by finding the temperature where

Δ G = 0

$Δ G = Δ H - T Δ S = 0$ ‍

Rearranging our equation to solve for

T

, we get:

\begin{aligned} Δ H - T Δ S & = 0 Move T Δ S term to other side \\ Δ H & = T Δ S Divide both sides by Δ S \\ T & = \frac{Δ H}{Δ S} = \frac{- 120 \frac{kJ}{mol-rxn}}{- 0.15 \frac{kJ}{mol-rxn \cdot K}} = 800 K \end{aligned}

Thus, at

T = 800 K

, we know that

Δ G = 0

and the reaction is at equilibrium.

Want to join the conversation?

Sort by:

natureforever.care
Posted 7 years ago. Direct link to natureforever.care's post “Well I got what the formu...”
Well I got what the formula for gibbs free energy is. but what's the nature of this energy and why is it called 'free'? It does free work is what textbooks say but didn't get the intuitive feel.
Button navigates to signup pageButton navigates to signup page
(22 votes)
Answer
- RogerP
  Posted 7 years ago. Direct link to RogerP's post “The word "free" is not a ...”
  The word "free" is not a very good one! In fact, IUPAC recommend calling it Gibbs energy or the Gibbs function, although most chemists still refer to it as Gibbs free energy.
  
  Gibbs originally called it available energy and that is a good term because it is the energy associated with a chemical reaction that is available (or you could say free) to do work, assuming constant T and P.
  Comment on RogerP's post “The word "free" is not a ...”
  (27 votes)
Ben Alford
Posted 7 years ago. Direct link to Ben Alford's post “Is there a difference bet...”
Is there a difference between the notation ΔG and the notation ΔG˚, and if so, what is it?
Button navigates to signup pageComment on Ben Alford's post “Is there a difference bet...”
(10 votes)
Answer
- dmelby
  Posted 6 years ago. Direct link to dmelby's post “STP is not standard condi...”
  STP is not standard conditions. Standard conditions are 1.0 M solutions and gases at 1.0 atm. Standard conditions does not actually specify a temperature but almost all thermodynamic data is given at 25C (298K) so many people assume this temperature.
  Comment on dmelby's post “STP is not standard condi...”
  (4 votes)
ila.engl
Posted 7 years ago. Direct link to ila.engl's post “Hey I´m stuck: The ∆G in ...”
Hey I´m stuck: The ∆G in a reaction is negative but the ∆H was positive and it is assumed that a change temperature doesn´ t significantly affect entropy and entalpy. What does this do to 1) spontanity 2) spontanity at high temp 3) value or sign of ∆S

i read it 3 times now but i´m still insecure - :(
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- awemond
  Posted 7 years ago. Direct link to awemond's post “This looks like a homewor...”
  This looks like a homework question, so I'll give you some hints to get you on the riht path rather than answering directly.
  
  3) We know that ∆G = ∆H - T∆S. Solving for ∆S, we have:
  
  ∆S=(∆H-∆G)/T.
  
  We know (from the question) that ∆G is negative and that ∆H is positive. Temperature is always positive (in Kelvin). From these values, we can know for certain whether ∆S is positive or negative (hint: remember that we are subtracting ∆G!).
  
  1) Knowing the sign of ∆G is enough to say whether the reaction is spontaneous or not under these conditions. If ∆G is negative (from the question), is the reaction spontaneous or non-spontaneous?
  
  2) Let's use ∆G = ∆H - T∆S again. Since ∆H and ∆S don't change significantly with temperature (given in the question), we can assume that they keep the same signs and values: i.e. ∆H is still positive and ∆S is still whatever sign you figured out above. As T increases, the T∆S component gets bigger. T is always positive, so if ∆S is positive then a bigger T∆S will make ∆G more negative (since we subtract T∆S). If ∆S is negative, then the negative signs (from the subtraction and the sign of ∆S) will cancel out, and so as T∆S gets bigger, ∆G will get more positive.
  Comment on awemond's post “This looks like a homewor...”
  (12 votes)
Phoebe Hall
Posted 8 years ago. Direct link to Phoebe Hall's post “In the subject heading, '...”
In the subject heading, 'When is ΔG is negative?', is it a typo that it says
'When the process is endothermic, ΔHsystem > 0, and the entropy of the system decreases, ΔSsystem>0, the sign of ΔG is positive at all temperatures. Thus, the process is never spontaneous' shouldn't the entropy be < 0? if there is a decrease in entropy?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- Mohamed Mahrous
  Posted 8 years ago. Direct link to Mohamed Mahrous's post “I think you are correct. ...”
  I think you are correct. Change in entropy must be smaller than zero, for the entropy to decrease. It is a typo. You can cross-check from the figure.
  Button navigates to signup page
  (5 votes)
Jasgeet Singh
Posted 7 years ago. Direct link to Jasgeet Singh's post “The Entropy change is giv...”
The Entropy change is given by Enthalpy change divided by the Temperature. Then how can the entropy change for a reaction be positive if the enthalpy change is negative?
Button navigates to signup pageButton navigates to signup page
(4 votes)
Answer
- tyersome
  Posted 7 years ago. Direct link to tyersome's post “Great question! Figuring...”
  Great question! Figuring out the answer has helped me learn this material.
  
  One way to define entropy is Q/T (where Q is the heat associated with a reversible process {also note that ∆H is only equal to Q when P is constant}).
  
  For spontaneous and thus irreversible reactions, the ∆S is the same as for a reversible reaction (because S is a state variable it doesn't depend on how we got from one condition to another). In contrast, ∆H is not equal to Q (because this is not a reversible reaction).
  
  A later video helps explain this:
  https://www.khanacademy.org/science/chemistry/thermodynamics-chemistry/gibbs-free-energy/v/more-rigorous-gibbs-free-energy-spontaneity-relationship
  Comment on tyersome's post “Great question! Figuring...”
  (5 votes)
Kaavinnan Brothers
Posted 7 years ago. Direct link to Kaavinnan Brothers's post “Hi all, Sal sir said we ...”
Hi all, Sal sir said we would prefer the reaction to proceed in a particular direction (the direction that makes our product!), but it's hard to argue with a positive ΔG! ( located before summary at other applications of del G) .can anybody please explain?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- RogerP
  Posted 7 years ago. Direct link to RogerP's post “If ΔG is positive, then t...”
  If ΔG is positive, then the only possible option is to vary the temperature but whether that would work depends on whether the reaction is exo- or endothermic and what the entropy change is.
  Button navigates to signup page
  (3 votes)
1448169
Posted 8 years ago. Direct link to 1448169's post “how do i see the sign of ...”
how do i see the sign of entropy when both reactant and product have the same phase
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Stephen R. Collier
  Posted 8 years ago. Direct link to Stephen R. Collier's post “We have to look up the ΔS...”
  We have to look up the ΔS for the whole reaction in a table (or test the reaction ourselves... I'd rather look it up!). The value will be either positive or negative. If the reaction can result in a phase change then we might be lucky enough to find a list that has the reaction with reactant and products in the phases we need.
  
  Otherwise we could calculate the change in energy and the use the specific heat equations to see if the phase would change. The example above with melting ice looks a little different because the reaction was a phase change (ice to water) instead of the usual combining or splitting of molecules.
  Button navigates to signup page
  (2 votes)
izzahsyamimi042
Posted 5 years ago. Direct link to izzahsyamimi042's post “can an exothermic reactio...”
can an exothermic reaction be a not spontaneous reaction ?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Andrew M
  Posted 5 years ago. Direct link to Andrew M's post “Sure. Paper doesn't ligh...”
  Sure. Paper doesn't light itself on fire, right?
  Comment on Andrew M's post “Sure. Paper doesn't ligh...”
  (3 votes)
estella.matveev
Posted 6 years ago. Direct link to estella.matveev's post “Hi, could someone explain...”
Hi, could someone explain why exergonic reactions have a negative Gibbs energy value? I get it in terms of doing the calculations by looking at the graphs, but don't get it in terms of particles gaining or losing energy.
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Oliver McCann
  Posted 5 years ago. Direct link to Oliver McCann's post “According to the laws of ...”
  According to the laws of thermodynamics, ever spontaneous process will result in an increase in entropy and thus a loss in "usable" energy to do work. When an exergonic process occurs, some of the energy involved will no longer be usable to do work, indicated by the negative Gibbs energy.
  Button navigates to signup page
  (1 vote)
anoushkabhat2016
Posted 6 years ago. Direct link to anoushkabhat2016's post “Is the reaction H2O(l) to...”
Is the reaction H2O(l) to H20(s) spontaneous or non spontaneous? The entropy, S, is positive when something goes from a solid to liquid, or liquid to gas, which is increasing in disorder. However, in this equation, water is going from a liquid to solid, so S is negative, and in the Gibbs free reaction equation, S must be positive for a reaction to be spontaneous. How do we determine, without any calculations, the spontaneity of the equation?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Betty :)
  Posted 4 years ago. Direct link to Betty :)'s post “Using that grid from abov...”
  Using that grid from above, if it's an exothermic reaction (water is releasing heat into its surroundings in order to turn into ice), we know it's on the left column. The entropy of liquid water is higher than ice (water as a solid state)so therefore it is not always going to be spontaneous. Putting into the equation, ΔH<0 because it's exothermic, and ΔS<0 because entropy is decreased. Therefore, the reaction is only spontaneous at low temperatures (TΔS). If you think about its real-world application, it makes sense. Liquid water will turn into ice at low enough temperatures.
  Button navigates to signup page
  (2 votes)

Chemistry library

Course: Chemistry library > Unit 13

Key points

Spontaneous processes

Gibbs free energy and spontaneity

Calculating change in Gibbs free energy

When is ΔG‍ negative?

Example 1‍ : Calculating ΔG‍ for melting ice

Other applications for ΔG‍ : A sneak preview

Summary

Try it!

Want to join the conversation?

When is $Δ G$ ‍ negative?

Example $1$ ‍: Calculating $Δ G$ ‍ for melting ice

Other applications for $Δ G$ ‍: A sneak preview