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# Gibbs free energy example

Determining if a reaction is spontaneous by calculating the change in Gibbs free energy. Also calculates the change in entropy using table of standard entropies.  Created by Sal Khan.

## Want to join the conversation?

• This is really amazing! But I still don't understand how energy being released in the reaction can cause it to become spontaneous, because, Sal, you said in the earlier video that when the change in entropy is negative the reaction isn't spontaneous....please help?!?!!!!
• As I understand, if energy is being released, the system doesn't require much energy to do the reaction, so it is easier for the reaction to occur.
• So I don't see how you got -890.3 KJ for the heat change... I looked up the heat of formations of everything. CH4; -74.8 KJ, O2: 0 CO2: -393.5 KJ, and H2O: -241.8.... so you would do heat of products minus heat of reactants.

-241.8*2+-393.5= -877.1
-74.8+0=-74.8
-877.1--74.8= -802.2
• That is because you looked up the heat of formation for water as a gas. If you used the heat of formation for water as a liquid ~ -285.8 KJ (see the equation above in the video), you would get -890.3 KJ.
• What if we are not allowed to look up the data on the tables. We get other reactions that we balance to find ΔH. Could you make a video using this method? I'm taking an AP class in high school and the table way won't be on the exam but the balance the equation way will be.
• I took AP chemistry last year and, if I understand your question, you need to add the equations together until you get a balanced equation that matches yours. You must also add the Enthalpies.
• Great video! I do have a question about why water is listed in the liquid state, though. Wouldn't it be in gas form if you were actually burning methane? Is this an artificial construct by stating it's occurring at 25 C? What happens to the heat produced by the exothermic nature of the process?
• It doesn't belong in the comment section, & there is no way for anyone to answer him from the comments. If you'll notice I quoted his name and votes!
• Why does oxygen have an enthalpy value of 205kJ? I thought that Sal said in a previous video that enthalpy of an element in its standard state is 0?
• Those numbers were not enthalpies, they are the standard molar entropies at 298 K and 101.3 kPa. Also, you misread the units.

The standard molar entropy of O₂ at the above conditions is 205.1 J/(mol∙K)
• Hey... Just in relation to the first reaction, I seem to be getting totally different numbers for change of formation for CH4, CO2 and H2O. Where do the numbers 186, 213 and 69 come from....? It's probably a very silly question, apologies, I am just stuck at this... I thought that to calculate the change of enthalpy you needed to add the products' heat of formation C+D and subtract reactants heat of formation -A-B from it?
• Numbers 186, 213 and 69 are entropy values to calculate delta S, not enthapy change of formation. For delta H he already did all the calculations and didn't inlude the original values.
• If the temperatures on the surface of the sun are that high the why do we see reactions occurring on the surface of the sun (hydrogen atoms coming together to make helium)?
(1 vote)
• First of all the conversion of hydrogen to helium occurs deep inside a star not on the surface. Second, these reactions are nuclear and not chemical. It is the extreme temperatures in the sun that allow the fusion reactions to occur.

Also at the temperature of the surface of the sun chemicals will be ripped apart and the atoms will be stripped of electrons becoming plasma.
• So after the entropy intuition, I have a little trouble with the standard entropy - how can it be defined? Shouldnt it depend also on volume of the system?
• It is standart -molar- entropy, I think there's the rub. It is the certain amount of molecules. And in liquids and solids volume is a constant, in gas it think they use standart molar volume of ideal gas.
(1 vote)
• At you gave your response for change in H. How did you get -890.3 KJ? I'm a missing a simple detail or should i go watch another one of your videos?