This presentation is the property of its rightful owner.
1 / 10

# Entropy, T he U niverse and F ree Energy PowerPoint PPT Presentation

Entropy, T he U niverse and F ree Energy. For any spontaneous process: ∆ S universe > 0 Because: ∆ S universe = ∆ S system +∆ S surroundings. How do changes in a system’s enthalpy and entropy affect ∆ S universe ?.

Entropy, T he U niverse and F ree Energy

An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

## Entropy, The Universe and Free Energy

• For any spontaneous process:

∆Suniverse > 0

Because:

∆Suniverse= ∆Ssystem +∆Ssurroundings

### How do changes in a system’s enthalpy and entropy affect ∆Suniverse ?

• In nature , ∆Suniverse tends to be positive for reaction processes under the following conditions:

• The reaction or process is exothermic, which means ∆Ssystem is a negative value

(the heat released by the system increases Temp of the surroundings, thus increasing the entropy.)

2. The entropy of the system increases, so ∆Ssystem is positive

### Free Energy

• Gibbs free energy (Gsystem) is commonly called free energy.

• Free energy is energy that is available to do work, thus free energy is useful energy.

### Gibbs Free Energy

• Gibbs Free Energy can be calculated by:

∆Gsystem = ∆Hsystem - T∆Ssystem

(where T is in Kelvin)

If ∆Gsystem is negative the process is spontaneous

If ∆Gsystem is positive then the process is nonspontaneous

Under standard conditions it is written:

∆G⁰system = ∆H⁰system - T∆S⁰system(T=298K and Pressure=1atm)

### Let’s look at the reaction for the formation of ammonia…..

N2(g) + 3H2(g)  2NH3(g)

∆H⁰system= -91.8kJ

∆S⁰system=-197 J/K

Note that Entropy decreases because four moles of gaseous molecules react to make only two moles of gaseous molecules. Therefore the entropy is negative which tends to make a reaction nonspontaneous.

Yet, The reaction is exothermic(∆H⁰system= -91.8kJ) which tends to make the reaction spontaneous.

So which one is it?

CONFLICT!!!!!!!!

### We need to use∆G⁰system = ∆H⁰system - T∆S⁰system

N2(g) + 3H2(g)  2NH3(g)

∆H⁰system= -91.8kJ

∆S⁰system=-197 J/K

∆G⁰system= -91.8 kJ – (298K)(-197J/K)

∆G⁰system= -91800J + 58700 J

∆G⁰system= -33100 J

∆G⁰systemis negative therefore, the reaction is spontaneous under standard conditions.

### Recall….

• If ∆G⁰system is negativethe reaction is spontaneous

• If ∆G⁰system is positive the reaction is nonspontaneous

• If ∆G⁰system= 0 , we say that the products and reactants are present in the state of chemical equilibrium. In other words, the rate of the forward reaction equals the rate of the reverse reaction.

### Coupled Reactions

• Sometimes a series of reactions must occur in order to allow for a nonspontaneous reaction to occur. For example, many of the chemical reactions in living organisms are nonspontaneous reactions.(∆G⁰system is positive) Yet, the reason they occur so readily is because they occur together with other spontaneous reactions.

• Free energy released by one or more spontaneous reactions drives the nonspontaneous reactions.

### EXIT TICKET

• Why is the sign of ∆Hfusion positive?

• Why is the sign of ∆Sfusion positive?

• Solve for ∆G for the ice melting at the two temperatures.

• Using your results, explain why ice melts at 274K but not at 272K