Entropy t he u niverse and f ree energy
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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 ?.

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Entropy t he u niverse and f ree energy

Entropy, The Universe and Free Energy

How do changes in a system s enthalpy and entropy affect s universe
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
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

  • 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
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?


We need to use g system h system t s system
We need to use ammonia…..∆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…. ammonia…..

  • 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
Coupled Reactions ammonia…..

  • 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
EXIT TICKET ammonia…..

  • 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