Entropy changes and Gibbs free energy

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  1. Will a chemical reaction happen spontaneously?
  2. The Gibbs free energy equation.
  3. What is entropy?
  4. Calculating entropy changes - example.
  5. Entropy and thermodynamics.
  6. Entropy change of the surroundings
  7. Is a reaction spontaneous?
  8. Gibbs free energy and equilibrium.
  1. Use the Gibbs free energy equation to predict the required temperature for a reaction to be feasible.
    A chemical reaction is known to have Ξ”H \Delta H = +14 kJ mol-1 and Ξ”S \Delta S = +44.5 J K-1 mol-1. At what temperature (K) does this reaction become feasible?
    1. Calculate the entropy changes of the system and surroundings for a reaction.
      1. A chemical reaction at 298K between compounds A and B to produce C and D is described below:

        2A (s) + B (aq)  \,   \, C (g) + D (aq)

        Molar entropy data for the substances is below:


        Standard molar entropy Ξ”S \Delta S om (J K-1 mol-1)

        A (s)


        B (aq)


        C (g)


        D (aq)


        Use this data to calculate the entropy change (Ξ”S \Delta S ) for the reaction
      2. This reaction is exothermic, with a Ξ”H \Delta H value of -75.2 kJ mol-1. What is the entropy change of the surroundings?

        Use this information and the Gibbs free energy equation to explain the general feasibility of this reaction.
    Topic Notes

    In this lesson, we will learn:

    • The 2nd law of thermodynamics, as it relates to entropy’s role in chemical reactions.
    • To calculate the standard entropy change of reactions.
    • To understand the factors that affect the Gibbs free energy change of a reaction.
    • To use the equation for Gibbs free energy and find if a reaction is feasible.
    • How Gibbs free energy relates to a reaction at equilibrium.
    • How to calculate Gibbs free energy changes at non-standard conditions.

    • In Calculations using Born-Haber cycles, we calculated the enthalpy of solution for the dissolving of some ionic compounds. Some solution enthalpies are endothermic; why do those compounds still dissolve at room temperature?
      It’s because enthalpy is not the only factor that determines whether a chemical process happens at certain conditions. Another factor called entropy is also important in determining whether a chemical process is feasible.

    • This is shown in the equation for Gibbs free energy:

      Ξ”G \Delta G ° == Ξ”H \Delta H ° βˆ’TΞ”S- T \Delta S °

      Ξ”G {\Delta} G ° = the standard change in Gibs free energy
      Ξ”H \Delta H ° = the standard enthalpy change
      T = temperature, in Kelvin (K)
      Ξ”S \Delta S ° = the standard change in entropy measured in J K-1 mol-1

      The standard change in Gibbs free energy must be a negative value for a reaction to be spontaneous at those conditions.

      The equation shows that entropy, enthalpy and temperature are all conditions that affect whether a reaction will happen. We will look at a quick guide to this with some calculations after introducing what entropy is.

    • In a sentence: entropy is a measure of disorder in a system.
    • It is about the number of different energetic states a system can have – how many ways could a system, and all the particles in it, distribute its energy?

      • Shuffling cards is a good practical way to think of entropy.
        If you use a small part of a deck, you can shuffle and you might get some card numbers or suits in order. The more of a full deck you use when shuffling (or multiple decks), the more random it gets. It is just probability - there are just far more random possible card combinations than ordered ones. You see this when buying a brand-newordered deck of cards. Once you start using and shuffling them, they just get more and more random. Have you ever shuffled a deck and got a perfect order back from it?

      In chemistry, entropy is about energy being distributed, which is closely related to the movement of particles. Gases have very energetic particles moving in different directions; each gas particle collision will redistribute some energy between particles. Compare this to a solid substance, where particles are not moving as much, moving with less energy and possibly being bonded together. With this said, gas substances have much greater entropy than solids do.
      It is about the number of possible arrangements of energy in a system, which gives a few general points for chemical reactions:
      • Reactions that increase the number of moles of gas will increase entropy. This is as simple as there being more highly energetic particles in the system, each of their potential collisions causing a different distribution of energy.
      • Changes of state from solids towards gases will increase in entropy. Taking a substance (e.g. water) from solid to liquid to gas would give the particles increasingly more ways to arrange themselves in space, because the increasing and varying amounts of energy the particles have allows for a lot of different possible arrangements. This is why the standard entropy of ice is lower than water, which is lower than steam.
      Given this, entropy changes are integral to chemical reactions like enthalpy, and particularly when a change in gas moles happens.

    • You can calculate an entropy change (β–³ \triangle S) for a reaction just like an enthalpy change.

    • Ξ”S=βˆ‘S( \Delta S = \sum S(products)βˆ’βˆ‘S() - \sum S(reactants)

      Compounds have a given standard molar entropy (Som) so with compounds changing from reactants to products a reaction will give an entropy change, Ξ” \Delta S.
      For example, the equation for the combustion of methane is below, with the molar entropy data1 for all the compounds involved:

      CH4 (g) + 2O2 (g)  \,   \, CO2 (g) + 2H2O (g)


      Standard molar entropy Ξ”S \Delta S 0m (J K-1 mol-1)

      CH4 (g)


      O2 (g)


      CO2 (g)


      H2O (g)


      Ξ”S=βˆ‘S( \Delta S = \sum S(products)βˆ’βˆ‘S() - \sum S(reactants)

      Ξ”S \Delta S = (213.74 + 2(188.83)) – (186.26 + 2(205.13)) = -5.12 J K-1

      The combustion of methane shows a slight decrease in entropy. This is rare for combustion reactions; most combustion reactions result in an increase in the number of moles of gas which results in higher entropy.

    • Enthalpy forms part of the laws of thermodynamics. The second law of thermodynamics states that there is an increase in total entropy of the universe with any spontaneous change.
      • As you can see above with the combustion of methane, you can still get reactions with a decrease in entropy that are possible, but that entropy decrease is only measuring the reaction substances. We would call this the entropy of the system. So where is the increase in entropy?
        The combustion of methane is a massively exothermic reaction that disperses heat energy to the surroundings, making it more energetic. This effect means, the net entropy change of the universe is an increase.
      When looking at problems in thermodynamics, you can think of the universe as being made of two parts:
      • The system you are studying, such as the substances in a reaction or the matter in a container
      • The surroundings which is everything outside the system.
      When spontaneous changes happen, the universe overall has increased in entropy:

      Suniverse = Ssystem + Ssurroundings

    • It is possible to calculate the entropy change of the surroundings after a chemical reaction, not just the system containing the reaction. This is important to show the overall entropy change:

    • β–³S \triangle S surroundings =βˆ’Ξ”HT= \large -\frac{ \Delta H} {T}

      Using this we can show, for example, that although the combustion of methane has a negative entropy change for the system, it gives a large increase in entropy of the surroundings.
      Make sure you convert units correctly! Like Ξ”S \Delta S system, Ξ”S \Delta S surroundings is JOULES per kelvin per mole, but Ξ”H \Delta H is KILOJOULES per mole.
      For the combustion of methane, where Ξ”H \Delta H c = -890 kJ mol-1 1:

      Ξ”S \Delta S surroundings =βˆ’(βˆ’890β€‰βˆ—β€‰1000)J molβˆ’1298 K= \large -\frac{(-890\, * \,1000) J \,mol^{-1}} {298\, K}

      Ξ”S \Delta S surroundings = +2986.58 J K-1 mol-1

      Using the formula Ξ”S \Delta S universe = Ξ”S \Delta S system + Ξ”S \Delta S surroundings:

      Ξ”S \Delta S universe = (-5.12 J K-1 mol-1) + (2986.58 J K-1 mol-1) = +2981.46 J K-1 mol-1

      As you can see, although the system was slightly lower entropy, the surroundings were substantially higher, resulting in an overall increase in entropy that is consistent with the second law.

    • With entropy and enthalpy both important in seeing if a reaction will happen or not, we can use the Gibbs free energy equation to picture four situations and tell if a reaction is spontaneous or not:

    • Ξ”G \Delta G == Ξ”Hβˆ’TΞ”S \Delta H - T \Delta S

      Ξ”S \Delta S is POSITIVE

      Ξ”S \Delta S is NEGATIVE

      Ξ”H \Delta H is POSITIVE

      The reaction is spontaneous at high temperatures.

      (larger TΞ” \Delta S term will turn Ξ” \Delta G negative)

      The reaction is never spontaneous.

      (TΞ” \Delta S change will never yield a negative Ξ” \Delta G value)

      Ξ”H \Delta H is NEGATIVE

      The reaction is always spontaneous.

      (TΞ” \Delta S change will never yield a positive Ξ” \Delta G value)

      The reaction is spontaneous at low temperatures.

      (larger TΞ” \Delta S will turn Ξ” \Delta G positive)

    • You can think of standard Gibbs free energy as a β€˜driving force’ of a reaction. It tells you which direction a reaction will go in order to reach equilibrium.
      We can relate the equilibrium constant K and the standard Gibbs free energy change Ξ”G0 \Delta G^{0} :

    • Ξ”G0=βˆ’RTln(K) \Delta G^{0} = -RTln(K)

      Where R is the gas constant and T is temperature (in Kelvin).
      It can also be expressed as:

      K=eβˆ’Ξ”G0RTK = e^{-{\frac{\Delta G^{0}} {RT} } }

      With the first expression, you can see how K predicts the sign and value of the standard Gibbs free energy change:
      • For reactions where K > 1 (meaning at equilibrium, there is more product than reactant), Ξ”G0\Delta G^{0} is negative. So, at standard conditions the reaction favours the products.
      • For reactions where K < 1 (at equilibrium there is more reactant than product), Ξ”G0\Delta G^{0} is positive. So, at standard conditions the reaction favours the reactants.

      The Ξ”G\Delta G driving force is finite. Eventually a system will reach equilibrium, where Ξ”G=0\Delta G = 0.
      When a real reaction begins, we probably aren’t at equilibrium, so we need to use the reaction quotient Q. We saw in The equilibrium constant that Q is calculated exactly like K; it is the ratio of products to reactants in a reaction mixture. The moment a reaction begins, Q starts changing because amounts of reactants and products start changing.

    • Remember the 0^{0} in Ξ”G0\Delta G^{0} means standard conditions: equal amounts of product and reactant in the mixture. Most real reactions don’t begin with any product in them, so reactions normally take place at non-standard conditions. There will probably be only one instant when the amounts of product and reactant are equal. In reality, ?G is constantly changing as the reaction heads toward equilibrium. We need an instantaneous Gibbs free energy equation for non-standard conditions:

    • Ξ”G=Ξ”G0+RTIn(Q)\Delta G = \Delta G^{0} + RTIn (Q)

      We can use a simple example to show how Ξ”G\Delta G is found at non-standard conditions. For the reaction:

      N2 + 3H2 β€‰β‡Œβ€‰\, \rightleftharpoons \, 2NH3

      At 300°C, where PN2 = 3 atm, PH2 = 6 atm and PNH3 = 1.1 atm

      We can begin by finding the reaction quotient Q with the given values for partial pressures:

      Q=ProductsReactants=Q = \frac{Products} {Reactants} = [1.1]2[3][6]3=\large \frac{[1.1]^{2}} {[3] [6]^{3} } = 1.87 × 10-3

      We can find a literature2 K value at 300°C which is 4.34*10-3 and use this to calculate the standard Gibbs free energy change using the K value:

      Ξ”G0=βˆ’RTIn(K)\Delta G^{0} = - RTIn (K)

      Ξ”G0=βˆ’(8.314Jmol K)(573.15 K)βˆ—In(4.34βˆ—10βˆ’3)=+25922 J molβˆ’1\Delta G^{0} = -(8.314 \frac{J} {mol \, K} ) (573.15 \, K) * In (4.34 * 10^{-3} ) = + 25922 \, J \, mol^{-1}

      All this Ξ”G0\Delta G^{0} means is that if Q = 1 i.e. there was equal concentrations of products and reactants (standard conditions), there would be a spontaneous drive for the reverse reaction.

      Finally, we can use Q and Ξ”G0\Delta G^{0} to find Ξ”G\Delta G at our nonstandard conditions:

      Ξ”G=Ξ”G0+RTIN (Q)\Delta G = \Delta G^{0} + RTIN \, (Q)

      Ξ”G\Delta G (+25922) + (8.314)(573.15)InIn(1.87 * 10-3) = -29908 J molJ \, mol -1

      This calculation of -29907 J mol-1 or -29.9 kJ mol-1 means that at our current partial pressures, there is a driving force towards the products. You can also see this by directly comparing K with Q. K is greater than our Q value, so our reaction will make more products in order to reach equilibrium because that will increase Q until it is equal to K.

      The point of these calculations is to be able to predict the direction of a reaction based on the nonstandard Ξ”G\Delta G we find.

    • There are many instances where a desired product comes from a reaction which is thermodynamically unfavourable – Ξ”G\Delta G is positive.
      That doesn’t mean we can’t access the product. We can do two things:
      • We can apply energy from an external source e.g. light or electric current. Remember, Ξ”G\Delta G is calculated for the system – is the system self-driving? If external energy is applied, we can push it into action with that β€˜push’.
        • Photosynthesis is a great example of this. Respiration, the opposite of photosynthesis, is favourable by enthalpy and entropy change. Photosynthesis is therefore unfavourable on both terms. This is why it requires sunlight for the process to occur.
      • We can couple an unfavourable reaction with a favourable one. This means combining reactions that have intermediates in common, so a reaction pathway exists for both.
        We cannot cheat the laws of thermodynamics though! The favourable reaction (Ξ”G\Delta G < 0) has to be more favourable than the unfavourable (Ξ”G\Delta G > 0) one is not. This way, the combined Ξ”G\Delta G values should be less than zero so the whole process is feasible. It’s just combining two systems and their free energies.

    • Despite being feasible (having negative Ξ”G\Delta G), there are many processes which don’t occur at standard conditions. Some proceed so slowly that it appears they are not happening. These are typical of processes under kinetic control, which means the current kinetic energy of the substance is preventing the reaction from taking place – the activation energy has not yet been reached.
      This does not mean the reaction isn’t feasible or at equilibrium! It simply means that temperature needs to increase for it to begin, until the activation energy barrier is overcome.
      Whenever a process is known to have a favourable Ξ”G\Delta G but does not proceed at the current conditions, it is fair to say it is under kinetic control.

    • 1 Source for enthalpy and entropy terms used: ATKINS, P. W., & DE PAULA, J. (2006). Atkins' Physical chemistry. Oxford, Oxford University Press.

      2 Source for equilibrium constant: Brown, Theodore L.; LeMay, H. Eugene, Jr; Bursten, Bruce E (2006). "Table 15.2".?Chemistry: The Central Science?(10th ed.). Upper Saddle River, NJ: Pearson.