How to determine the temperature at which a reaction is feasible

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A spontaneous or feasible process must be accompanied by a positive change in total entropy (entropy of the universe). Whether this is the case is almost always down to the temperature – hence ice melts at 23°C but not at -14°C.

Before we go any further, I’m assuming that you are confident in your understanding of the concept of entropy and that these equations look familiar:

ΔSsys = ∑ Sproducts – ∑ Sreactants

ΔStotal = ΔSsurr + ΔSsys

ΔSsurr = -ΔH / T

ΔStotal = (-ΔH / T) + ΔSsys

NOTE: not all exam specifications require you to be able to determine the feasibility of a reaction by calculating total change in entropy – some jump straight from the concept of entropy change of system to Gibbs free energy, ΔG, hence I’m going to show you both methods of determining feasibility.

ΔG = ΔH – TΔSsys

Personally, I’m not sure how you can genuinely understand Gibbs free energy without an understanding of entropy change of surroundings and total entropy change, so I strongly recommend you read up on it all 🧐.

Calculating the temperature at which a reaction becomes feasible

The decomposition of calcium carbonate into calcium oxide and carbon dioxide is an endothermic reaction, however if we know ΔSsys and ΔrH for the reaction, we can figure out the temperature at which this reaction becomes feasible or spontaneous.

CaCO3(s) ⇾ CaO(s) + CO2(g) ΔSsys = +161 JK-1mol-1 ΔrH = +178 kJ mol-1

  1. ΔStotal = (-ΔH / T) + ΔSsys and a reaction is feasible when ΔStotal = 0 since this is when the reaction is at equilibrium (Kc = 1).

0 = (-178000 / T) + 161 (remembering to convert ΔrH into J mol-1)

-161 = -178000 / T

T = 1106K

There are lots of practice exam Q&A below 👇.

  1. ΔG = ΔH – TΔSsys and ΔG = 0 at the minimum temperature required for the reaction to occur

0 = ΔH – TΔSsys

TΔSsys = ΔH

T = ΔH / ΔSsys = +178 / + 0.161 = 1106K remembering to convert ΔSsys into kJ K-1 mol-1

You can find some practice exam Q&A for calculating the feasibility of a reaction using Gibbs energy here!

Entropy exam practice questions

You will need standard molar entropy data from the table below:

    1. (a) Calculate ΔSsys for the reaction between methane and steam.

     CH4(g)  +   H2O(g)  ⇾  CO(g)   +   3H2(g)       ΔrH = +206 kJ mol-1

    (b) Explain how the sign of your answer to part (a) is predicted by the equation.

    (c) Calculate the minimum temperature required for the forward reaction to be feasible.

    –

    1. CaCO3(s)  ⇾   CaO(s)   +   CO2(g)    ΔrH = +180 kJ mol-1

    (a) Explain the term ‘entropy’ and predict, with reason, whether you would expect ΔSsys for the reaction above be positive or negative.

    (b)  Calculate a value for ΔSsys .

    (c)  Calculate ΔStotal at 298K.

    (d)  Calculate the temperature at which ΔStotal is zero and comment on the significance of this temperature.

    –

    3. Calculate the minimum temperature for the following reaction to occur:

    2NaHCO3(s)   ⇾   Na2CO3(s)   +   H2O(l)  +  CO2(g)                        ΔrH = +91 kJ mol-1

    –

    4. H2(g)   +   CO2(g)    ⇌   H2O(g)   +  CO(g)     ΔrH = +40 kJ mol-1

    (a)  Calculate ΔSsys for the forward reaction.

    (b)  Calculate the temperature at which ΔStotal = 0. 

    (c)  Comment on the equilibrium position when ΔStotal = 0.

    –

    5. What is the entropy change, ΔS⦵,in J K-1 mol-1 for the reaction below?

    CO(g) +  2H2(g)  ⇾  CH3OH(l)

    –

    6. Consider the reaction between hydrogen and carbon sulfide:

    4H2(g)  +  CS2(g)   ⇾  CH4(g)  +  2H2S(g)      ΔSsys = -164 J K-1 mol-1

    (a)  Why does the reaction have a negative entropy change?

    (b)  Calculate a value for the standard molar entropy for H2, showing your working.

    –

    7.  Calculate the entropy change for the reaction between ammonia and oxygen:

    4NH3(g)   +  5O2(g)   ⇾    4NO(g)   +   6H2O(g)

    –

    8.  Predict, with reason, whether the entropy of the following reactions increases or decreases:

    (a)  CO2(g)  ⇾  CO2(s)

    (b)  NaCl(s)  +  (aq)  ⇾  NaCl(aq)

    (c)  N2(g)  +  3H2(g)  ⇾  2NH3(g)

    –

    9.  Determine the temperature at which the back reaction becomes feasible for the reaction below:

    2SO2(g)   +  O2(g)   ⇌  2SO3(g)  ΔrH = -288.4 kJ mol-1

    Answers

    1.

    2. (a) Entropy is a measure of the disorder of the molecules and their associated energy quanta – the more ways they can be arranged, the higher / more positive the entropy. ΔSsys for this reaction will be positive as a gas is produced.

    (b)  ΔSsys = ∑Sproducts – ∑Sreactants

               = (40.0 + 214) – 93.0 = +161 JK-1mol-1

    (c)  ΔStotal =  ΔSsurr +  ΔSsys 

                =  (-180000 / 298) + 161 = -443 JK-1mol-1

    (d)  0 = (-180000 / T) + 161

         T = -180000 / -161  =  1118K

         1118K is the temperature at which the system is at equilibrium and the reaction becomes feasible.

    3. The minimum temperature is when ΔStot = 0

        0  =  (-ΔH / T)  +  ΔSsys 

        ΔSsys =  ∑Sproducts – ∑Sreactants  =  (135 + 70 + 214) – (2 x 102)  

                                        = +215 JK-1mol-1

         0 =  (-91000 / T) + 215

       T =  -91000 / -215  =  423K

    EXAM TIP: State symbols are important – we need the value for liquid water, not gaseous, in this calculation!

    4. (a)   ΔSsys =  (189.0 + 197.6) – (130.6 + 214.0)

                    =  42.0 JK-1mol-1 

    (b)   0  = (-ΔH / T)  +  ΔSsys 

             =  (-40000 / T)  +  42

          T  =  950K  (answer is quoted to 2 sig. fig. as ΔH is only given to 2 sig. fig.)

    (c)  This is the temperature at which the reaction is at equilibrium and Kc=1.

    5. ΔSsys⦵ = 239.7 – (197.6 + (2 x 130.6)  

               =  -219.1 JK-1mol-1 

    6. (a)  The reaction has a negative  ΔSsys because 5 molecules of gas become 3 molecules of gas and the system becomes less disordered / random.

    (b)  ΔSsys =  ∑Sproducts – ∑Sreactants  

        -164  =  (186.2 + (2 x 206)) – (4x + 238)

        -164  =  598.2  –  4x  – 238

        -164  =  360.2  –  4x

        4x  =  524.2

        SH2⦵ = 131 JK-1mol-1 

    7.  ΔSsys =  ∑Sproducts – ∑Sreactants  

               =  [(4 x 211) + (6 x 189.0)] – [(4 x 193) + (5 x 205)]

              = 181 JK-1mol-1 

    8. (a)  Decrease – gaseous reactant becomes solid product (fewer quanta of energy, less disorder)

        (b)  Increase – solid ionic lattice becomes hydrated ions in aqueous solution (increase in disorder as ions are free to move in aqueous solution)

        (c)  Decrease – 4 molecules of gas become 2 molecules of gas (decrease in disorder as fewer product molecules)

    9. ΔStotal = 0 when the reaction is at equilibrium and becomes feasible

           0  =  (-ΔH* / T) + ΔSsys

    EXAM TIP: the question is asking for the temperature when the BACK reaction is feasible so ΔH = +288.4 kJ mol-1

      ΔSsys =  ∑Sproducts – ∑Sreactants  

           = ((248.1 x 2) + 205)  –  (95.6 x 2)  =  +510.0 JK-1mol-1 

     0 =  (-288400 / T)  +  510.0

     T =  (-288400 / -510)  =  565K

    (In reality, since this is the temperature at equilibrium it is irrelevant whether we work with the back or the forward reaction)