How to determine the temperature at which a reaction is feasible

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 at which we are carrying out the reaction, hence ice melts at 23°C but not at -14°C.

We know how to calculate the entropy change of a system, ΔS^⦵_sys :

ΔS^⦵_sys = ∑ S^⦵_products – ∑ S^⦵_reactants

We know that the total entropy change, ΔS_total, must include both the entropy change of the system and of the surroundings, ΔS^⦵_surr :

ΔS^⦵_total = ΔS^⦵_surr + ΔS^⦵_sys

And we know that the entropy change of the surroundings depends on the enthalpy change of the reaction (is heat being released to the surroundings in an exothermic reaction or taken from it in the case of an endothermic reaction?) and the temperature at which the reaction is taking place:

We can combine these equations to give us a way to calculate the total entropy change of a reaction so that we can determine whether or not the reaction is spontaneous at the given temperature.

ΔS^⦵_total = (-Δ_rH^⦵ / T) + ΔS^⦵_sys

Let’s look at some typical calculations …

1. Calculate ΔS^⦵_surr for the reaction between calcium and oxygen to form calcium oxide at 25°C given that the enthalpy change for the reaction is -1270 kJ mol^-1.

2. Calculate ΔS^⦵_total for the reaction between sulfur and hydrogen to form hydrogen sulfide at 0°C, given that the enthalpy change for the reaction is – 21 kJ mol^-1.

Substance	S⦵ / JK-1mol-1
S(s)	31.8
H2(g)	130.6
H2S(g)	205.7

There are some practice exam Q&A below 👇.

Calculating the temperature at which an endothermic reaction becomes feasible

In an exothermic reaction, (- Δ_rH^⦵ / T) will always be positive since Δ_rH^⦵ is negative, so ΔS^⦵_surr is always positive.

Conversely, in an endothermic reaction (- Δ_rH^⦵ / T) will always be negative since Δ_rH^⦵ is positive, so ΔS^⦵_surr is always negative. How negative depends on the temperature, which means that an endothermic reaction may become spontaneous if the temperature is high enough and ΔS^⦵_sys is sufficiently positive.

ΔS^⦵_total for any reaction is zero when that reaction has reached an equilibrium position. If we can calculate the temperature at which ΔS^⦵_total = 0 JK^-1mol^-1 (the temperature at which equilibrium is achieved), we will know the temperature at which the reaction becomes feasible.

Let’s look at an example …

The decomposition of calcium carbonate into calcium oxide and carbon dioxide is an endothermic reaction. Calculate the temperature at which this reaction becomes feasible or spontaneous given that ΔS^⦵_sys = +161 JK^-1mol^-1 and Δ_rH = +178 kJ mol^-1.

CaCO_3(s) ⇾ CaO_(s) + CO_2(g)

ΔS^⦵_total = (-Δ_rH^⦵ / T) + ΔS^⦵_sys and a reaction is feasible when ΔS^⦵_total = 0 since this is when the reaction is at equilibrium.

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

-161 = -178000 / T

T = 1106K

There are some practice exam Q&A below 👇.

---

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 here how to determine the temperature at which an endothermic reaction becomes feasible if we are using the following formula:

ΔG^⦵ = Δ_rH^⦵– TΔS^⦵_sys

(Personally, I’m not sure how you can genuinely understand Gibbs free energy without an understanding of entropy change of surroundings, so I would recommend you get your head around everything here 🧐).

ΔG^⦵ = Δ_rH^⦵ – TΔS^⦵_sys and at the minimum temperature required for the reaction to occur, ΔG^⦵ = 0.

0 = Δ_rH^⦵ – TΔS^⦵_sys

TΔS^⦵_sys = Δ_rH^⦵

T = Δ_rH^⦵ / ΔS^⦵_sys = +178 / + 0.161 = 1106K

(remember that with this formula we need to convert ΔS^⦵_sys into kJ K^-1 mol^-1 and leave Δ_rH^⦵ in kJ mol^-1)

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

---

Practice questions

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

1. (a) Calculate ΔS^⦵_total at 298K for the combustion of methane, given that Δ_rH^⦵ = -890 kJ mol^-1 .

CH_4(g) + 2O_2(g) ⇾ CO_2(g) + 2H₂O_(g)

(b) Calculate ΔS^⦵_total at 273K for the formation of sulfur trioxide, given that Δ_rH^⦵ = -99 kJ mol^-1 .

SO_2(g) + ½O_2(g) ⇾ SO_3(g)

–

2. (a) Calculate ΔS^⦵_sys for the reaction between methane and steam.

 CH_4(g)  +   H₂O_(g)  ⇾  CO_(g)   +   3H_2(g)       Δ_rH^⦵ = +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.

–

3. CaCO_3(s)  ⇾   CaO_(s)   +   CO_2(g)    Δ_rH^⦵ = +180 kJ mol^-1

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

(b)  Calculate a value for ΔS^⦵_sys .

(c)  Calculate ΔS^⦵_total at 298K.

(d)  Calculate the temperature at which ΔS^⦵_total is zero and comment on the significance of this temperature.

–

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

2NaHCO_3(s)   ⇾   Na₂CO_3(s)   +   H₂O_(l)  +  CO_2(g)                        Δ_rH^⦵ = +91 kJ mol^-1

–

5. H_2(g)   +   CO_2(g)    ⇌   H₂O_(g)   +  CO_(g)     Δ_rH = +40 kJ mol^-1

(a)  Calculate ΔS^⦵_sys for the forward reaction.

(b)  Calculate the temperature at which ΔS^⦵_total = 0. 

(c)  Comment on the equilibrium position when ΔS^⦵_total = 0.

–

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

2SO_2(g)   +  O_2(g)   ⇌  2SO_3(g)  Δ_rH^⦵ = -288.4 kJ mol^-1

Answers

1. (a)

(b)

2.

3. (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. ΔS_sys for this reaction will be positive as a gas is produced.

(b)  ΔS^⦵_sys = ∑S^⦵_products – ∑S^⦵_reactants

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

(c)  ΔS^⦵_total =  ΔS^⦵_surr +  ΔS^⦵_sys 

            =  (-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.

---

4. The minimum temperature is when ΔS^⦵_tot = 0

    0  =  (-Δ_rH^⦵ / T)  +  ΔS^⦵_sys 

    ΔS^⦵_sys =  ∑S^⦵_products – ∑S^⦵_reactants  =  (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!

5. (a)   ΔS^⦵_sys =  (189.0 + 197.6) – (130.6 + 214.0)

                =  42.0 JK^-1mol^-1 

(b)   0  = (-Δ_rH / T)  +  ΔS^⦵_sys 

         =  (-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 K_c=1.

6. ΔS^⦵_total = 0 when the reaction is at equilibrium and becomes feasible

       0  =  (-Δ_rH^⦵ / T) + ΔS^⦵_sys

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

  ΔS^⦵_sys =  ∑S^⦵_products – ∑S^⦵_reactants  

       = ((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)