Understanding entropy

Why does water freeze when you place it in a freezer? Certainly not because the freezer sucks the heat out of it. 

Entropy (S) is the true driving force of all spontaneous processes, chemical reactions included. Entropy is a property of matter, just like density – think of it as a quantity associated with the randomness or disorder of molecules and energy in a system. 

In this context, the ‘system‘ is everything under investigation which includes the chemical reaction or physical change itself (reactants and products), the solvent in which the reaction is taking place (usually water), the test tube or beaker, the air inside the test tube or beaker and if we have any other equipment involved such as a spatula for stirring or a thermometer measuring the temperature, we include that as well.

Spontaneous processes, once started, happen without a continued outside intervention and are always the result of an increase in the disorder of molecules and in the random distribution of energy in the universe (this its essentially the 2nd Law of Thermodynamics). 

ΔS =  S_final – S_initial  = positive value for any spontaneous change in a system

E.g.   Imagine a glass bulb containing a gas connected to a second glass bulb in which there is a vacuum. 

E.g.    What if we place hot metal block in close contact with a cold metal block?

Now we need to get a little more specific – for a spontaneous process the entropy of the Universe (ΔS_universe) must increase, not the system in isolation. 

E.g.     If we mix gaseous ammonia with gaseous hydrogen chloride we see clouds of white smoke of ammonium chloride 

ΔS is negative, the entropy of the system has decreased, and yet this is still a spontaneous process …. how can we explain what is going on?

This is a highly exothermic reaction releasing heat energy as the new strong bonds form between the NH₄⁺ and Cl^– ions, and this energy is moving from the system (our reaction) to the surroundings (the air). The molecules in the air absorb this heat, the temperature of the surroundings increases. Not only is there more energy to distribute, we should remember that now the air molecules have more energy they are moving around faster, and their disorder increases as well. However we look at it, there is an increase in the entropy of the surroundings. 

ΔS_universe  =  ΔS_surroundings  +  ΔS_system

Clearly, ΔS_surroundings must be sufficiently positive to compensate for  ΔS_system being negative as ΔS_universe is positive overall.

Recap:  

• in an exothermic process the system gives out heat to the surroundings and so ΔS_surr is positive
• In an endothermic process the system absorbs heat from the surroundings and so ΔS_surr is negative
• ΔS_universe is often written as  ΔS_total (we’ll use this term moving forward)

 Let’s return to the original question. 

H₂O_(l)   ⇾   H₂O_(s)          ΔS_sys is negative and ΔH = – 6010 Jmol^-1

This makes sense.  Energy is released as strong hydrogen bonds form and lock water molecules into a lattice of ice (water freezing is an exothermic process), and ΔS_sys decreases as in ice the water molecules have less energy to distribute and they themselves are not free to move – the system is becoming more ordered, less random. 

We know from everyday experience that water forming ice is a spontaneous process, but only at temperatures less than 0°C.

ΔS_total   =   ΔS_surr   +   ΔS_sys

ΔS_total  is only positive at temperatures < 0°C; ΔS_surr  is positive because this is an exothermic process and energy is being released to the surroundings; ΔS_sys is negative.

So what are we saying? The heat energy released is only sufficiently increasing the entropy of the surroundings to the point where ΔS_total is positive (overcoming the decrease in ΔS_sys) if those surroundings are cold enough (< 0°C)  to begin with. 

Imagine you have £10 in your bank account. If I give you £10 your overall worth has doubled  – the addition of that £10 has had a significant impact. However, if you had £10000 in your bank account and I gave you £10, your overall worth has not increased noticeably at all. 

It is the same with energy. If the surroundings are cold, a certain addition of energy has a big effect on increasing the entropy of those surroundings but if the surroundings are hot, adding the same fixed amount of energy leads to a far less significant increase in their entropy. 

When working under constant pressure (which is pretty much always as chemists) the heat absorbed by the surroundings, ΔH_surr^⦵, is identical to enthalpy change of the reaction, Δ_rH^⦵.

And so we can pull all of this together to give us an equation for ΔS^⦵_surr …

You can explore this further in the next post on entropy 👈.

Recap: 

• we can calculate the entropy change of the surroundings if we know the enthalpy change for the reaction and the temperature
• units of entropy =  J / K per mole  =  JK^-1mol^-1

But how do we calculate the entropy change of the system or reaction?

Luckily the absolute entropies of many substances, including phase changes, have been determined and we can simply look them up.

Substance	S⦵ / J k-1 mol-1
C(s, graphite)	5.7
Cu(s)	33.0
Ca(s)	41.4
Pb(s)	64.8
H2O(l)	69.9
H2O(g)	188.7
H2(g)	130.6
O2(g)	205.0
CO(g)	197.6
CH4(g)	186.2
CH3OH(l)	239.7
N2(g)	191.6
NH3(g)	192.3
NO(g)	211.0
CaO(s)	39.7
CS2(g)	238.0
H2S(g)	206.0

S^⦵ is the standard molar entropy. The standard conditions are the same as for enthalpy (101 kPa, 298K and the substance is in its standard physical state at this temperature and pressure).

In general, we can see that gases have higher standard molar entropies than liquids, which have higher entropies than solids. Molecules of compounds with fewer atoms have lower standard molar entropies than those with more atoms. Similarly, elements with a higher molar mass will have a higher standard molar entropy than those with a lower molar mass.

The entropy change of a system during a chemical reaction is simply the difference between the total standard molar entropies of the products and the reactants.

ΔS_sys^⦵  =  ∑ S^⦵_products   –   ∑ S^⦵_reactants

E.g. calculate ΔS^⦵_sys for the reaction between hydrogen and nitrogen to produce ammonia

3H_2(g) + N_2(g) ⇾ 2NH_3(g)

∑ S^⦵_reactants = (130.6 x 3) + (191.6) = 583.4 J k^-1 mol^-1

∑ S^⦵_products = (192.3 x 2) = 384.6 J k^-1 mol^-1

ΔS^⦵_sys  = 384.6 – 583.4 = – 198.8 J k^-1 mol^-1

For most reactions, we can make a qualitative prediction as to whether ΔS^⦵_sys is positive or negative based on the generalisations outlined above.

E.g. if we dissolve ammonium nitrate, NH₄NO_3(s), in water to form a solution, NH₄NO_3(aq), we would expect an increase in entropy and ΔS^⦵_sys would be positive. In solution the NH₄⁺ ions and NO₃^– ions are no longer arranged in a lattice with strong ionic bonds but are able to move freely. There is clearly more disorder in the system as there are far more ways to arrange the ions.

E.g. if we burn magnesium ribbon in oxygen (Mg_(s) + ½O_2(g) ⇾ MgO_(s)) we would expect a decrease in entropy as a gaseous reactant becomes a solid product and two reactants become a single product.

Practice questions

1. State and explain whether there would be a positive or negative change in entropy for each of the following processes:

(a)  the freezing of water, H₂O_(l) ⇾ H₂O_(s)

(b)  calcium carbonate reacting with nitric acid, CaCO_3(s) + 2HNO_3(aq) ⇾ Ca(NO₃)_2(aq) + H₂O_(l) + CO_2(g)

(c)  forming ozone from molecular oxygen, O_2(g) + O^• ⇾ O_3(g)

2. Explain why solutes such as sodium chloride dissolving in water are accompanied by a positive entropy change.

3. Write an equation for the reaction between aqueous barium nitrate and aqueous sodium sulphate, including state symbols. Explain the change in entropy for this reaction.

4. Write an equation to represent the standard atomisation of bromine, including state symbols. Explain the change in entropy for the process.

5. Which of the following reactions is accompanied by a negative ΔS^⦵_sys?

(a) 2H₂O_2(l)   ⇾   2H₂O_(l)  +  O_2(g)

(b) 2Mg_(s)   +   O_2(g)   ⇾   2MgO_(s)

(c) SrCO_3(s) + H₂SO_4(aq)  ⇾   SrSO_4(aq)  +  H₂O_(l)  +  CO_2(g)

(d) Fe_(s)  +  2HCl_(aq)   ⇾   FeCl_2(aq)  +  H_2(g)

6. Predict whether the entropy of the following reactions increases or decreases:

(a)  CO_2(g)  ⇾  CO_2(s)

(b)  NaCl_(s)  +  (aq)  ⇾  NaCl_(aq)

(c)  N_2(g)  +  3H_2(g)  ⇾  2NH_3(g)

7. Calculate ΔS^⦵_sys for the following reactions using the data in the table above:

(a) C_(s) + H₂O_(g) ⇾ CO_(g) + H_2(g)

(b) 2Ca_(s) + O_2(g) ⇾ 2CaO_(s)

(c) CO_(g) +  2H_2(g)  ⇾  CH₃OH_(l)

(d) 4NH_3(g)   +  5O_2(g)   ⇾    4NO_(g)   +   6H₂O_(g)

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8. Consider the reaction between hydrogen and carbon sulfide:

4H_2(g)  +  CS_2(g)   ⇾  CH_4(g)  +  2H₂S_(g)      ΔS_sys = -164.0 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.

Answers

1. (a)  negative; in ice the molecules are locked into a regular crystalline lattice so there are fewer ways to arrange the molecules and the energy – the system is less disordered

(b)  positive; a solid and a solution are reacting to form a gas (CO₂) and a solution of ions so the products are more disordered than the reactants 

(c)  negative; a molecule of ozone is formed from a molecule of oxygen and an oxygen radical – there is less disorder as a result

2. Aqueous ions are more disordered than ions in a solid crystalline lattice.

3. Ba(NO₃)_2(aq)   +   Na₂SO_4(aq)   ⇾   BaSO_4(s)   +   Na₂(NO₃)_(aq)

There is a decrease in entropy / negative entropy change because a solid precipitate is formed and so there is less disorder / fewer ways to arrange the particles and the energy.

4. ½Br_2(g)   ⇾   Br_(g)  (doubling up is not a correct answer – you need to know your definitions!); there is an increase in entropy / a positive entropy change because there is more disorder and more ways of arranging the atoms / particles and energy. 

5. B (all the other reactions produce a gas so entropy of system increases / is positive)

6. (a) and (c) decrease, (b) increase

7. (a) (197.6 + 130.6) – (5.7 + 188.7) = + 133.8 J k^-1 mol^-1 (always put the sign in)

(b) (2 x 39.7) – [(2 x 41.4) + 205.0] = – 208.4 J k^-1 mol^-1

(c) (239.7) – [197.6 + (2 x 130.6)]  =  – 219.1 JK^-1mol^-1 

(d) [(4 x 211.0) + (6 x 188.7)] – [(4 x 192.7) + (5 x 205.0)] = +180.4 JK^-1mol^-1 

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

(b)  ΔS_sys =  ∑S_products – ∑S_reactants  

-164.0  =  (186.2 + (2 x 206.0)) – (4𝓧 + 238.0)

    -164.0  =  598.2  –  4𝓧  – 238.0

    -164.0  =  360.2  –  4𝓧

    4𝓧  =  524.2

    S^⦵H₂ = 131.1 JK^-1mol^-1