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. 

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 =  Sfinal – Sinitial  = 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 (ΔSuniverse) 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 

 NH3(g)      +      HCl(g)      ⇾    NH4Cl(s)

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. 

ΔSuniverse  =  ΔSsurroundings  +  ΔSsystem

Clearly, ΔSsurroundings must be sufficiently positive to compensate for  ΔSsystem being negative as ΔSuniverse is positive overall.

Recap:  

  • in an exothermic process the system gives out heat to the surroundings and so ΔSsurr is positive
  • In an endothermic process the system absorbs heat from the surroundings and so ΔSsurr is negative
  • ΔSuniverse is often written as  ΔStotal

 Let’s return to the original question. 

H2O(l)   ⇾   H2O(s)          ΔSsys 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 ΔSsys 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.

ΔSuniv   =   ΔSsurr   +   ΔSsys

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

So what are we saying? The heat energy released is only sufficiently increasing the entropy of the surroundings to the point where ΔSuniv is positive (overcoming the decrease in ΔSsys) 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 change is identical to enthalpy change ΔH.

Units of entropy =  J / K per mole  =  JK-1mol-1

Recap: 

  • we can calculate the entropy change of the surroundings if we know the enthalpy change for the reaction and the temperature.

And finally, how to 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. 

ΔSsys  =  ∑ Sproducts   –   ∑ Sreactants

(there is a worked example of this calculation in the video)

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

(b)  calcium carbonate reacting with nitric acid

(c)  forming ozone from molecular oxygen

  1. Explain why solutes such as sodium chloride dissolving in water are accompanied by a positive entropy change.
  1. 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.
  1. Write an equation to represent the standard atomisation of bromine, including state symbols. Explain the change in entropy for the process.
  1. Which of the following reactions is accompanied by a negative ΔSsys?

(a) 2H2O2(l)   ⇾   2H2O(l)  +  O2(g)

(b) 2Mg(s)   +   O2(g)   ⇾   2MgO(s)

(c) SrCO3(s) + H2SO4(aq)  ⇾   SrSO4(aq)  +  H2O(l)  +  CO2(g)

(d) Fe(s)  +  2HCl(aq)   ⇾   FeCl2(aq)  +  H2(g)

6. Predict 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)


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)   +   Na2SO4(aq)   ⇾   BaSO4(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. ½Br2(g)   ⇾   Br(g)  (doubling up is not a correct answer); 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

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