IB syllabus > energetics (sl) > 6.5 

These notes were written for the old IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

15.3 - Entropy


15.3.1: State and explain the factors that increase the entropy of a system.


Entropy has been described as nature's arrow. It is the natural tendency of everything towards disorder.

There are two factors that contribute to the degree of disorder of a system.

Of these three factors, by far the most important are the number of particles and the available energy (temperature)

Gases have the greatest entropy as their particles are free to move randomly

An increase in disorder can result from the mixing of different types of particles, changes of state (increased distance between particles), increased movement of particles or increased numbers of particles. An increase in the number of particles in the gaseous state usually has a greater influence than any other possible factor.

 


15.3.2: Predict whether the entropy change (ΔS) for a gives reaction or process is positive or negative.


The entropy of reactants can be gauged by considering the number of moles of gas on the left hand side of a reaction. The entropy of the products can be considered in the same way. To decide whether the entropy increases or decreases we look at whether the numberof moles of gas increases or decreases in the course of the reaction:

Example: The Haber process

N2 + 3H2 2NH3

On the left hand side there are 4 moles of gas and on the right hand side there are 2 moles of gas. The moles of gas have decreased from 4 to 2, therefore the total entropy has decreased.


15.3.3: Calculate the standard entropy change of a reaction (ΔSº) using standard entropy values (Sº).


Calculation of standard entropy change is similar to the assessment carried out in the previous section, only this time useing the actual numbers given for the absolute entropy ogf each substance.

Example: The Haber process

N2 + 3H2 2NH3

The absolute entropy values are:

  • nitrogen: 192 J K-1 mol-1
  • hydrogen: 131 J K-1 mol-1
  • ammonia: 193 J K-1 mol-1

On the left hand side there is 1 mole of nitrogen and 3 moles of hydrogen, hence the total absolute entropy on the left hand side = 192 + 3(131) = 585 J K-1

On the right hand side there are 2 moles of ammonia, hence the total absolute entropy on the right hand side = 2(193) = 386 J K-1

Hence the entropy has decreased from 585 to 386 = -199 J K-1

As can be seen the operation that is carried out is:

Entropy change = sum of the products entropy - sum of the reactants entropy


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