Enthalpy of Solution - Calculations (Edexcel International A Level Chemistry)

Revision Note

Philippa

Author

Philippa

Last updated

Enthalpy of Solution - Calculations

  • Questions in this topic typically ask you to calculate the hydration enthalpy of one of the ions, given the lattice enthalpy, enthalpy of solution and hydration enthalpy of the other ion.
  • This can be done by constructing an appropriate energy cycle and using Hess's Law to find the unknown energy value
  • The energy cycle above shows that there are two routes to go from the gaseous ions to the ions in an aqueous solution:
      • Route 1: going from gaseous ions → ionic solid → ions in aqueous solution (this is the indirect route)
      • Route 2: going from gaseous ions  →  ions in aqueous solution (this is the direct route)

  • According to Hess’s law, the enthalpy change for both routes is the same, such that:

ΔhydH= ΔlattH + ΔsolH

  • Each ion will have its own enthalpy change of hydration, ΔHhyd, which will need to be taken into account during calculations
    • The total ΔhydHis found by adding the ΔhydHvalues of both anions and cations together

Worked example

Constructing an energy cycle for KCl

Calculate the enthalpy of hydration of the chloride ion given the following data:

ΔlattH [KCl] = -711 kJ mol-1

ΔsolH[KCl] = +26 kJ mol-1

ΔhydH[K+] = -322 kJ mol-1

Answer

Step 1: Draw the energy cycle and make ΔhydH[Cl-] the subject of the formula:

Worked Example -Energy Cycle KCl, downloadable AS & A Level Chemistry revision notes

Step 2: Substitute the values to find ΔhydH[Cl-]

ΔhydH[Cl-]   = (-711) + (+26) - (-322) = -363 kJ mol-1

Alternative Diagram

  • You can also draw a Born-Haber cycle as an alternative approach to the same problem

Energy level diagram:

Chemical Energetics - Energy Level Diagram KCl (1), downloadable AS & A Level Chemistry revision notesChemical Energetics - Energy Level Diagram KCl (2), downloadable AS & A Level Chemistry revision notes

Worked example

Constructing an energy cycle and energy level diagram of MgCl2

Construct an energy cycle to calculate the ΔhydHof magnesium ions in magnesium chloride, given the following data:

ΔlattH [MgCl2] = -2592 kJ mol-1

ΔsolH [MgCl2] = -55 kJ mol-1

ΔhydH [Cl-] = -363 kJ mol-1

Answer

Step 1: Draw an energy cycle:

Worked Example - Energy Cycle MgCl2, downloadable AS & A Level Chemistry revision notes

Step 2: Substitute the values to find ΔHhyd [Mg2+]

ΔHhyd[Mg2+]  = (-2592) + (-55) - (2 x -363) = -1921 kJ mol-1

Alternative route to find ΔhydH[Mg2+

  • Here is the same solution using a Born-Haber cycle

Chemical Energetics - Energy Level Diagram MgCl2, downloadable AS & A Level Chemistry revision notes

Examiner Tip

It doesn't matter whether you use Hess cycles or Born-Haber style cycles to solve these problems as long as the information is correctly labelled and the direction of the arrows matches the definitions.Exam problems in this topic often show diagrams with missing labels which you have to complete and find unknown values.The key to success in energy cycle calculations is not to panic, but have a careful step-by-step approach, show your workings and use brackets to separate mathematical operations from the enthalpy changes.

Enthalpy of Solution - Predictions

  • When an ionic compound dissolves in water, there are two changes that need to take place
    • The lattice structure needs to be broken down and the ions must become hydrated
  • The breaking down of the lattice structure is endothermic (equivalent to the reverse lattice energy) it also results in an increased number of moles of particles present, therefore entropy will increase
  • The hydration of the ions is an exothermic process but results in water molecules becoming more ordered as they arrange themselves around the cations and anions 
  • This increasing in order decreases the entropy of the water 
  • Therefore to explain why some ionic compounds are soluble and some are insoluble we must consider both entropy and enthalpy changes involved 
  • As we have seen previously                                                           

ΔStotal = Δ Ssystem+ Δ Ssurroundings

      And since

Δ Ssurroundingsbegin mathsize 14px style equals negative fraction numerator increment subscript s o l end subscript H over denominator T end fraction end style

 The expression can become

Δ Stotal= Δ Ssystem begin mathsize 14px style negative fraction numerator increment subscript s o l end subscript H over denominator T end fraction end style

  • Therefore the solubility of an ionic solid depends upon three factors
    • The entropy of the system, ΔSsystem
    • The enthalpy change of solution, ΔsolH
    • The temperature, in K, of the water, T

If we consider ammonium nitrate, NH4NO3 (s), at 298 K

NH4NO3 (s) → NH4+ (aq) + NO3- (aq)

  • The enthalpy change of solution, Δsol= +25.8 kJ mol-1
  • The temperature, in K, of the water, T = 298 K

ΔSsystemS[NH4+ (aq)] + S[NO3- (aq)] - S[NH4NO3 (s)]

= +113.4 + 146.4 - 151.1

= +108.7 J K-1 mol-1 

begin mathsize 16px style increment S subscript italic s italic u italic r italic r italic o italic u italic n italic d italic i italic n italic g italic s italic space end subscript equals space minus sign fraction numerator increment subscript italic s italic o italic l end subscript H over denominator italic T end fraction end style

 = begin mathsize 16px style negative space fraction numerator plus 25800 over denominator 298 end fraction end style

= -86.6 J K-1 mol-1

  • Therefore the total entropy is: 

ΔStotal = Δ Ssystem+ Δ Ssurroundings

= +108.7 + (-86.6)

= +22.1 J K-1 mol-1

  • If the entropy change is positive, as it is for dissolving ammonium nitrate in water at 298 K, then this is thermodynamically spontaneous 
  • The activation energy (Ea) for this reaction is very low, we can conclude that at 298 K, ammonium nitrate is soluble in water

You've read 0 of your 10 free revision notes

Unlock more, it's free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Philippa

Author: Philippa

Expertise: Chemistry

Philippa has worked as a GCSE and A level chemistry teacher and tutor for over thirteen years. She studied chemistry and sport science at Loughborough University graduating in 2007 having also completed her PGCE in science. Throughout her time as a teacher she was incharge of a boarding house for five years and coached many teams in a variety of sports. When not producing resources with the chemistry team, Philippa enjoys being active outside with her young family and is a very keen gardener.