These are word documents, describing the different methods of Hess Law!

Exam questions (in word format)

Enthalpy changes questions

# Experimental Measurement of Enthalpy Change

Measurement of enthalpies of combustion

Measuring enthalpy changes for reactions in solution.

Hess' Law and the calculation of enthalpy change

Calculations using Hess' Law

Need to understand the difference between the two terms heat and temperature. While they are related, they are not the same thing!

TEMPERATURE

• Measures the kinetic energies of molecules present in a substance.
• Independent of the amount of substance present.

HEAT

• Measure of the total energy of a substance.
• Depends on the amount of substance present.

So a bucket full of water at 50ºC would have the same temp. as a 250ml beaker of water at the same temperature, but the heat content of the bucket would be bigger.

We need a new term.

Specific Heat Capacity, is the amount of heat required to raise the temperature of 1g of a substance by 1K.
· Differs from substance to substance.
· Water = 4.2 J g-1 K-1
· Ethanol = 2.4 J g-1 K-1

This new term gives us a new equation to learn:

Heat transfer = mass x specific heat capacity x temperature change

Heat transfer = m c  DT

Units are absolutely essential. Specific Heat Capacity is normally given symbol c in units J g-1 K-1. As a result m must be in grams and DT can either be in ºC or K.

Calorimetry was invented to study the heat transfer to water in this way using a calorimeter.

## Measurement of enthalpies of combustion.

This is basically a method identical to that alcohol's investigation!

• Known mass of water placed in beaker (150g)
• Temperature noted (23 ºC)
• Final temperature noted (43ºC)
• Mass of ethanol needed to obtain heat increase (0.9g)

Therefore

Heat gained by water in calorimeter = m x c x D

= 150g x 4.2 J g-1 K-1 x 20ºC

= 12,600 J Heat produced by burning 0.9g of ethanol = 12.6kJ

A simple calorimeter.

Since 0.9g is 0.9/46 mol ethanol:

Heat produced by burning 1 mol of ethanol = 12.6 kJ / (109/46) kJ mol-1= 644kJ
Therefore DHc [CH3CH2OH] = -644kJ mol-1

Answer is much smaller than the accepted value of -1371 kJ mol-1.

• Heat is lost to surroundings despite precautions
• Some heat is given to the calorimeter (the glass beaker). We could allow for this if we calibrated the calorimeter.
• Incomplete combustion of reactants, leading to formation of soot and carbon monoxide.

For accurate methods look up in Fullick p114, look up and make a brief note including diagram - important to know what is inside the bomb.

## Measuring enthalpy changes for reactions in solution.

This is normally called molar enthalpy of solution, or in this case we can be more specific and call it molar enthalpy change of neutralisation.

Apparatus needed:

• Polystyrene cup - vacuum flask used in real life.
• Accurate thermometer (± 0.2 ºC)
• 1.0 mol dm-3 HCl (aq) & 1.1 (or greater) mol dm-3 NaOH (aq).
• Large beaker and insulation (optional).

Method:

• Pipette accurately 50 cm3 of the hydrochloric acid into the insulated polystyrene beaker.
• Note temperature.
• Add - stirring continuously - 50 cm3 NaOH. Observe the thermometer and record maximum temperature.

Calculation:

• Assume density to be close to that of water 100 cm3 hence total mass = 100g.
• Specific Heat Capacity = 4.2 J K-1 g-1.
• Temperature rise (DT) = (30.1 - 23.5)°C = 6.5°C

Heat absorbed by solution = m x c x DT
= 100g x 4.2 J K-1 g-1 x 6.5°C
= 2730 J

No. moles HCl = 1.0 mol dm-3 ´ (50cm3/1000cm3)
= 54,600 J

DHN = -54.6 kJ mol-1

This compares quite nicely with the accepted figure of -57.1 kJ mol-1

Go through temperature correction method - if necessary - for slower reactions where there will be a greater heat loss. This is the method we did in class with the graph which actually gets some pretty impressive results!

## Hess' Law and the calculation of enthalpy change

Most enthalpy changes cannot be measured directly by experimental methods. An indirect method must be used.

These calculations are based on The first Law of Thermodynamics

### Energy cannot be created or destroyed. It can only be converted from one form to another.

From this two important deductions are made:

1. If the enthalpy change of a reaction is known, then the enthalpy change of the reverse reaction is the same but with the sign changed

C (graphite, solid) + O2(g) ®  CO2(g) DH = -393 kJ mol-1

CO2(g) ®  C (graphite, solid) + O2(g) DH = +393 kJ mol-1

### NB The positive sign for an endothermic change is necessary! DO NOT FORGET IT!

2. Hess' Law which states the following:

### The enthalpy change for a reaction is independent of the route by which the reaction is achieved, but depends only on the initial and final stages.

This needs to be learnt like a parrot!

Hess' Law allows us to calculate enthalpy changes that would otherwise cause a problem. This is a key issue remember this as you are struggling through yet another calculation

## Calculations using Hess' Law

There are as many different techniques as there are chemists.

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