title: lesson 2 measuring enthalpy changes learning objectives: – understand the technique of...

Title: Lesson 2 Measuring Enthalpy Changes

Learning Objectives:– Understand the technique of calorimetry, including the

assumptions underpinning it

– Calculate enthalpy changes from experimental data

– Complete a calorimetry experiment

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Which is true for a chemical reaction in which the products have a higher enthalpy than the reactants?

Reaction ∆HA. endothermic positiveB. endothermic negativeC. exothermic positiveD. exothermic negative

True or False?

1) Exothermic reactions have a negative ΔH

2) Endothermic reactions have a negative ΔH

3) Breaking bonds is an exothermic reaction

4) In endothermic reactions, the products have less energy than the reactants

5) Standard temperature is 275K

6) Standard pressure is 100kPa

7) Standard enthalpy of combustion only involves fuels

8) Photosynthesis is an exothermic reaction

9) Endothermic reactions absorb energy

10) In E = mcΔT, c is the number of moles of compound used

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7.3 MEASURING ENTHALPY CHANGES

HEAT & TEMPERATURE - What is the difference?

Temperature• Average kinetic energy of the particles• Independent of the number of particles

Heat• Total energy of all particles• Energy of every particle is included• Dependent on the number of particles• Heat always flows from high low temp

Q: What has more heat, luke warm swimming pool or a red hot nail? Why?


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Heat changes can be calculated from temperature changes If the same amount of heat is added to two different objects,

the ΔT will not be the same, as the average kinetic energy of the particles will not increase by the same amount.

Smaller number of particles will experience a larger temperature increase.

The increase in temperature when an object is heated will depend on:

- Mass of object- Heat added- Nature of substance

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Specific heat capacity, C The specific heat capacity of a substance is

the amount of energy required to raise the temperature of one gram (g) by one Kelvin (K).

Specific heat capacity is different for different substances:

Substance Specific Heat CapacityJ g-1K-1

Water 4.18

Ethanol 2.44

Air 1.00

Iron 0.450

Copper 0.385

DIFFERENT TYPES OF ΔH

Standard enthalpy change of reaction (ΔHr)

Enthalpy change when a reaction occurs in the molar quantities shown in the balanced chemical equation under standard conditions

Standard enthalpy change of formation (ΔHf)

Enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions

e.g. 2C (s) + 3H2 (g) + ½O2 (g) C2H5OH (l)

Standard enthalpy change of combustion (ΔHc)

Enthalpy change when 1 mole of a substance is completely burned in oxygen under standard conditions in standard states


USING THE EQUATION q = mcΔT

No instrument measures heat directly

To measure enthalpy change we transfer the heat to a known mass of another substance or material, usually water

q (or ∆H) = mcΔT


q = enthalpy or heat change (J)

m = mass (in g) of solution in calorimeter (either directly or indirectly heated) this is also same as volume in cm3

ΔT = change in temp of the solution (K)

c = specific heat capacity of solution (4.18 Jg-1K-1)

∆H can easily be converted to a molar value by dividing it by the number of moles of reactant (Jmol-1)

Assumptions:

m is just the total mass of water usedThis is valid as the mass of water used is much greater than the mass of any of the other substances

C is just the specific heat capacity of water, ignoring the reactantsThis is valid as the specific heat capacity of water is much higher than most other substances, so they absorb very little of the heat

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Relationship between different objects

Use heat capacity (c) Defined as the heat needed to increase the temperature of

an object by 1K.

A swimming pool has a larger heat capacity than a kettle.

Note the different between specific heat capacity and heat capacity!

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Complete the test yourself questions Page 191 Questions 1-4 Check your answers on page 562

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Solutions

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Enthalpy changes and the direction of change

An exothermic reaction can be compared to a person falling off a ladder. Both changes lead to a decrease in stored energy.

Products are more stable than the reactants.

• However, stability is relative. E.g. Hydrogen peroxide is stable compared to its elements but unstable compared to its decomposition products water and oxygen…

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Falling up the ladder… We do not expect a person to fall up a ladder

but endothermic reactions do occur…

Endothermic reactions are less common and occur when there is an increase in disorder of the system, e.g. owing to a formation of a gas…

HOW CAN WE CALCULATE ENTHALPY CHANGES?


Simple calorimeters Bomb calorimeter

There are two ways of measuring enthalpy changes:

• Measuring enthalpy changes of combustion

• Measuring enthalpy changes of reaction

KEY IDEASSpecific heat capacity (c) – the amount of energy needed to raise the temperature of 1g of substance by 1K. For water and solutions it is 4.2 Jg-1K-1

KEY IDEASTwo step calculations.1)Work out the energy transferred2)Work out the energy supplied per mole.


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Calorimetry – Enthalpy of combustion

Calorimetry is used to measure the amount of heat released/absorbed in a reaction.

The reaction is used to heat some water, and the temperature change measured

If we know the mass of water used, the specific heat capacity of the water and the temperature change, we can calculate the heat change.

If we are being accurate, we should also take into account the heat capacity of the calorimeter itself as this also heats up.

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Enthalpy of combustion experiment


The bomb calorimeter

CALCULATING ENTHALPY CHANGE OF COMBUSTION

STEP 1:

Calculate the heat lost or gained using q = mcΔTConvert Joules to kilojoules

STEP 2:

Calculate number of moles of one of fuel that caused the enthalpy change, from the mass that reacted use n = mass ÷ Molar mass

STEP 3:

Calculate ΔHc (in kJmol-1) using actual heat change (q in kJ) and number of moles of fuel that burned (n)

Use equation: ΔHc = q ÷ n

CALCULATING ENTHALPY CHANGE OF COMBUSTION

E.g. 1.16g of a fuel is burned in oxygen raising the temp of 100g of water from 295.3K to 357.8K. Mr of the fuel is 58. Calculate ΔHc

Step 1:

q = mcΔT = 100 x 4.18 x (357.8 – 295.3) = 26125J = 26.125kJ

Step 2:

n = mass ÷ Mr = 1.16 ÷ 58 = 0.02 moles of fuel

Step 3:

ΔHc = q ÷ n = 26.125 ÷ 0.02 = -1306 kJmol-1

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The IB data booklet value is -1367 kJ/mol…

Can you think of 3 reasons why the experimental value is less than the literature value?

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The difference between the values can be accounted for by any of the following factors:

Not all the heat produced by the combustion reaction is transferred to the water. Some is needed to heat the calorimeter and some has passed to the surroundings.

The combustion is unlikely to be complete due to the limited oxygen available.

The experiment was not performed under standard conditions.

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Solutions

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Enthalpy changes are determined by simple calorimetry experiments.

Known volumes, masses (m) and concentrations are used. The initial (pre-reaction) temperature is taken (t1

OC). The reaction is allowed to proceed and then the final temperature is recorded (t2

OC).

Overall temperature change ΔT is t2OC- t1

OC

ENTHALPY CHANGES OF REACTION


Enthalpy of neutralization experiment

CALCULATING ENTHALPY CHANGE OF REACTION


STEP 1:

Calculate the heat lost or gained using q = mcΔTConvert Joules to kilojoules

STEP 2:

Calculate number of moles (n) of one of the reactants that caused the enthalpy change, from the mass that reacted use n = mass ÷ molar mass OR n = (concn x volume) ÷ 1000

STEP 3:

Calculate ΔHr (in kJmol-1) using actual heat change (q in kJ) and number of moles (n)

Use equation: ΔHr = (q ÷ n) x number of moles reacting in balanced

equation



E.g. 30g of ammonium chloride (NH4Cl) is dissolved in water. Temperature decreases from 298K 296K. Total mass of the solution was 980g. Calculate the standard enthalpy change of reaction

Balanced equation: NH4Cl (s) NH4+ (aq) + Cl- (aq)

Molar mass, M, of NH4Cl = 14 + (4 x 1) + 35.5 = 53.5 gmol-1



Step 1:

q = mcΔT = 980 x 4.18 x (298-296) = 8192.8J = 8.1928kJ

Step 2:

n = mass ÷ molar mass = 30g ÷ 53.5gmol-1 = 0.5607 moles of NH4Cl

Step 3:

The balanced reaction involves 1 mole of NH4Cl, so:

ΔHr = (q ÷ n) x 1 = 8.1928 ÷ 0.5607 = 14.6 kJmol-1

Remember: IF temp increases need to add a minus to q value!


Calorimetry calculation examples

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Another example

When 200 cm3 of 1.00 mol dm-3 sodium hydroxide solution was added to 200 cm3 of a 0.400 mol dm-3 solution of sulphuric acid, the temperature rose from 24.5oC to 30.0oC and a neutral solution was obtained. Determine the enthalpy change when one mole of sulphuric acid is fully neutralised by sodium hydroxide.

Determine ∆H: ∆H = -m.C∆.T ∆H = -(200+200) x 4.18 x (30.0-24.5) = -9,196 J

Determine n(H2SO4): n(H2SO4) = conc. x vol. = 0.400 x (200/1000) = 0.0800 mol

Determine Molar ∆H: Molar ∆H = ∆H / n(H2SO4) = -9196 / 0.0800 = 114950 J = -115 kJ mol-1


Using experimental results

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Sources of error and assumptions for measuring enthalpy in a polystyrene cup

Heat is lost from the system as soon as the temperature rises above the temperature of the surroundings (around 20oC)

Maximum recorded temperature is lower than the true value obtained in a perfectly insulated system.

Excess zinc powder is added to a calorimeter of a known volume of copper sulphate solution

We can make allowances for heat loss by extrapolating the cooling line back to when the reaction started…

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To proceed with our calculations we have made the following assumptions…

1. No heat loss from the system2. All heat goes from reaction to the water3. The solution is dilute: V(CuSO4) = V(H2O)

4. Water has a density of 1.00gcm-3

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The solution is dilute: V(CuSO4) = V(H2O)

Water has a density of 1.00gcm-3

Concentration

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Calorimetry calculation problems

SUMMARY QUESTIONS


1. 0.05 mol of a compound dissolves in water, causing the temperature to increase from 298K to 301K. Total mass is 220g. Calculate the enthalpy change in kJmol-1. Assume c = 4.18 Jg-1K-1

2. A calorimeter, containing 200g of water, was used to calculate ΔHc of pentane (C5H12) (Mr = 72). 0.5 g of pentane was burnt, increasing the water temp by 29K. a) calculate ΔHc; b) Suggest reasons why your value may be different to the value given in a data book

3. ΔHc of octane is -5512 kJmol-1, its Mr is 114. Some octane was burnt in a calorimeter with 300g of water. The temperature increase was 55K. Calculate the mass of propane burnt.

SELF ASSESS


1. 0.05 mol of a compound dissolves in water, causing the temperature to increase from 298K to 301K. Total mass is 220g. Calculate the enthalpy change in kJmol-1. Assume c = 4.18 Jg-1K-1

q = mcΔT = 220 x 4.18 x (301 – 298) = 2758.8 J = 2.7588 KJ

ΔH = q ÷ n = 2.7588 ÷ 0.05 = -55.2 kJmol-1

REMEMBER: the minus symbol as temp went up!

SELF ASSESS


2. A calorimeter, containing 200g of water, was used to calculate ΔHc of pentane (C5H12) (Mr = 72). 0.5 g of pentane was burnt, increasing the water temp by 29K. a) calculate ΔHc; b) Suggest reasons why your value may be different to the value given in a data book

q = mcΔT = 200 x 4.18 x 29 = 24244 J = 24.244 KJ

n = m ÷ Mr = 0.5g ÷ 72 = 0.00694 moles of fuel

ΔH = q ÷ n = 24.244 ÷ 0.00694 = -3490 kJmol-1

b) Some heat from combustion will be transferred to the surroundings and not the water. May not have been complete combustion. May not have been standard conditions. Inaccuracies in the measuring equipment

SELF ASSESS


3. ΔHc of octane is -5512 kJmol-1, its Mr is 114. Some octane was burnt in a calorimeter with 300g of water. The temperature increase was 55K. Calculate the mass of propane burnt.

q = mcΔT = 300 x 4.18 x 55 = 68970 J = 68.97 KJ

ΔH = q ÷ n rearranges to n = q ÷ ΔH = 68.97 ÷ 5512 = 0.0125 mol

n = mass ÷ Mr rearranges to mass = n x Mr

= 0.0125 x 114= 1.43g octane

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Calorimetry in Practice

In this experiment you will determine the enthalpy change for the reaction of magnesium with sulphuric acid.

To do this accurately, you will first need to determine the heat capacity of the calorimeter.

Follow the instructions here

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Key Points

∆H = -m.C.∆T Units: Joules, J

Assumptions: Specific heat capacity of solutions is the same as

that of water Total mass is the same as the volume of solution

used

title: lesson 2 measuring enthalpy changes learning objectives: – understand the technique of...

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