Industrial Applications of Chemical reactions

Unit 26Enthalpy Changes and Hess’s Law

Principles of energetics

• Law of conservation of energy• Energy cannot be created or destroyed but only

changed from one form into another.• Chemical changes • Chemical reactions are accompanied by energy

changes which are usually in the form of heat. Where heat is released to the surroundings is called EXOTHERMIC and when heat is taken in from the surroundings is called ENDOTHERMIC

Enthalpy change

• It is impossible to measure the total heat energy of a system but we can measure changes in the heat energy when a reaction occurs.

• This is called the enthalpy change. It can be positive or negative.

Enthalpy Change• Usually there is energy change involved in a chemical

reaction somehow. In chemistry this change in heat energy is represented as enthalpy change using the following symbol:

• So that this figure is the same everywhere and fair, it is taken at standard conditions which are 298K (250C),100 kPa pressure (1atmosphere) and 1 moldm-3.

Standard conditions

Accurate comparisons of enthalpy changes should be measured under standard conditions and these are set as being 25oC (298K) and one standard atmosphere pressure with any solutions involved at a concentration of 1.00moldm-3 and with all other reacting species in their standard states.

Two examples:

For the first rection, the enthalpy change is negative this is because the reaction is exothermic meaning it gives out heat. The reason it is negative is energy is leaving the reaction.The second reaction has a positive enthalpy change, this is because it is endothermic and heat is entering the system.

Enthalpy of Reaction, Combustion, and Formation

• ΔHr is standard enthalpy of reaction this is the energy change of a reaction in the quantities that are expressed. It uses the symbol r.

• ΔHc is the standard enthalpy of combustion is the enthalpy change when 1 mole of substance is completely burned in excess oxygen under standard conditions, all reactants in their standard states. The enthalpies of combustion can be determined using a calorimeter which is explained below; it uses the symbol c.

• ΔHf is the standard enthalpy of formation is the enthalpy change when 1 mole of compound is produced from its elements in standard conditions, all products and reactants in the standard state. The standard state is important, it means the way that the element is, at standard conditions (see above). So you would have H2(g) NOT H on its own. Therefore you sometimes need to do fractions of elements, as with the standard enthalpy of formation of ethanol:

• C (s) + 2H2 (g) + ½O2 (g) CH3OH (l)

Using Calorimetry to measure enthalpy change

• This is a method we can use to find out what the enthalpy changes of reactions are. Typically you do this using some sort of alcohol heating up some water. To do it correctly, you would have it in a closed system so no heat escapes, but this is not usually possible in a school or college lab. Once the water has been heated up you must use the following equation:

specific heat capacity is the energy required to raise the temperature of one gram of substance by 1 kelvin (or celcius depending on what you're using). It is measured in Jg-1K-1 and water has a specific heat capacity of 4.18 Jg-1K-1.

Problem:

• 1g of ethanol is burned to heat 100g of water, raising its temperature by 42K. Calculate the standard enthalpy of combustion.

• First we calculate the energy transferred:Q = m c ∆T

Q = 100g x 4.18 Jg-1K-1 x 42K

Q = 17556 J or 17.6 kJ

To find the kJ per mole divide the energy released by the number of moles of ethanol:

Number of moles = mass/Mr = 1/46 =

So ∆Hѳc = -17.6/0.02174 =

Note: the figure obtained experimentally is lower than the textbook value as energy is lost from the system during the experiment.

(12*2) +Mr ethanol (C2H5OH) = (1*6) + (16*1)

= 46

0.02174

-809.6 kJmol-1

Standard enthalpy change of neutralisation, ∆Hѳneut

• This is the enthalpy change that takes place when 1mol of hydrogen ions (H+(aq)) is neutralised under standard conditions:

H+ (aq) + OH- (aq) H2O (l) ∆Hѳneut = -57.5 kJmol-1

• The method of determining ∆Hѳneut involves mixing

equal volumes of 1 mol dm-3 solutions of a strong acid(e.g. hydrochloric acid) and a strong base (e.g. sodium hydroxide) and measuring the maximum temperature rise produced.

• The heat produced (Q) can then be determined using the formula:

Q (J) = total mass of solution(g) x 4.2 (JK-1g-1) x ∆T (oC)The value in kJ mol-1 can then be calculated.e.g. If 25cm3 of 1M HCl is reacted with 25cm3 1M NaOH

and a temperature rise of 6.7oC is obtained:Q = 50 x 4.2 x 6.7 = 1407 J = 1.4kJNo. moles HCl in 25cm3 of 1M solution = 25/1000 = 1/40 moles. So 1.4k J is

produced by producing 1/40 moles of water. To work out how much energy would be evolved if we produced 1 mole of water we simply multiply our answer by 40:

∆Hѳneut = 1.4 x 40 = -56kJmol-1

Problem

• In an experiment 50.0 cm3 of 1.00 moldm-3 HCl was exactly neutralised by 2.00 moldm-3 KOH. The initial temperature of both solutions was 15.3oC and the final temperature was 24.3oC. Calculate the molar enthalpy change of neutralisation of HCl.

Answer KOH + HCl KCl + H2O

ΔH = mCΔT

ΔH = 75.0 x 4.18 x 9.0

= 2882J = 2.882kJ

The temperature rise shows the reaction must be exothermic hence ΔH = -2.882kJ

n = M x V = 50.0 x 1.00

1000 1000

= 0.005mol

Hence molar enthalpy change for HCl = -2.822 x 10.005

= -56.4kJmol-1

Problems

1. Iron metal will react with copper sulphate solution in an exothermic reaction

Fe(s) + CuSO4(aq) FeSO4(aq) + Cu(s)

When 0.5g of powdered iron is added to 100cm3 of 0.200 mol dm-3 copper sulphate solution the temperature rises by 3.3oC

(a) calculate the enthalpy change of reaction(b) how many moles of iron were used(c) what is the molar enthalpy change for the reaction

ANSWER

ΔH = mCΔT

= 100 x 4.18 x 3.3

= - 1400J = - 1.4kJ

(a)

n = 0.50

55.85

(b)

(c)

ΔH = 1.4 0.009

= 0.009mol

= -156 kJ mol-1 for iron

Problem• 2. When 1.00g of methanol was completely burned under a

container of water the temperature of the water changed from 21.0oC to 45.5oC. The mass of water in the container was 220g.

• (a) how much energy was absorbed by the water?• (b) how much energy is given out by the methanol?• (c) is the combustion of methanol endothermic?• (d) how many moles of the methanol were used?• (e) calculate the molar enthalpy change for the reaction • when methanol is completely burned.

ANSWER

ΔH = 220 x 4.18 x 24.5 = +22.5kJ(a)

> 22.5kJ as some heat will be lost to the surrounding air and container(b)

Exothermic(c)

(d) 1.00 = 0.0312 mol32.0

(e) At least 22.5 = -721 kJ

0.0312

Standard Enthalpy of Formation, ∆Hѳf

• This is the enthalpy change that occurs when 1 mol of a compound is formed from its elements in their standard states and under standard conditions.

• E.g. • C (graphite) + 2H2(g) CH4 (g) ∆Hѳ

f = -75kJmol-1

• It is not possible to measure the standard enthalpy change of formation directly. So they have to be calculated indirectly from enthalpy changes of combustion using enthalpy cycles.

Note: The standard enthalpy of formation of an element in its standard state is zero.

Use of standard enthalpy of formation

• The standard molar enthalpy of reaction ΔHθr

can be calculated if the standard molar enthalpy of formation ΔHθ

f of the reactants and products are known.

• i.e. ΔHθr = ΔHθ

f(products)- ΔHθf(reactants)

Hess’s Law

This law states that the overall enthalpy change that accompanies a chemical reaction is independent of the route by which the reaction takes place, provided the initial and final states are the same. .

• In practical terms this means that you can do other reactions to find the enthalpy change for the main one.

Hess’s Law (continued)

For any given overall chemical reaction, the overall energy change is the same whether the reaction takes place in a single step or via a series of steps.

A ΔH1 DΔH2 ΔH4 ΔH1= ΔH2 + ΔH3 + ΔH4 B C ΔH3

Hess’ Law Triangle

• Hess’ law is useful for working out reactions which are either hard or impossible to do in the laboratory.

• Suppose we wanted to measure ΔHθf for

carbon monoxide, CO. The reaction:-• C(graphite) + ½O2(g) CO(g)

• is very hard to control. However the following combustions are known:-

C(graphite) + O2(g) CO2(g) ΔHc = -394kJmol-1

CO(g) + ½ O2(g) CO2(g) ΔHr = -283kJmol-1

All the information is then placed into the Hess’ triangle as shown below:-

C(graphite) + ½ O2(g) CO(g)

CO2(g)

½O2(g)O2(g)

ΔH1

ΔH2ΔH3

Hence ΔH1 = ΔHof = ΔH3 - ΔH2 = -394 - (-283) = 111kJmol-1

PROBLEM

Calculate the standard molar enthalpy of formation for butane:-

4C(graphite) + 5H2(g) C4H10(g)

given the following enthalpy changes of combustion below:-

C4H10(g) + 6½O2(g) 4CO2(g) + 5H2O(g) ΔHc = -2877kJmol-1

C(graphite) + O2(g) CO2(g) ΔHc = -394kJmol-1

H2(g) + ½O2(g) H2O(l) ΔHc = -286kJmol-1

ANSWER

Setting up the triangle, remembering to take into account the amounts present in each of the equations:-

4C(graphite) + 5H2(g) C4H10(g) ΔH1

4CO2(g) + 5H2O(l)

4 x ΔH3 5 x ΔH4ΔH2

From Hess’ Law:-

ΔH1 = ΔHof = ( 4 x ΔH3 ) + ( 5 x ΔH4 ) - ΔH2

= ( 4 x -394 ) + ( 5 x -286 ) - ( -2877 )

= -1576 -1430 + 2877 = 129kJmol-1

Stability of metal oxides

• The standard molar enthapies of formation of the metal oxides show how stable they are and hence the ease of extraction of their corresponding metals. The more exothermic (negative) is the value then the more stable the oxide and hence the more difficult it is to extract the metal.

• Ag2O (-31kJmol-1) Fe2O3 (-824kJmol-1)

• Cr2O3 ( -1128kJmol-1) Al2O3 (-1669kJmol-1)

Extraction of the metal

• Ag2O is broken up on heating

• Fe2O3 needs to be heated with carbon or carbon monoxide

• Cr2O3 needs to be heated with another metal

• Al2O3 can only be extracted by electrolysis

Use of enthalpy cycles These can also be used to find different enthalpy changes, not just those of formation.The Enthalpy Cycle below provides an alternative route for the combustion of 1 mole of methanol. The methanol is first converted to the elements from which it was formed (the reverse of the enthalpy of formation), and these are then burned to form the products (enthalpy of combustion).

Example problem• Calculate the standard enthalpy change for the

reaction:• C2H4 (g) + H2 (g) C2H6 (g)

Given the following standard enthalpy changes of combustion are as follows:

∆Hѳc (C2H4(g) = -1411 kJmol-1

∆Hѳc (H2(g)) = -286 kJmol-1

∆Hѳc (C2H6(g)) = -1560 kJmol-1

You are given the equation so write it down with ∆H above the arrow:

2CO2 (g) + 2H2O (l) + H2O (l)

-1411 -286 -1560

Find two routes around the equation without going against the flow of any arrows.Using Hess’s law:∆H + –1560 = -1411 + -286∆H = -1411 -286 +1560 =

C2H4 (g) + H2 (g) C2H6 (g)∆H

-137kJmol-1

Bond Enthalpy• To break a bond needs an input of energy – an

endothermic change.• When a bond is made energy is evolved – an

exothermic change.The amount of energy needed to break a bond is

called the bond energy or bond enthalpy.For a simple diatomic molecule (containing only

two atoms) the energy needed to break one mole of the bond is called the bond dissociation energy.

• For a compound with more than two atoms e.g. CH4 an average value is taken – this is called the mean bond enthalpy. These are often used in calculations.

Average bond enthalpies

Knowing the average molar bond enthalpies and that bond breaking is endothermic (positive) and bond forming is exothermic (negative) we can calculate the overall enthalpy change for a reaction.

e.g. CH4(g) + 2O2(g) CO2(g) + 2H2O(g)

By drawing out the full graphical formula we can see which bonds need to be broken and which bonds need to be formed.

H C H

H

H

+ 2O O O C O + 2 H O H

Bonds broken

4 x C H = (4 x +413) = +1652

2 x O O = (2 x + 498) = + 996

Total = + 2548kJ

Bonds formed

2 x C O = (2 x -805) = -1610

4 x O H = (4 x 464) = -1856

Total = -3466kJ

Hence:- ΔHr = +2548 -3466 = -918kJmol-1

PROBLEM

Calculate the molar enthalpy of combustion of ethanol:-

C2H5OH(g) + 3O2(g) 2CO2(g) + 3H2O(g)

Bonds broken

1 x C C = +347

5 x C H = +2065

1 x C O = +358 1 x O H = +464

Bonds formed

3 x O O = +1494

Total = +4728kJmol-1

4 x C O = -3320 6 x O H = -2784

Total = -6004kJmol-1

Hence ΔHc = +4728 -6004 = -1276kJmol-1