4: Determination of enthalpy changes

Question 1

heat loss in a chemical system=

Answer 1

heat gain by surroundings

Question 2

heat gain in a chemical system=

Answer 2

heat loss by surroundings

Question 3

In the equation Q=mcT Joules:

Explain what each individual variable means.

Answer 3

m- the mass of the surroundings involved in the heat exchange

c - the specific heat capacity of the surroundings

Change in T - the temperature change of the surroundings: Change in T = T final - T initial

Question 4

Give an example of a simple calorimeter

Answer 4

The simplest calorimeter is a plastic coffe cup,.

Question 5

An excess of magnesium is added to a 100cm³ of 2.00 mol dm³ CuSO₄(aq). The temperature increases from 20.0*C to 65.0*C. Find the enthalpy change of reaction for the following equation: Mg(s) + CuSO₄(aq) —–> MgSO₄(aq) + Cu(s)

• Specific heat capacity of solution c = 4.18 J g^-1 K^-1
• density of solution = 1.00 g cm^-3

Answer 5

1) Find energy change : 100cm3 of soltuion has a mass of 100g.

Temperature change= (65-20) = +45*C

Heat gained by surroundings Q=mcT = 100 x 4.18 x 45.0 = 18810 J

This heat has been released to the chemical system = -18810J

2) Find out the amount,in mol, reacted: c x V / 1000 => 2 x 100/ 1000 = 0.200mol.
3) Scale the quantities to matht he molar quantities in the equation.

For 0.200 mol CuSO4, H = -18810J

1 mol of CuSO4 = 5x0.200mol

For 1 mol of CuSO4, H = 5 x -18810

= -94050 J

4) Write down the equation with enthalpy change.

Mg(s) + CuSO4 —-> Mg SO4 + Cu(s) H = -94.05 KJ mol^-1

Question 6

Define Specific heat capacity,c.

Answer 6

Specific heat capacity,c, is the energy required to raise the temperature of 1g of a substance by 1*C.

Question 7

Quick tip:

Answer 7

In an equation to represent Change in enthalpy of combustion you must not put a balancing number in front of the substance being burnt. If you do, the more than 1 mol would have been combusted.

Question 8

Draw a calorimeter suitable for the determination of the enthalpy of combustion of fuel.

Answer 8

Question 9

What do you need to do in an experiment to determine the change in combustion of a liquid fuel?

Answer 9

1) Burn a known mass of a substance.
2) To heat a known mass of water
3) Measure the temperature change in the water.

Question 10

What would be needed and needed to measure to determine the enthalpy of combustion for a liquid fuel?

Answer 10

• You measure a volume of water into the beaker
• 1cm³ of water weighs 1g, so the mass of water is easy to deduce.
• The burner containing the fuel is weighed
• The initial temperature of the water is taken
• The burner is lit and the water is heated until the temperature has risen by a reasonable amount. The maximum temperature is taken and the temperature change T can then be determined.
• The flame is extinguished and the burner re-weighed to find the mass of fuel that has been burnt.

Question 11

During a combustion , 1.50 g of propan-1-ol, CH3CH2CH2OH, heated 250cm3 of water by 45*C.

Find the enthalpy change of combustion of propan-1-ol

• specific heat capacity of water, c= 4.18 J g-1 K-1
• density of water = 1g cm-3

Find the heat change during the experiment.

Answer 11

1) 250cm3 water weighs 250g

So, heat gained by the water , Q = 250 x 4.18 x 45J = 47025J

The same quantity of heat will have been lost by the fuel,propan-1-ol, and oxygen during combustion.

So heat loss from chemicals = -47.025 KJ.

2) Find the amount in mol of propan-1-ol that reacted.

Molar mass, M, of CH3CH2CH2OH = 60.0 g mol^-1

So, amount, n, of CH3CH2CH2OH that reactd = 1.5/60.0 = 0.025 mol

3) Work out the heat loss in KJ mol^-1

So, 1 mol of CH3CH2CH2OH loses 47.025/0.025 = 1881 KJ.

So, enthalpy of combustion of CH3CH2CH2OH = -1881KJ mol^-1

Question 12

Give two reasons why a calculated value of a change in enthalpy of combustion might differ from the book value?

Answer 12

• There may have been incomplete combustion
• There may have been heat loss to the surroundings.

Question 13

Draw a bomb calorimeter

Answer 13

Question 14

Define bond enthalpy

Answer 14

Bond enthalpy is the average enthalpy change that takes place when breaking by homolytic fission 1 mol of a given bond in the molecules of a gaseous species.

Question 15

What information does a bond enthalpy tell you?

Answer 15

You get the information about the strength of a chemical bond from its bond enthalpy.

Question 16

Why will the same bonds in different molecules or formula units vary?

Answer 16

The bond strength will vary across different environments in which it is found.

Question 17

Whats happens in terms of bonds in an endothermic reaction?

Answer 17

Energy is first needed to break bonds in the reactants. Bond breaking is an endothermic process and requires energy.

Question 18

Describe an exothermic process in terms of bonds.

Answer 18

Energy is releases as new bonds are formed in the products. Bond making is an exothermic process and releases energy.

Question 19

How is a reaction exothermic overall?

Answer 19

In an exothermic reaction, the bonds that are formed are stronger than the bonds that are broken.

Question 20

How is a reaction endothermic overall?

Answer 20

In an endothermic reaction, the bonds that are formed are weaker than the bonds that are broken.

Question 21

How do you work out the energy required to break the bonds in the reaction?

Answer 21

Energy required to break bonds=Σ(bond enthalpies of bonds broken)

Question 22

How do you work out the energy released when bonds are made in a reaction?

Answer 22

Energy released when bonds are made = -Σ(bond enthalpies of bonds are made)

Question 23

How do you work out change in enthalpy of reaction knowing the average bond enthalpies?

Answer 23

Change in H = Σ(bond enthalpies of bonds broken)-Σ( bond enthalpies of bonds made)

Question 24

Using average bond enthalpies: Work out the change in in enthalpy of CH₄(g) + 2O₂(g) —–> CO₂(g) + H₂O(g)

Average bond enthalpies: C-H: +413, O=O: 497, C=O, 805, O-H: 463

Answer 24

Bonds broken = 4(C-H) + 2(O=O)

Bonds made = 2(C=O) + 4(O-H)

So enthalpy change = (4x413) + (2x497) - (2x805) + (4x463) = -816 KJ mol^-1

Question 25

Define Hess’ law

Answer 25

Hess’law states that, if a reaction can take place by more than one route and the initial and final conditions are the same, the total enthalpy change is the same for each route.

Question 26

Define an enthalpy cycle

Answer 26

An enthalpy cycle is a diagram showing alterantive routes between reactants and products which allows the indirect determination of an enthalpy changes using Hess’ law.

Question 27

Give some reasons why its not always possible to measure the enthalpy change of a reaction directly?

Answer 27

• high activation energy
• a slow rate of reaction
• more than one reaction taking place.

Question 28

Explain Hess’ law of A=B+C

Answer 28

A being route 1: reactants —–> to products

B+C being route 2: reactants—-> intermidiate—-> product

By Hess’ law, the total enthalpy change is the same for each route.

So, A-=B+C

Question 29

What is Hess’law an extension of?

Answer 29

Hess’ law is an extension of the law of conservation of energy.

Question 30

Why would the following reactions enthalpy change not be able to be measured directly? :

3C(s) + 4H₂(g) ——> C₃H_8(g)

Answer 30

• It is impossible to have a reaction that could take place between carbon and hydrogen to form just one hydrocarbon.

Question 31

Calculate the enthalpy change indirectly using the enthalpy changes of combustion of the three chemicals in the equation:

3C(s) + 4H₂(g) ——-> C₃H₈(g)

C=-394

H₂=-286

C₃H₈=-2219

Answer 31

Change in enthalpy = -107

Question 32

Draw the enthalpy cycle of enthalpy of formation.

Answer 32

Question 33

Draw the enthalpy of combustion cycle diagram.

Answer 33

Question 34

Do some enthalpy of formation and enthalpy of combustion questions.

Answer 34

Tada