Unit 5 A3: Energy changes in industry - HCC Applied Sciencehccappliedscience.weebly.com/uploads/8/3/3/6/83363112/unit_5_a3... · L3 Applied Science Unit 5: Physical Chemistry 1 Unit

L3 Applied Science Unit 5: Physical Chemistry 1

Unit 5 – A3: Energy changes in industry

1. ENTHALPY CHANGES

1.1 Introduction to enthalpy and enthalpy changes 2

1.2 Enthalpy profile diagrams 2

1.3 Activation energy 3

1.4 Standard conditions 5

1.5 Standard enthalpy changes 5

2. CALCULATING ENTHALPY CHANGES

2.1 Determination of standard enthalpy changes 8

2.2 Calculating enthalpy changes directly from experiment 8

2.3 Comparison of experimental values with standard enthalpy values 12

2.4 Calculating enthalpy changes indirectly 12

2.5 Using bond enthalpies to calculate enthalpy changes indirectly 13

2.6 Using Hess’ law to calculate enthalpy changes indirectly 17

2.7 Finding the enthalpy change from enthalpy of combustion data 18

2.8 Finding the enthalpy change from enthalpy of formation data 21

Mixed questions 24


For exothermic reactions, heat is given

out to the surroundings because the

chemicals lose energy.

Therefore ∆H is negative.

The enthalpy of the products is less than

the enthalpy of the reactants.

1. ENTHALPY CHANGES

1.1 Introduction to enthalpy and enthalpy changes

During a chemical reaction, bonds are broken and made. Within these bonds is chemical energy, which

can be transferred to another form of energy (heat, light, sound etc) during the reaction. The amount of

energy that leaves a chemical system is the exact same amount that goes into the surroundings (conservation

of energy).

Enthalpy, H, is the heat content (thermal energy) that is stored in a chemical reaction. We cannot directly

measure the enthalpy of the reactants or the products. However, we can measure the energy absorbed or

released to the surroundings during a chemical reaction.

An enthalpy change, ∆H, is the heat exchanged with the surroundings during a chemical reaction at

constant pressure. This energy exchange is given units of kJ mol-1. A chemical reaction will either release

heat (exothermic reaction, ∆H is negative) or absorb heat (endothermic reaction, ∆H is positive).

Enthalpy change: ∆H = ∆U + p∆V, (where U = internal energy of system, p = pressure and V = volume)

Q. How can we measure the heat lost to the surroundings for a chemical reaction? How do we know if the

reaction is exothermic or endothermic from an experiment?

Exothermic: a reaction in which energy leaves the system to the surroundings

Endothermic: a reaction in which energy enters the system from the surroundings

1.2 Enthalpy profile diagrams

We can use enthalpy profile diagrams to show what happens to enthalpies during a reaction. We can

determine whether the reaction is exothermic or endothermic by comparing the enthalpy of the reactants

with the enthalpy of the products (looking at the enthalpy change, ∆H).

i) Exothermic reactions:

CO2 (g) + 2 H2O (g)

∆H = - 890 kJ mol-1

CH4 (g) + 2 O2 (g)

En

thal

py

H

Progress of reaction


ii) Endothermic reactions:

You must label the reactants, products, ∆H and Ea (see later) on an enthalpy profile diagram!

Note that ∆H = Hproducts - Hreactants

Q 1. A student performs two reactions. Reaction A has a positive ∆H value and reaction B has a negative

∆H value. Draw enthalpy profile diagrams for these reactions.

1.3 Activation energy

We need to put in energy, the activation energy, to break the first bond and start a chemical reaction. E.g.

for combustion of fuels, we supply the activation energy with a spark.

Activation energy is the minimum energy required to start a reaction

We can show the activation energy, Ea, on an enthalpy profile diagram:

CaCO3 (s)

En

thal

py

H CaO (s) + CO2 (g)

∆H = + 178 kJ mol-1

For endothermic reactions, heat is taken

in from the surroundings because the

reacting chemicals gain energy.

Therefore ∆H is positive.

The enthalpy of the products is greater

than the enthalpy of the reactants.


En

thal

py

H


En

thal

py

H


Reaction A Reaction B


The direction of the arrows is important. Activation energy is positive and the arrow points upwards. For

exothermic reactions, ∆H is negative and the arrow points down. For endothermic reactions, ∆H is positive

and the arrow points up!

Q 2. Draw an enthalpy profile diagram for the following reaction:

CO (g) + NO2 (g) → CO2 (g) + NO (g) ∆H = -226 kJ mol-1 Ea = +134 kJ mol-1

CHALLENGE: The enthalpy change for the reverse direction = ..............................

CHALLENGE: Activation energy for the reverse direction = ..............................

Q 3. Classify the reactions below as being either exothermic or endothermic.

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ∆H = -890 kJ mol-1 ......................................................

C6H12O6 (aq) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) ∆H = -2801 kJ mol-1 ....................................................

CaCO3 (s) → CaO (s) + CO2 (g) ∆H = +178 kJ mol-1 .....................................................

6 CO2 (g) + 6 H2O (l) → C6H12O6 (aq) + 6 O2 (g) ∆H = +2801 kJ mol-1 ...................................................

Ea (positive)

∆H (is positive)

Products

Reactants

Reactants

En

thal

py

H


En

thal

py

H


Exothermic Endothermic

Products

∆H (is negative)

Ea (positive)


1.4 Standard conditions

The enthalpy change of a reaction varies depending on the conditions that are present (e.g. temperature,

pressure). Enthalpy changes for reactions are therefore measured under the same conditions. These are

known as the standard conditions, which are:

Pressure of 100 kPa (or 1 atmosphere)

Temperature of 298 K (or 250C)

∆HƟ represents an enthalpy change measured under these standard conditions.

Under these standard conditions, a substance will be in its standard state (the physical state of the substance

at 1 atmosphere pressure and 250C).

Q 4. Complete the following standard states:

Magnesium has the standard state Mg (s) (under standard conditions of 100 kPa and 298 K).

Hydrogen has the standard state ....................

Water has the standard state ....................

1.5 Standard enthalpy changes

The standard enthalpy change of formation, ∆HfƟ, is the enthalpy change when one mole of a compound

is made from its elements (the elements being in their standard states, reaction under standard conditions)

e.g. H2 (g) + ½ O2 (g) → H2O (l) ∆HfƟ = -286 kJ mol-1

Note – the enthalpy of formation of an element is defined as 0 kJ mol-1 (If we are forming one mole of the

element H2 (g) from the element H2 (g) then there is no chemical change!).

The standard enthalpy change of combustion, ∆HcƟ, is the enthalpy change when one mole of a substance

is completely combusted (in excess oxygen under standard conditions)

e.g. C2H6 (g) + 3½ O2 (g) → 2 CO2 (g) + 3 H2O (l) ∆HcƟ = -1560 kJ mol-1

The standard enthalpy change of hydration, ∆HhydƟ, is the enthalpy change when one mole of isolated

gaseous ions is dissolved in water forming one mole of aqueous ions under standard conditions.

e.g. K+ (g) + aq → K+ (aq) ∆HhydƟ = -322 kJ mol-1


Q 5. For the following reactions carried out under standard conditions, identify the enthalpy change as

either ∆HcƟ, ∆Hf

Ɵ or ∆HhydƟ.

Cl- (g) + aq → Cl- (aq) ...............

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2O (l) ...............

2 C (s) + 2 H2 (g) → C2H4 (g) ...............

Sr2+ (g) + aq → Sr2+ (aq) ...............

2 Na (s) + C (s) + 1.5 O2 (g) → Na2CO3 (s) ...............

C5H12 (g) + 8 O2 (g) → 5 CO2 (g) + 6 H2O (l) ...............

Q 6. Write equations to show the enthalpy of formation of:

Propane .........................................................................................................................................

Butane .........................................................................................................................................

Ethanol ........................................................................................................................................

Calcium carbonate ......................................................................................................................

Q 7. Write equations to show the enthalpy of combustion of:

Propane .........................................................................................................................................

Butane .........................................................................................................................................

Ethanol ........................................................................................................................................

Methanol .....................................................................................................................................

Q 8. Write equations to show the enthalpy of hydration of:

Calcium .........................................................................................................................................

Sodium ........................................................................................................................................

Bromide .....................................................................................................................................

Sulfur .....................................................................................................................................


Q 9. Past paper question (June 2012)


2. CALCULATING ENTHALPY CHANGES

2.1 Determination of standard enthalpy changes

How do we determine ∆H for a reaction? There are direct and indirect ways of doing this:

Directly from experiment: Carry out an experiment and use a thermometer to determine the

temperature change of the surroundings. Then use the data from the experiment to calculate ∆H. This

can be done for a reaction that has taken place either in solution or for a combustion reaction.

Indirectly using either Hess’s law or bond enthalpy data: ∆H is not determined from experiment,

but is calculated by using known data from a book.

2.2 Calculating enthalpy changes directly from experiment

Reaction taking place in in solution Combustion reaction

A simple way of calculating the enthalpy change for a

reaction taking place in solution is to perform the

reaction in a polystyrene cup (that acts as an insulator).

For an exothermic reaction, heat is released to the

surroundings (the water) and you will see a

temperature increase on the thermometer.

For an endothermic reaction heat is taken in from

the surroundings (the water) and you see a

temperature decrease on the thermometer.

Substitute the temperature rise and the mass of the

surroundings (water) into the equation below.

Place a spirit burner containing the fuel (weighed)

under a beaker of water with known volume and

temperature.

Burn the fuel and measure the temperature rise of

the water.

Weigh the spirit burner again at the end of the

reaction to find the mass burned.

Substitute the temperature rise and the mass of the

surroundings (water) into the equation below.

Q = m x c x ∆T

This equation gives you the heat exchanged with the surroundings in J.

You must then calculate ∆H in kJ mol-1; see the example calculations below.

Heat exchanged with

surroundings (J)

Mass of

surroundings

(usually water, g)

Specific heat

capacity of

surroundings

(J g-1 K-1)

Change in

temperature of

surroundings

(K or 0C)


Worked example for a reaction taking place in solution:

25 cm3 of 1.0 mol dm-3 HCl was added to 25 cm3 of a 1.0 mol dm-3 NaOH solution in a polystyrene cup.

The temperature rose by 6.90C.

The specific heat capacity of water is 4.18 J g-1 K-1; the density of water is 1.00 g cm-3.

NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l) ∆Hr = ?

i) Calculate the energy released, in kJ, during this reaction.

Total mass of surroundings = 50g (water has a density of 1.00 g cm-3)

Q = m c ∆T

= (25 + 25) x 4.18 x 6.9

= 1442.1 J = 1.4421 kJ

ii) Calculate the amount, in moles, of HCl that caused the temperature change.

Moles = concentration x volume (in dm3)

= 1.0 x (25/1000)

= 0.025 mol

iii) Calculate the enthalpy change of the reaction. Give your answer to three significant figures.

Scale the quantities to match the molar quantities in the equation.

0.025 mol released 1.4421 kJ

1.0 mol would release (1.0/0.025) x 1.4421 = 57.684 kJ

The temperature rose, this was an exothermic reaction, the final answer should be negative!

∆Hr = -57.7 kJ mol-1 to 3 S.F.

Worked example for a combustion reaction:

A student burns 0.64 g of methanol in order to determine the enthalpy of combustion. The energy released

was used to heat 100 cm3 of water from to 190C to 390C.


i) Calculate the energy released during the combustion of methanol.

Mass of surroundings = 100g, NOT 0.64g!

Q = m c ∆T

= 100 x 4.18 x 20

= 8360 J = 8.36 kJ

ii) Calculate the amount, in moles, of methanol burnt.

Mr of CH3OH = 32

Moles = mass / molar mass

= 0.64 / 32

= 0.02 mol

iii) Calculate the enthalpy of combustion of methanol.

Scale up to 1 mole for enthalpy of combustion (check the definition for ∆HcƟ)

0.02 mol released 8.36 kJ

1.0 mol would release (1.0/0.02) x 8.36 = 418 kJ

The temperature rose, this was an exothermic reaction, the final answer should be negative!

∆Hc = -418 kJ mol-1


Q 10. 5.5 g of BaCO3 was added to 100 cm3 of a 2 mol dm-3 HCl solution.

The temperature rose from 200C to 32.70C.


BaCO3 (s) + 2 HCl (aq) → BaCl2 (aq) + H2O (l) + CO2 (g)

(i) Calculate the energy released, in kJ, during this reaction.

Give your answer to three significant figures.

(ii) Calculate the amount, in mol, of BaCO3 that caused the temperature change.

Give your answer to four significant figures.

(iii) Calculate the enthalpy change of the reaction.

Include the sign in your answer. Give your answer to three significant figures.

Q 11. A student burns 0.86 g of ethanol in order to determine the enthalpy of combustion. The energy

released was used to heat 100 cm3 of water from to 190C to 370C.


i) Calculate the energy released during the combustion of ethanol.


ii) Calculate the amount, in mol, of ethanol burnt.


iii) Calculate the enthalpy of combustion of ethanol.



Q 12. Past paper question (January 2011)

Sodium ammonium thiocyanate, NH4SCN, reacts with solid barium hydroxide, B(OH)2, as shown in the

equation below.

2 NH4SCN (s) + Ba(OH)2 (s) → Ba(SCN)2 (s) + 2 H2O (l) + 2 NH3 (g)

A research chemist carries out an experiment to determine the enthalpy change of this reaction.

In the experiment, 15.22 g of NH4SCN is reacted with a slight excess of Ba(OH)2. The reaction absorbs

energy, cooling the 50.0 g of water from 21.90C to 10.90C.

i) Calculate the energy absorbed, in kJ, during this reaction.

The specific heat capacity of water = 4.2 J g-1 K-1.

Energy = ........................................ kJ [2]

ii) Calculate the amount, in moles, of NH4SCN used by the research chemist.

Amount = ........................................ mol [1]

iii) Calculate the enthalpy change of reaction.

Include the sign in your answer. Give your answer to two significant figures

∆Hr = ........................................ kJ mol-1 [3]


2.3 Comparison of experimental values with standard enthalpy values

From a data book, ∆Hc of propan-1-ol is given as -2021 kJ mol-1. The value obtained from an experiment

is -1881 kJ mol-1. There is usually a difference between experimental values and standard enthalpy change

of combustion values that you find in a text book. Why?

Heat may have been lost to the surroundings

There may have been incomplete combustion / an incomplete reaction

Non-standard conditions may have been used

A more sophisticated piece of apparatus that allows accurate measurements of energy changes is called a

bomb calorimeter. This piece of apparatus ensures complete combustion and also reduces heat losses to

the surroundings.


Suggest two reasons why standard enthalpy changes of combustion determined experimentally are less

exothermic than the calculated theoretical values.

..........................................................................................................................................................................

.................................................................................................................................................................... [2]

2.4 Calculating enthalpy changes indirectly

It is not always possible to measure the enthalpy change directly from experiment for the following reasons:

The activation energy of the reaction is too high

There is a slow reaction rate

More than one reaction may be taking place (and so more than one product is formed)

We can, however, calculate these enthalpy changes indirectly by using other known enthalpy values. This

can be done by one of two methods that shall now be discussed in more detail:

a) Using bond enthalpy data.

b) Constructing enthalpy cycles and using Hess’s law.


2.5 Using bond enthalpies to calculate enthalpy changes indirectly

Bond enthalpy is the enthalpy change when one mole of (gaseous covalent) bonds is broken

(homolytically)

e.g. H – H (g) → 2 H (g) ∆H = +436 kJ mol-1

Bond enthalpies tell you how much energy is needed to break different bonds. Bond enthalpy values are

positive as you need to put in energy to break a bond.

Breaking bonds absorbs energy and is an endothermic process.

Forming bonds releases energy and is an exothermic process.

What determines whether a reaction is exothermic or endothermic overall?


Many organisms use the aerobic respiration of glucose, C6H12O6, to release useful energy. The overall

equation for aerobic respiration is the same as for the complete combustion of C6H12O6.

i) Write the equation for the aerobic respiration of C6H12O6.

.................................................................................................................................................................... [1]

ii) Explain, in terms of bond breaking and bond forming, why this reaction is exothermic.

..........................................................................................................................................................................

.................................................................................................................................................................... [2]

Energy is needed to

break bonds in

reactants

Energy is released as

new bonds are

formed in products


Bond enthalpies are average values; for example, you can find C–H bonds in almost every organic

molecule. The C–H bond strength will vary across the different environments it is found in. The bond

enthalpy of a C–H bond given in an exam, +415 kJ mol-1, is an average value.

We can use average bond enthalpy values to work out the enthalpy change of reactions involving gases

using the equation (which you must learn):

∆H = Ʃ (bond enthalpies of bonds broken) - Ʃ (bond enthalpies of bonds made)

Look through the worked example below to see how this equation is used.

Worked example on using bond enthalpy data to calculate enthalpy changes indirectly:

Use the average bond enthalpies in the table to calculate the enthalpy change of combustion of propane.

C3H8 + 5 O2 → 3 CO2 + 4 H2O

Bond Average bond enthalpy / kJ mol-1

C – H +413

C – C +347

O = O +498

C = O +805

O – H +464

Draw displayed formulae for each molecule in the reaction so that you can easily see all of the bonds:

Bonds broken bonds formed

8 x C – H 8 x 413 6 x C = O 6 x 805

2 x C – C 2 x 347 8 x O – H 8 x 464

5 x O = O 5 x 498 ______

6488 8542

∆H = Ʃ (bond enthalpies of bonds broken) - Ʃ (bond enthalpies of bonds made)

= 6488 – 8542

= -2054 kJ mol-1


Q 15. Calculate the enthalpy of combustion of ethanol using the bond enthalpy data given below.

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


C – H +413

C – C +347

O = O +498

C = O +805

O – H +464

C – O + 358

Q 16. Calculate the enthalpy of combustion of butane using the bond enthalpy data given below.

C4H10 (g) + 6.5 O2 (g) → 4 CO2 (g) + 5 H2O (l)


C – H +413

C – C +347

O = O +498

C = O +805

O – H +464




2.6 Using Hess’ law to calculate enthalpy changes indirectly

Hess’ law: for any chemical change, the enthalpy change is the same regardless of the route taken.

Worked examples: Introduction to using Hess’ cycles; work through the examples below:

A B

C

From Hess’ law: ∆HAB = ∆HAC + ∆HCB

-200 = -120 + -80

-200 = -200

A B

C


A B

C


A B

C


∆H = -150 kJ mol-1 ∆H = -100 kJ mol-1

∆H =

∆H = -150 kJ mol-1 ∆H = -100 kJ mol-1

∆H =

∆H = -150 kJ mol-1 ∆H = -100 kJ mol-1

∆H =

∆H = -120 kJ mol-1 ∆H = -80 kJ mol-1

∆H = -200 kJ mol-1

Going from B to C is

exothermic (negative ∆H),

so going in the opposite

direction, from C to B

would be endothermic

(positive ∆H )

We do not know ∆HAB,

but we can calculate this

using Hess’ law.

The enthalpy change is

the same, regardless of

which route you take to

go from A to B (-200).

Remember to

always reverse the

sign (+ / -) if going

against the arrow


We can use Hess’ law to calculate the:

a) enthalpy change of formation from enthalpy of combustion data

b) enthalpy change of reaction from enthalpy of formation data.

2.7 Finding the enthalpy change from enthalpy of combustion data

Many compounds can be reacted with oxygen and the enthalpy change of combustion measured accurately.

These enthalpy changes can be used to calculate the enthalpy changes for other reactions.

Worked example for calculating the enthalpy change of formation when given ∆Hc data:

Use the data given below to calculate the enthalpy change of formation for methane.

Substance ∆Hc / kJ mol-1

C (s) -393

H2 (g) -286

CH4 (g) -890

1. Write an equation to show the enthalpy change of formation.

2. Construct an enthalpy triangle with arrows pointing down towards ‘combustion products’ if given

enthalpy of combustion data.

3. Use Hess’ law to calculate ∆Hf.

C (s) + 2 H2 (g) CH4 (g)

Combustion products

(CO2 (g) and 2 H2O (l))

From Hess’ law: ∆Hf = -965 + 890 = -75 kJ mol-1

Note - there is also a shortcut that you can use without having to draw out a Hess’ cycle:

∆Hf = Ʃ ∆Hc (reactants) - Ʃ ∆Hc (products)

For the example above: ∆Hf = -965 – (-890) = -75 kJ mol-1

∆Hc = -393 + (2 x -286)

= -965 kJ mol-1

∆Hc = -890 kJ mol-1

∆Hf Direction of the arrows

points down towards

the combustion

products

Reverse the sign

(+890) when going

against an arrow in the

opposite direction


Q 18. Calculate the enthalpy change of formation of ethane using the data given below.


C (s) -393

H2 (g) -286

C2H6 (g) -1560

Q 19. Calculate the enthalpy change of formation of ethanol using the data given below.


C (s) -393

H2 (g) -286

C2H5OH (g) -1367



i) Add state symbols to the equation to show each species in its standard state.

6 C (..........) + 7 H2 (..........) → C6H14 (..........)

[1]

ii) It is very difficult to determine the standard enthalpy change of formation of hexane directly. Suggest a

reason why.

..........................................................................................................................................................................

.................................................................................................................................................................... [1]

iii) The standard enthalpy of formation of hexane can be determined indirectly. Calculate the standard

enthalpy change of formation of hexane using the standard enthalpy changes of combustion below.


C -394

H2 -286

C6H14 -4163

answer = ........................................ kJ mol-1 [3]


2.8 Finding the enthalpy change from enthalpy of formation data

Chemists have measured enthalpy changes of formation for many compounds accurately. These enthalpy

changes can be used to calculate the enthalpy changes for other reactions.

Worked example for calculating the enthalpy change of combustion when given ∆Hf data:

Use the data given below to calculate the enthalpy change of combustion for ethane.

Substance ∆Hf / kJ mol-1

C2H6 (g) -84.7

CO2 (g) -394

H2O (l) -286

1. Write an equation to show the enthalpy change of combustion.

2. Construct an enthalpy triangle with arrows pointing up from the ‘elements’ if given enthalpy of

formation data.

3. Use Hess’ law to calculate ∆Hc.

C2H6 (g) + 3.5 O2 (g) 2 CO2 (g) + 3 H2O (l)

Elements

(2 C (s), 3 H2 (g) and 3.5 O2 (g))

From Hess’ law: ∆Hr = +84.7 + (-1646) = -1561.3 kJ mol-1

Note - there is also a shortcut that you can use without having to draw out a Hess’ cycle:

∆Hr = Ʃ ∆Hf (products) - Ʃ ∆Hf (reactants)

For the example above: ∆Hr = -1646 – (-84.7) = -1561.3 kJ mol-1

∆Hf = (2 x -394) + (3 x -286)

= -1646 kJ mol-1

∆Hf = -84.7 + (3.5 x 0)

= -84.7 kJ mol-1

∆Hc Direction of

arrows points

up from the

elements

Reverse the sign

(+84.7) when going

against an arrow in

the opposite

direction

Enthalpy change of

formation for

elements = 0


Q 21. Calculate the enthalpy change of combustion of methanol using the data given below.


CH3OH (l) -239.1

CO2 (g) -394

H2O (l) -286

Q 22. Calculate the enthalpy change of combustion of ethanol using the data given below.


CH3CH2OH (l) -278

CO2 (g) -394

H2O (l) -286


Q 23. Use the information below to calculate ∆Hf for ethane, given that ∆Hf for ethene is +52 kJ mol-1.

C2H4 (g) + H2 (g) → C2H6 (g) ∆Hr = -137 kJ mol-1


Hess’ law can be used to calculate enthalpy changes of reaction. The equation for the reaction that gives

the enthalpy change of formation, ∆Hf, of N2O (g) is as follows.

N2(g) + ½ O2 (g) → N2O (g)

i) It is not possible to measure the enthalpy change of formation of N2O directly. Suggest why it is not

possible.

..........................................................................................................................................................................

.................................................................................................................................................................... [1]

ii) The data below can be used to calculate the enthalpy change of formation, ∆Hf, of N2O (g). Calculate

∆Hf for N2O (g).

reaction ∆Hr / kJ mol-1

C (s) + N2O (g) → CO (g) + N2 (g) -193

C (s) + ½ O2 (g) → CO (g) -111

∆Hf = ........................................ kJ mol-1 [2]


MIXED QUESTIONS: Careful! For the questions below, decide which method to use to determine the enthalpy change!

Q 25. A student burns 0.27 g of hexane in order to determine the enthalpy of combustion. The energy

released was used to heat 100 cm3 of water from to 180C to 380C.


i) Calculate the energy released during the combustion of hexane.


ii) Calculate the amount, in mol, of hexane burnt.


iii) Calculate the enthalpy of combustion of hexane.

Include the sign in your answer. Give your answer to four significant figures.

Q 26. Calculate the enthalpy change for the reaction using the bond enthalpy data given in the table.

N2H4 (l) + 2 F2 (g) → N2 (g) + 4 HF (g)


N – N +163

N – H +388

N = N +944

F – F +158

H – F +562


Q 27. 0.327 g of Zn powder was added to 55.0 cm3 of excess CuSO4 at 22.80C.

The temperature rose to 32.30C.


Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s)

(i) Calculate the energy released, in kJ, during this reaction.


(ii) Calculate the amount, in mol, of Zn that caused the temperature change.

(iii) Calculate the enthalpy change of the reaction.


Q 28. Calculate the enthalpy change of formation of butane using the data given below.


C (s) -393

H2 (g) -286

C4H10 (g) -2877


Q 29. A student mixed 100.0 cm3 of a 1.50 mol dm-3 hydrochloric acid solution with 100.0 cm3 of a 1.50

mol dm-3 sodium hydroxide solution. Both solutions were at 19.670C initially and the highest temperature

reached by the reaction mixture was 34.060C.


(i) Construct the equation for the reaction.

……………………………………………………………………………………………………………......

(ii) Calculate the enthalpy change of the reaction.



Q 30. Calculate the enthalpy change for the reaction using the bond enthalpy data given in the table.

C6H14 (g) → 3 C2H4 (g) + H2 (g)


C – H +413

C – C +347

C = C +610

H – H +436

Q 31. Calculate the enthalpy change of formation of benzene using the data given below.


C (s) -393

H2 (g) -286

C6H6 (g) -3267


Q 32. Calculate the enthalpy change for the reaction using the data given below.


CaCO3 (s) -1207

CO2 (g) -394

CaO (s) -635

CaCO3 → CaO + CO2

Q 33. Calculate the enthalpy change for the reaction using the data given below.


CO (g) -111

CO2 (g) -394

Fe2O3 (s) -822

3 CO (g) + Fe2O3 (s) → 2 Fe + 3 CO2



You could be asked

to determine the

enthalpy change of

reaction from an

unfamiliar looking

enthalpy cycle

[3]

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Unit 5 A3: Energy changes in industry - HCC Applied Sciencehccappliedscience.weebly.com/uploads/8/3/3/6/83363112/unit_5_a3... · L3 Applied Science Unit 5: Physical Chemistry 1 Unit