3.2.1 (g) Hess' Law & Enthalpy Cycles

Syllabus

(g) Hess’ law for construction of enthalpy cycles and calculations to determine indirectly:

(i) an enthalpy change of reaction from enthalpy changes of combustion

(ii) an enthalpy change of reaction from enthalpy changes of formation

(iii) enthalpy changes from unfamiliar enthalpy cycles.

{Definition of Hess’ law not required.}

{Unfamiliar enthalpy cycles will be provided.}

{Application of the principle of conservation of energy to determine enthalpy changes.}

What does this mean?

What is Hess' Law

Physicists will tell you that the First Law of Thermodynamics states that energy cannot be created or destroyed, merely converted from one form to another.

In other words, in a chemical reaction the total amount of energy must remain the same even if it is converted from potential energy in bonds to heat or any other form of energy.

This may be referred to as Conservation of Energy.

It is also another way of stating Hess' Law which is usually written as "Hess's law states that the change of enthalpy in a chemical reaction is independent of the pathway between the initial and final states."

In the reaction above Reactants A can become Products B directly (route 1)

Or they can pass through some intermediate stage and then become Products B (route 2)

Hess' Law means that ∆H is the same for both routes.

The syllabus now states that you won't be asked to write the definition. Shame, it was easy marks.

Using Hess' Law

∆H_Combustion data

∆H_Combustion is well established established for almost every substance known.

So if we were asked to find ∆H_Reaction for C₂H₄ +H₂ → C₂H₆ we could just carry out the reaction.

But all three reactants and products have known ∆H_Combustion

So we could save time by drawing a triangle like the one above.

Notice that the arrows point from both the reactants and the reaction-products towards the CO₂ and H₂O.

Clearly this is because these are the final products of burning either C₂H₄ and H_{2 or} C₂H₆ .

You don't gain any extra marks for working out that they both make 2 CO₂ + 6 H₂O.

So you might equally write just CO₂ + H₂O. Or "Combustion Products". Or "Burnt Stuff". Or nothing at all.

The examiner will have to provide the ∆H_Combustion needed - though (s)he may give you some that you don't need too.

In this case ∆H_c(C₂H₄) = -1393 kJ/mol, ∆H_c(C₂H₆) = -1561 kJ/mol, ∆H_c(H₂) = -286 kJ/mol.

So, burning C₂H₄ +H₂ will release -1393 + - 286 = -1679 kJ/mol

And burning C₂H₆ will release -1561 kJ/mol

Route 2 involves following the left-hand arrow (leave the sign alone) but going against the right-hand arrow (change the sign).

So the enthalpy change for Route 2 is -1679 + 1561 = -118 kJ/mol

And Hess' Law states that Route 1 = Route 2 so ∆H_reaction= -118 kJ/mol.

Example 2

In this case we are going to use ∆H_c to calculate an enthalpy change that we cannot measure directly.

It's impossible to measure ∆H_formation (CH₄) directly because combining Carbon And Hydrogen is likely to make lots of other products as well as Methane.

Examiners sometimes ask you that.

∆H_combustion (CH₄)= -890 kJ/mol, ∆H_combustion (C) = -394 kJ/mol & ∆H_combustion (H₂)= -286 kJ/mol

On the diagram above H_c2 = ∆H_combustion (C) + 2 x ∆H_combustion (H₂) = -394 = (2 x -286) = -966 kJ/mol

And H_c1 = ∆H_combustion (CH₄)= -890 kJ/mol.

Hess' Law gives us that: ∆H_f (CH₄) = H_c2 - H_c1 = -966 --890 = -966 + 890 = -76 kJ/mol

∆H_Formation data

In this case we would probably be asked to calculate ∆H_reaction given ∆H_formation (C₂H₄), ∆H_formation (HCl) & ∆H_formation (C₂H₅Cl).

The bottom of this triangle is the elements that the compounds are formed from.

So this time the arrows point up.

Again, it really doesn't matter what you write at the bottom.

"Elements" would do just as well - or nothing at all.

You could follow the blue arrows. Or we could apply Hess' Law directly, taking note of which arrows we are following (leave sign alone) and which we are going against (change sign)

Or you could say that: ∆H_reaction = - ∆H_formation (C₂H₄) - ∆H_formation (HCl) + ∆H_formation (C₂H₅Cl)

∆H_reaction = -(+52.2) -(-92.3) + (-109)

= -52.2 + 92.3 -109

= - 68.9 kJ/mol

Unfamiliar enthalpy changes

You'll get a lot of triangles.

The syllabus implies that anything else would be provided and that you would simply need to apply your knowledge of Hess' Law

So in the somewhat more complicated diagram (right) we would need to look carefully at the arrows and if any one enthalpy change was missing we could calculate it from:

∆H₁ = T - ∆H₃ - ∆H₂

∆H2 = - ∆H₁ + T - ∆H₃

∆H3 = - ∆H₂ - ∆H₁ + T

T = ∆H₁ + ∆H₂ + + ∆H₃

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Exam-style Questions

1. (a) Define the term standard enthalpy of formation, ∆Hfο

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(b) Use the data in the table to calculate the standard enthalpy of formation of liquid methylbenzene, C7H8

7C(s) + 4H2(g) → C7H8(l)

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2. (a) Define the term standard enthalpy of combustion, ∆Hcο

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(b) Use the mean bond enthalpy data from the table and the equation given below to calculate a value for the standard enthalpy of combustion of propene. All substances are in the gaseous state.

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(c) State why the standard enthalpy of formation, ∆Hfο, of oxygen is zero.

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(d) Use the data from the table below to calculate a more accurate value for standard enthalpy of combustion of Propene.

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(e) Explain why your answer to part (b) is a less accurate value than your answer to part (d).

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(Total 12 marks)

3. Below are some standard enthalpy changes including the standard enthalpy of combustion of Nitroglycerine, C3H5N3O9

½ N2(g) + O2(g) ® NO2(g) DH_f = +34 kJ mol–1

C(s) + O2(g) ® CO2(g) DH_f = –394 kJ mol–1

H2(g) + ½O2(g)®H2O(g) DH_f = –242 kJ mol–1

C3H5N3O9(l) + 11/4 O2(g) ®3CO2(g) + 5/2 H2O(g) + 3NO2(g) DH_f = –1540 kJ mol–1

(a) Standard enthalpy of formation is defined using the term standard state.What does the term standard state mean?

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(b) Use the standard enthalpy changes given above to calculate the standard enthalpy of formation of Nitroglycerine.

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(c) Calculate the enthalpy change for the following decomposition of Nitroglycerine.

C3H5N3O9(l) ® 3CO2(g) + 5/2 H2O(g) + 3/2 N2(g) + 1/4 O2(g)

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(d) Suggest one reason why the reaction in part (c) occurs rather than combustion when a bomb containing nitroglycerine explodes on impact.

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(e) An alternative reaction for the combustion of hydrogen, leading to liquid water, is given below.

H2(g) + ½O2(g) ® H2O(l) DH_R = –286 kJ mol–1

Calculate the enthalpy change for the process H2O(l) ® H2O(g) and explain the sign of DH in your answer.

Calculation............................................................................................................................................................

Explanation for sign of DH........................................................................................................................................

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(Total 12 marks)

Answers

1. (a) Enthalpy change when 1 mol of compound 1

Is formed from it’s elements 1

All substances in their standard state 1

(b) DH = åDHοc (reactants) – åDHοc (products) 1

= (7x – 394) + (4 x – 286) – (– 3909) 1

= + 7 kJmol–1 1

2. (a) enthalpy change/ heat energy change when 1 mol of a substance 1

is completely burned in oxygen 1

at 298K and 100 kPa or standard conditions 1

(b) ∆H = ∑ bonds broken – ∑ bonds formed 1

= (6 × 412) + 612 +348 + (4.5 × 496) – ((6 × 743) + (6 × 463)) 1

= – 1572 kJ mol–1 1

(c) by definition ∆Hf is formation from an element 1

(d) ∆Hc = ∑ ∆Hf products -∑ ∆Hf reactants or cycle 1

= (3 × – 394) + (3 × -242) – (+20) 1

= − 1928 kJmol–1 1

(e) bond enthalpies are mean/average values 1

from a range of compounds

1

3. (a) normal physical state (1)

at 100kPa and 298K (1)

(b) DH_f = SAH_c reactants - SAH_c prods (1) or cycle

DH_f = 3 × DH_f (CO2) + 2×5 × DH_f (H2O) + 3 × DH_f (NO2)

(i.e number ratios) –DH_c (N.G) (1)

= 3× –394 + 2×5 × –242 + 3 ×34 –(–1540) (1)

= – 145 (1)

(c) DH = 3×DH_f (CO2) + 2×5 × DH_f (H2O) –DH_f (N.G) (1)

= 3 × –394 + 2×5 × –242 – (–145) (1)

= –1642 (1)

(or DH = DH_c (N.G) –3 × DH_c ( ½N2) (1)

= –1540 –3 × 34 (1) = –1642 (1) ) 3

(d) does not require Oxygen (1)

(or lower Ea) 1

(e) +44 (1)

energy needed to break bonds (1) 2

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