Cambridge AS & A Level9701

Enthalpy change, ΔH

Chemistry 9701 Chapter Notes

What this chapter covers

Enthalpy change, ΔHHess’s law
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In this chapter

  1. 1Exothermic & Endothermic Reactions
  2. 2Standard Enthalpy Change Definitions
  3. 3Measuring Enthalpy Change: Calorimetry
  4. 4Hess's Law and Enthalpy Cycles
  5. 5Calculating Enthalpy Change Using Bond Energies

1. Exothermic & Endothermic Reactions

In any chemical reaction, energy is transferred. We call the heat energy change at constant pressure the enthalpy change, represented by the symbol ΔH (delta H). Enthalpy (H) is the total energy content of a system. We can't measure H directly, but we can measure the change, ΔH. If a reaction releases heat into the surroundings, it is exothermic. The products have less enthalpy than the reactants, so ΔH is negative. Examples include combustion and neutralisation. If a reaction absorbs heat from the surroundings, it is endothermic. The products have more enthalpy than the reactants, so ΔH is positive. An example is thermal decomposition. We can visualise these changes using enthalpy profile diagrams, which plot enthalpy against the progress of the reaction. These diagrams also show the activation energy (Ea), which is the minimum energy required to start the reaction.

ΔH = H(products) - H(reactants)

Key term

Enthalpy Change (ΔH): The heat energy exchanged with the surroundings during a chemical reaction at constant pressure.

Examiner insight

Examiners award marks for correctly labelling the axes, showing reactants higher than products for exothermic reactions (and vice versa for endothermic), and clearly indicating ΔH with its correct sign.

Common pitfall

Forgetting that for exothermic reactions, ΔH is negative because the system loses energy to the surroundings. A temperature rise in the surroundings means a negative ΔH for the reaction.

Worked example 14 marks

The combustion of methane is an exothermic reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = -890 kJ mol⁻¹. Draw a labelled enthalpy profile diagram for this reaction.

  1. 1

    Step 1: Draw the axes. The y-axis is 'Enthalpy (H)' and the x-axis is 'Reaction Progress' or 'Reaction Pathway'.

  2. 2

    Step 2: Since the reaction is exothermic (ΔH is negative), the reactants have more energy than the products. Draw a horizontal line for the reactants (CH₄ + 2O₂) higher up on the y-axis.

  3. 3

    Step 3: Draw a horizontal line for the products (CO₂ + 2H₂O) lower down on the y-axis.

  4. 4

    Step 4: Connect the reactant and product lines with a curve that goes up first (representing activation energy) and then comes down to the product level.

  5. 5

    Step 5: Label the activation energy (Ea) as the 'hump' from the reactant level to the peak of the curve.

  6. 6

    Step 6: Draw a vertical arrow pointing downwards from the reactant level to the product level. Label this arrow 'ΔH = -890 kJ mol⁻¹'.

Worked example 24 marks

The thermal decomposition of calcium carbonate is endothermic: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ mol⁻¹. Sketch a labelled enthalpy profile diagram for this reaction.

  1. 1

    Step 1: Draw and label the axes: 'Enthalpy (H)' on the y-axis and 'Reaction Progress' on the x-axis.

  2. 2

    Step 2: Since the reaction is endothermic (ΔH is positive), the products have more energy than the reactants. Draw a line for the reactant (CaCO₃) lower down on the y-axis.

  3. 3

    Step 3: Draw a line for the products (CaO + CO₂) higher up on the y-axis.

  4. 4

    Step 4: Connect the lines with a curve, showing the activation energy barrier.

  5. 5

    Step 5: Draw a vertical arrow pointing upwards from the reactant level to the product level and label it 'ΔH = +178 kJ mol⁻¹'. Also label the activation energy (Ea).

Recap

  • Exothermic reactions release heat, have a negative ΔH, and products are at a lower energy level than reactants.
  • Endothermic reactions absorb heat, have a positive ΔH, and products are at a higher energy level than reactants.
  • Enthalpy profile diagrams show the energy pathway from reactants to products.
  • Activation energy (Ea) is the energy barrier that must be overcome for a reaction to start.
  • The sign of ΔH indicates the direction of heat flow: negative is heat out (exo), positive is heat in (endo).

Quick check

  1. A cold pack for sports injuries works by dissolving ammonium nitrate in water. Is this process exothermic or endothermic? What is the sign of ΔH?2 marks
  2. What do the y-axis and x-axis represent on an enthalpy profile diagram?2 marks

2. Standard Enthalpy Change Definitions

To compare enthalpy changes fairly, we must use a common set of conditions called standard conditions. These are a pressure of 100 kPa (or 10⁵ Pa) and a temperature of 298 K (25 °C). When an enthalpy change is measured under these conditions, it is called a standard enthalpy change, indicated by the symbol ⦵ (pronounced 'standard'). The substances involved must also be in their normal physical state at these conditions (e.g., H₂O as a liquid, O₂ as a gas). There are several key types of standard enthalpy change, each with a precise definition that always refers to 'one mole' of a substance.

Key term

Standard Enthalpy Change of Formation (ΔH_f⦵): The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states under standard conditions.

Examiner insight

Examiners are very strict with definitions. Forgetting to specify 'one mole' of product for formation or 'one mole' of reactant for combustion is a common reason for losing marks.

Common pitfall

Writing definitions without including all three key parts: 'one mole', 'standard states', and 'standard conditions'. All are required for the marks.

Worked example 13 marks

Write the balanced chemical equation, including state symbols, that corresponds to the standard enthalpy change of formation of liquid ethanol, C₂H₅OH(l).

  1. 1

    Step 1: The definition is for the formation of ONE mole of the compound. So, the product must be 1C₂H₅OH(l).

  2. 2

    Step 2: The reactants must be the constituent elements in their standard states. The elements are Carbon, Hydrogen, and Oxygen.

  3. 3

    Step 3: The standard state for Carbon is solid graphite, C(s). The standard state for Hydrogen is a diatomic gas, H₂(g). The standard state for Oxygen is also a diatomic gas, O₂(g).

  4. 4

    Step 4: Write the unbalanced equation: C(s) + H₂(g) + O₂(g) → C₂H₅OH(l).

  5. 5

    Step 5: Balance the equation for one mole of product. We need 2 carbons, so 2C(s). We need 6 hydrogens (5+1), so 3H₂(g). We need 1 oxygen, so ½O₂(g).

  6. 6

    Step 6: Final equation: 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l).

Worked example 23 marks

The standard enthalpy change of combustion (ΔH_c⦵) of propane is -2219 kJ mol⁻¹. Write the equation that represents this value.

  1. 1

    Step 1: The definition is for the combustion of ONE mole of the substance. The substance is propane, C₃H₈(g).

  2. 2

    Step 2: Combustion means reacting completely with excess oxygen, O₂(g).

  3. 3

    Step 3: Complete combustion of a hydrocarbon produces carbon dioxide, CO₂(g), and water, H₂O(l) (water is liquid in its standard state at 298 K).

  4. 4

    Step 4: Write the unbalanced equation: C₃H₈(g) + O₂(g) → CO₂(g) + H₂O(l).

  5. 5

    Step 5: Balance the equation for one mole of propane. We need 3 carbons, so 3CO₂. We need 8 hydrogens, so 4H₂O. Now count oxygen atoms on the right: (3 × 2) + (4 × 1) = 10. So we need 5O₂ on the left.

  6. 6

    Step 6: Final equation: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l).

Recap

  • Standard conditions are 100 kPa pressure and 298 K (25 °C).
  • The symbol ⦵ indicates standard conditions.
  • ΔH_f⦵ is for forming 1 mole of a compound from its elements in their standard states.
  • ΔH_c⦵ is for completely burning 1 mole of a substance in excess oxygen.
  • The standard enthalpy of formation of any element in its standard state is zero.
  • ΔH_neut⦵ is the enthalpy change when 1 mole of water is formed from the reaction of an acid and an alkali.

Quick check

  1. What is the value for the standard enthalpy of formation of N₂(g)? Explain your answer.2 marks
  2. What are the standard conditions for enthalpy change measurements?2 marks

3. Measuring Enthalpy Change: Calorimetry

Calorimetry is the experimental technique used to measure the heat transferred in a chemical reaction. For reactions in solution, a simple calorimeter can be made from a polystyrene cup with a lid and a thermometer. The key principle is that the heat released or absorbed by the reaction (the 'system') is exchanged with the solution (the 'surroundings'). By measuring the temperature change (ΔT) of the solution, we can calculate the heat energy(q) transferred using the equation q = mcΔT. Here, 'm' is the mass of the solution (often approximated as the mass of water used), 'c' is the specific heat capacity of the solution (usually taken as 4.18 J g⁻¹ K⁻¹ for water), and 'ΔT' is the change in temperature. This gives 'q' in Joules. To find the molar enthalpy change (ΔH), you must convert q to kJ (divide by 1000), calculate the number of moles(n) of the limiting reactant, and then find ΔH = -q/n. The negative sign is crucial: if the solution's temperature increases (exothermic), ΔH for the reaction must be negative.

q = mcΔT

ΔH = -q / n

Key term

Specific Heat Capacity (c): The amount of heat energy required to raise the temperature of one gram of a substance by one Kelvin (or one degree Celsius).

Examiner insight

Candidates must show clear, logical working steps. This includes calculating q, calculating moles, and then combining them for ΔH with the correct units and sign. Marks are often lost for unit errors (J vs kJ) or sign errors.

Common pitfall

Forgetting to divide by the number of moles to get the final answer in kJ mol⁻¹, or getting the sign wrong. A rise in temperature of the water means the reaction is exothermic, so ΔH must be negative.

Fun fact

The energy content of food, measured in Calories (Cal), is determined by burning the food in a device called a bomb calorimeter. One food Calorie is actually a kilocalorie (1000 calories).

Worked example 15 marks

In an experiment, 50.0 cm³ of 1.00 mol dm⁻³ HCl was mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH in a polystyrene cup. The initial temperature was 21.5 °C and the final temperature was 28.2 °C. Calculate the standard enthalpy of neutralisation (ΔH_neut⦵) in kJ mol⁻¹.

  1. 1

    Step 1: Calculate the temperature change, ΔT. ΔT = T_final - T_initial = 28.2 °C - 21.5 °C = 6.7 °C (or 6.7 K).

  2. 2

    Step 2: Calculate the heat energy absorbed by the solution, q. Total volume = 50.0 + 50.0 = 100.0 cm³. Assume density is 1.0 g cm⁻³, so mass(m) = 100.0 g. Specific heat capacity(c) = 4.18 J g⁻¹ K⁻¹. q = mcΔT = 100.0 g × 4.18 J g⁻¹ K⁻¹ × 6.7 K = 2800.6 J.

  3. 3

    Step 3: Calculate the moles of water formed. Moles of HCl = conc × vol = 1.00 mol dm⁻³ × (50.0/1000) dm³ = 0.0500 mol. Moles of NaOH = 1.00 mol dm⁻³ × (50.0/1000) dm³ = 0.0500 mol. The reaction is HCl + NaOH → NaCl + H₂O, a 1:1 ratio, so 0.0500 moles of H₂O are formed. n = 0.0500 mol.

  4. 4

    Step 4: Calculate the enthalpy change per mole, ΔH. First, convert q to kJ: 2800.6 J / 1000 = 2.8006 kJ.

  5. 5

    Step 5: Use ΔH = -q/n. The reaction is exothermic (temp increased), so ΔH is negative. ΔH = -2.8006 kJ / 0.0500 mol = -56.012 kJ mol⁻¹.

  6. 6

    Step 6: Give the answer to an appropriate number of significant figures (3 s.f. based on data). ΔH_neut⦵ = -56.0 kJ mol⁻¹.

Recap

  • Calorimetry measures heat changes in reactions.
  • Use q = mcΔT to find the heat energy (q) absorbed by the surroundings.
  • The mass 'm' is the total mass of the solution being heated or cooled.
  • The specific heat capacity 'c' of water is 4.18 J g⁻¹ K⁻¹.
  • To find molar enthalpy change ΔH, divide q (in kJ) by the moles of the limiting reactant.
  • Remember to add a negative sign for exothermic reactions (where temperature increases).

Quick check

  1. Why is a polystyrene cup used in simple calorimetry experiments?1 mark
  2. If 2000 J of energy is released in a reaction involving 0.05 moles of a reactant, what is the enthalpy change in kJ mol⁻¹?2 marks

4. Hess's Law and Enthalpy Cycles

Some enthalpy changes are impossible or very difficult to measure directly. For example, we cannot measure the enthalpy of formation of methane because carbon and hydrogen do not readily react to form only methane. Hess's Law provides a way around this. It states that the total enthalpy change for a reaction is independent of the route taken. This means if we can find an alternative, indirect route from reactants to products where all the steps have known enthalpy changes, we can calculate the unknown enthalpy change. We do this by constructing an enthalpy cycle. The two most common types of cycles use either standard enthalpies of formation (ΔH_f⦵) or standard enthalpies of combustion (ΔH_c⦵).

For a formation cycle: ΔH_reaction = ΣΔH_f⦵(products) - ΣΔH_f⦵(reactants)

For a combustion cycle: ΔH_reaction = ΣΔH_c⦵(reactants) - ΣΔH_c⦵(products)

Key term

Hess's Law: States that the total enthalpy change for a chemical reaction is independent of the route by which the reaction takes place, provided the initial and final conditions are the same.

Examiner insight

Examiners look for a clearly drawn and labelled enthalpy cycle or a correct application of the relevant formula. Following the arrows correctly in a cycle is crucial; going 'against' an arrow reverses the sign of the enthalpy change.

Common pitfall

Mixing up the two Hess's Law formulas. A simple way to remember: 'Formation is Forward' (Products - Reactants), 'Combustion is Contrary' (Reactants - Products).

Worked example 13 marks

Calculate the standard enthalpy change for the hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g), using the following standard enthalpy of formation data: ΔH_f⦵[C₂H₄(g)] = +52 kJ mol⁻¹, ΔH_f⦵[C₂H₆(g)] = -85 kJ mol⁻¹.

  1. 1

    Step 1: This problem uses formation data. The elements involved are C(s) and H₂(g). We can construct a cycle where the elements form both the reactants and the products.

  2. 2

    Step 2: The direct route is the reaction itself: C₂H₄(g) + H₂(g) → C₂H₆(g), which is ΔH_r.

  3. 3

    Step 3: The indirect route is: Reactants → Elements → Products. Going from elements to reactants is the reverse of formation, so we use -ΔH_f⦵(reactants). Going from elements to products is formation, so we use +ΔH_f⦵(products).

  4. 4

    Step 4: Apply the formula: ΔH_r = ΣΔH_f⦵(products) - ΣΔH_f⦵(reactants). Note that the enthalpy of formation of an element like H₂(g) is zero.

  5. 5

    Step 5: Substitute the values: ΔH_r = (ΔH_f⦵[C₂H₆(g)]) - (ΔH_f⦵[C₂H₄(g)] + ΔH_f⦵[H₂(g)]).

  6. 6

    Step 6: Calculate: ΔH_r = (-85) - (+52 + 0) = -85 - 52 = -137 kJ mol⁻¹.

Worked example 24 marks

Using the standard enthalpy of combustion (ΔH_c⦵) data below, calculate the standard enthalpy of formation (ΔH_f⦵) of methane, CH₄(g). ΔH_c⦵[C(s)] = -394 kJ mol⁻¹, ΔH_c⦵[H₂(g)] = -286 kJ mol⁻¹, ΔH_c⦵[CH₄(g)] = -890 kJ mol⁻¹.

  1. 1

    Step 1: The target reaction is the formation of methane: C(s) + 2H₂(g) → CH₄(g). We need to find ΔH_f for this reaction.

  2. 2

    Step 2: This problem uses combustion data. We can construct a cycle where the reactants and products are both burned to form combustion products (CO₂ and H₂O).

  3. 3

    Step 3: The direct route is the formation reaction, ΔH_f. The indirect route is: Reactants → Combustion Products → Product. The path from reactants to combustion products involves burning the reactants, so we use ΣΔH_c⦵(reactants). The path from combustion products back to the product is the reverse of burning the product, so we use -ΔH_c⦵(product).

  4. 4

    Step 4: Apply the formula: ΔH_reaction = ΣΔH_c⦵(reactants) - ΣΔH_c⦵(products). Here, the 'reaction' is the formation of methane.

  5. 5

    Step 5: Substitute the values: ΔH_f⦵[CH₄] = (ΔH_c⦵[C(s)] + 2 × ΔH_c⦵[H₂(g)]) - (ΔH_c⦵[CH₄(g)]). Note the '2 ×' for the two moles of H₂.

  6. 6

    Step 6: Calculate: ΔH_f⦵[CH₄] = (-394 + 2 × (-286)) - (-890) = (-394 - 572) - (-890) = -966 + 890 = -76 kJ mol⁻¹.

Recap

  • Hess's Law allows calculation of unknown enthalpy changes via an indirect route.
  • An enthalpy cycle visually represents the direct and indirect routes.
  • For cycles using ΔH_f⦵ data, the formula is ΔH_r = ΣΔH_f(products) - ΣΔH_f(reactants).
  • For cycles using ΔH_c⦵ data, the formula is ΔH_r = ΣΔH_c(reactants) - ΣΔH_c(products).
  • Always check stoichiometry and multiply enthalpy values by the number of moles in the balanced equation.

Quick check

  1. State Hess's Law in your own words.2 marks
  2. For a Hess's Law calculation using enthalpies of combustion, is the general formula 'products - reactants' or 'reactants - products'?1 mark

5. Calculating Enthalpy Change Using Bond Energies

Chemical reactions involve breaking old chemical bonds and making new ones. Energy is needed to break bonds, so bond breaking is an endothermic process (+ΔH). Energy is released when new bonds are formed, so bond formation is an exothermic process (-ΔH). The bond enthalpy is the energy required to break one mole of a specific covalent bond in the gaseous state. We can estimate the enthalpy change of a reaction by summing the energy required to break all the bonds in the reactants and subtracting the energy released when forming all the bonds in the products. Because the strength of a bond (like C-H) can vary slightly in different molecules, we use average bond enthalpies from a data table. This method only works for reactions where all species are in the gaseous state.

ΔH_reaction = Σ(bond enthalpies of bonds broken) - Σ(bond enthalpies of bonds formed)

Key term

Average Bond Enthalpy: The average energy required to break one mole of a specified type of bond in the gaseous phase, averaged over a wide variety of different compounds.

Examiner insight

Students who draw out the displayed formulas of all reactants and products to carefully count the number and type of each bond are most successful. It is a very common mistake to miscount the bonds.

Common pitfall

Getting the calculation the wrong way around. It's always 'Energy IN (broken) MINUS Energy OUT (formed)'. Also, forgetting to account for stoichiometric coefficients (e.g., 2H₂O has four O-H bonds, not two).

Worked example 14 marks

Calculate the enthalpy change for the combustion of methane. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). Use the average bond enthalpies: C-H = 413, O=O = 498, C=O = 805, O-H = 464 (all in kJ mol⁻¹).

  1. 1

    Step 1: Identify and count all the bonds broken in the reactants. It's essential to draw out the molecules. Reactants: 1 × CH₄ and 2 × O₂. Bonds broken: 4 × (C-H) bonds and 2 × (O=O) bonds.

  2. 2

    Step 2: Calculate the total energy input for bond breaking. Energy IN = (4 × 413) + (2 × 498) = 1652 + 996 = 2648 kJ.

  3. 3

    Step 3: Identify and count all the bonds formed in the products. Products: 1 × CO₂ and 2 × H₂O. Bonds formed: 2 × (C=O) bonds and 4 × (O-H) bonds (2 O-H bonds per water molecule, and there are 2 water molecules).

  4. 4

    Step 4: Calculate the total energy released from bond formation. Energy OUT = (2 × 805) + (4 × 464) = 1610 + 1856 = 3466 kJ.

  5. 5

    Step 5: Apply the formula: ΔH = Σ(bonds broken) - Σ(bonds formed).

  6. 6

    Step 6: Calculate the final answer: ΔH = 2648 - 3466 = -818 kJ mol⁻¹.

Worked example 24 marks

The enthalpy change for the reaction N₂(g) + 3H₂(g) → 2NH₃(g) is -92 kJ mol⁻¹. Use the bond enthalpies for N≡N (945 kJ mol⁻¹) and H-H (436 kJ mol⁻¹) to calculate the average bond enthalpy of the N-H bond in ammonia.

  1. 1

    Step 1: Write down the formula: ΔH = Σ(bonds broken) - Σ(bonds formed).

  2. 2

    Step 2: Identify bonds broken: 1 × (N≡N) and 3 × (H-H).

  3. 3

    Step 3: Identify bonds formed: In one NH₃ molecule there are 3 N-H bonds. Since 2 moles of NH₃ are formed, there are 2 × 3 = 6 × (N-H) bonds formed. Let the N-H bond enthalpy be 'x'.

  4. 4

    Step 4: Substitute all known values into the formula: -92 = [(1 × 945) + (3 × 436)] - [6 × x].

  5. 5

    Step 5: Simplify the equation: -92 = [945 + 1308] - 6x = 2253 - 6x.

  6. 6

    Step 6: Rearrange to solve for x: 6x = 2253 + 92 = 2345. x = 2345 / 6 = 390.83 kJ mol⁻¹.

  7. 7

    Step 7: Round to an appropriate number of significant figures. The average bond enthalpy of an N-H bond is 391 kJ mol⁻¹.

Recap

  • Bond breaking is endothermic (requires energy, positive value).
  • Bond formation is exothermic (releases energy, negative value).
  • The overall enthalpy change is the sum of energy for bonds broken minus the sum of energy for bonds formed.
  • Always draw out the displayed formulas of molecules to count the bonds correctly.
  • This method is an estimation and requires all species to be in the gaseous state.
  • Remember to multiply bond enthalpies by the number of moles of that bond in the balanced equation.

Quick check

  1. Is the process of breaking a chemical bond endothermic or exothermic?1 mark
  2. How many bonds are broken and how many are formed in the reaction H₂(g) + Cl₂(g) → 2HCl(g)?2 marks

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Define the term 'standard enthalpy change of formation' (ΔH_f⦵).3 marks
  2. A reaction has an enthalpy change of +65 kJ mol⁻¹. Is the reaction exothermic or endothermic? Draw a clearly labelled enthalpy profile diagram for this reaction, showing the relative enthalpies of reactants and products, ΔH, and the activation energy, Ea.4 marks
  3. When 2.94 g of solid potassium hydroxide (KOH) was dissolved in 100.0 cm³ of water in an insulated beaker, the temperature of the solution rose by 7.0 °C. Calculate the enthalpy change of solution for KOH in kJ mol⁻¹. (Mr of KOH = 56.1; assume the specific heat capacity of the solution is 4.18 J g⁻¹ K⁻¹ and its density is 1.0 g cm⁻³).5 marks
  4. Calculate the standard enthalpy change of combustion of ethene, C₂H₄(g), using the bond enthalpies provided. C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g). Bond enthalpies (kJ mol⁻¹): C=C: 614, C-H: 413, O=O: 498, C=O: 805, O-H: 464.4 marks
  5. Use the following standard enthalpy changes of formation to calculate the standard enthalpy change for the combustion of liquid ethanol, C₂H₅OH(l). ΔH_f⦵[C₂H₅OH(l)] = -278 kJ mol⁻¹, ΔH_f⦵[CO₂(g)] = -394 kJ mol⁻¹, ΔH_f⦵[H₂O(l)] = -286 kJ mol⁻¹.4 marks
  6. Write the chemical equation, including state symbols, that corresponds to the standard enthalpy change of atomisation of bromine.2 marks
  7. Explain why enthalpy changes calculated using average bond enthalpies may differ from values determined by calorimetry.2 marks
  8. The standard enthalpy changes of combustion of carbon(s), hydrogen(g), and propane(g), C₃H₈, are -394 kJ mol⁻¹, -286 kJ mol⁻¹, and -2219 kJ mol⁻¹ respectively. Construct a suitable enthalpy cycle and use it to calculate the standard enthalpy change of formation of propane.5 marks
  9. In an experiment to determine the enthalpy of displacement, 1.20 g of magnesium powder was added to 50.0 cm³ of 1.00 mol dm⁻³ copper(II) sulfate solution in a polystyrene cup. The temperature increased by 25.1 °C. Determine the limiting reactant and calculate the enthalpy change for the reaction in kJ per mole of magnesium. (Ar of Mg = 24.3; c = 4.18 J g⁻¹ K⁻¹; assume solution density = 1.0 g cm⁻³).6 marks
  10. State Hess's Law and explain its importance in thermochemistry.3 marks

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