Chemical Energetics

Question 1

What is the enthalpy of a chemical system?

Answer 1

• The total energy content within a chemical system
• It is measured per mole
• Every chemical reaction has an enthalpy change associated with it
• A positive enthalpy change (+ΔH) implies an endothermic reaction and a negative enthapy change (-ΔH) implies an exothermic reaction

Question 2

What is activation energy?

Answer 2

• The energy required for successful collisions between reactants to occur
• It is the enthalpy difference between the reactants and the transition state
• Endothermic reactions have larger activation energies as they absorb larger amounts of energy from the environment

There will be a small decrease in enthalpy between the transition state and th products in an endothermic reaction, releasing a small amount of energy; however, the overall enthalpy change is still positive

Question 3

What are standard conditions?

Answer 3

• 298K (25 degrees Celsius)
• 101kPa (1atm)
• 1moldm⁻³ (if concentration is not given)

⁰ or ⦵ denotes standard conditions

Question 4

What is the enthalpy of reaction?

Answer 4

• The enthalpy change of a reaction, with all reactants and products present in the amounts stated in the balanced chemical equation
• Example: N₂(g) + 3H₂(g) → 2NH₃(g)
• For this reaction, the enthalpy of reaction is the enthalpy change when one mole of nitrogen reacts with three moles of hydrogen

Question 5

What is the enthalpy of formation?

Answer 5

• The enthalpy change when one mole of a substance is formed from its constituent elements with all substances in their standard states
• It is exothermic for most substances
• Example: 2Na(s) + 1/2O₂(g) → Na₂O(s)

Question 6

What is the enthalpy of combustion?

Answer 6

• The enthalpy change when one mole of a substance undergoes complete combustion in oxygen with all substances in standard state
• It is exothermic
• Example: H₂(g) + 1/2O₂(g) → H₂O(g)

Question 7

What is the enthalpy of neutralisation?

Answer 7

• The enthalpy change when one mole of water is formed in a reaction between an acid and alkali under standard conditions
• It is exothermic
• Example: 1/2H₂SO₄(aq) + NaOH(aq) → 1/2Na₂SO₄(aq) + H₂O(l)

Question 8

What is the first ionisation enthalpy?

Answer 8

• The enthalpy change when each atom of one mole of gaseous atoms loses an electron to form one mole of gaseous 1+ ions
• It is endothermic
• Example: Mg(g) → Mg⁺(g) + e⁻

Question 9

What is the second ionisation enthalpy?

Answer 9

• The enthalpy change when each ion of one mole of gaseous 1+ ions loses an electron to form one mole of gaseous 2+ ions
• It is endothermic
• Example: Mg⁺(g) → Mg²⁺(g) + e⁻

Question 10

What is the first electron affinity enthalpy?

Answer 10

• The enthalpy change when each atom in one mole of gaseous atoms gains one electron to form one mole of gaseous 1- ions
• It is generally exothermic
• Example: O(g) + e⁻ → O⁻(g)

Question 11

What is the second electron affinity enthalpy?

Answer 11

• The enthalpy change when each ion in one mole of gaseous 1- atoms gains one electron to form one mole of gaseous 2- ions
• It is endothermic
• Example: O⁻(g) + e⁻ → O²⁻(g)

Question 12

What is the enthalpy of atomisation?

Answer 12

• The enthalpy change when one mole of gaseous atoms is produced from an element in its standard state
• It is endothermic
• Example: 1/2I₂(s) → I(g)

Question 13

What is the hydration enthalpy?

Answer 13

• The enthalpy change when one mole of gaseous ions becomes hydrated (dissolved in water)
• It is exothermic
• Example: Mg²⁺ (g) + aq → Mg²⁺ (aq)

Question 14

What is the enthalpy of solution?

Answer 14

• The enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough so that the dissolved ions are separated and do not interact
• It can be exothermic or endothermic
• Example: MgCl₂(s) + aq → Mg²⁺(aq) + 2Cl⁻(aq)

Question 15

What is the bond dissociation enthalpy?

Answer 15

• The enthalpy change when one mole of covalent bonds is broken in the gaseous state
• It is endothermic
• Example: I₂(g) → 2I(g)

One mole of covalent bonds means 6.022 × 10²³ single covalent bonds, so, for example, dissociating both of the OH bonds in water would be bond dissociation enthalpy x2

Question 16

What is the lattice enthalpy of formation?

Answer 16

• The enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gas phase
• It is exothermic
• Example: Mg²⁺(g) + 2Cl⁻(g) → MgCl₂(s)

Question 17

What is the lattice enthalpy of dissociation?

Answer 17

• The enthalpy change when one mole of a solid ionic compound is broken into its constituent ions in the gas phase
• It is endothermic
• Example: MgCl₂(s) → Mg²⁺(g) + 2Cl⁻(g)

Question 18

What is the enthalpy of vaporisation?

Answer 18

• The enthalpy change when one mole of a liquid is turned into a gas
• It is endothermic
• Example: H₂O(l) → H₂O (g)

Question 19

What is the enthalpy of fusion?

Answer 19

• The enthalpy change when one mole of a solid is turned into a liquid
• It is endothermic
• Example: Mg(s) → Mg(l)

Question 20

How are bond energies used to calculate enthalpy change?

Answer 20

• This method is used when enthalpy change cannot be measured experimentally
• The bond energies of the reactants minus the bond energies of the products yields the overall enthalpy change for the reaction
• A higher bond enthalpy for the products, for example, would yield a negative enthalpy change, implying the reaction is exothermic

• Bond energies are often averages as they depend on which other atoms are present in a molecule
• Bond energies are always measured in the gaseous state to ensure comparability

Question 21

How can enthalpy change be calculated from experimental results?

Answer 21

• Calculate the total energy transfer associated with the reaction using the equation q = mcΔT
• m is the mass of the substance being heated and c is the specific heat capacity of the substance being heated
• Once q has been obtained, divide it by the number of moles of a reactant (provided it is not in excess) to find the enthalpy change of the reaction

Question 22

How can total temperature change be calculated using graphical extrapolation?

Answer 22

• Instrumental reading often misrepresent the maximum temperature reached in a calorimetry experiment as heat is lost after the reaction begins
• Record the temperature before mixing reactants
• Continue to record the temperature at regular intervals while the reaction is taking place and after it has taken place while the products are cooling down
• Draw a graph of temperature against time
• Extrapolate the line showing the temperature trends of the reactants before the reaction forwards and extrapolate the line showing the cooling of the products backwards
• Draw a straight vertical line connecting the two extrapolated lines at the point when the reactants were added to obtain a more accurate value for total temperature change

Question 23

What is Hess’s law?

Answer 23

The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place provided the initial and final conditions are the same

Question 24

How can Hess cycles be used to calculate reaction enthalpies?

Answer 24

• They are used to find enthalpy changes that cannot be obtained experimentally
• The enthalpy change of an alternative method of obtaining the products from the reactants is found
• This will be equal to the enthalpy change of the original method of obtaining the products from the reactants
• One method may contain more steps than the other
• Pay attention to the number of moles involved; the enthalpy of formation of a substance, for example, will only account for one mole

Hess cycles can be used to obtain bond energy values as well

Question 25

How would one solve this problem?

Answer 25

• Draw a Hess cycle with ethane’s constituent elements in their standard states (C (graphite) and H₂ (g)) at the top left, ethane (C₂H₆) at the top right, and the combustion products (CO₂ (g) and H₂O (l)) at the bottom.
• Connect both ethane and its elements to the combustion products using downward arrows. Label these arrows with their respective enthalpies of combustion, adjusting for the number of moles:
  – 2C (graphite): 2 × (–393.5 kJ mol⁻¹)
  – 3H₂ (g): 3 × (–285.8 kJ mol⁻¹)
  – C₂H₆ (g): –1559.7 kJ mol⁻¹
• Connect the elements to ethane with an arrow representing the enthalpy of formation of ethane — this is the unknown to be calculated.
• Apply Hess’s Law:
  [2(–393.5) + 3(–285.8)] – (–1559.7)
• Calculate the result:
  = –1644.4 + 1559.7
  = –84.7 kJ mol⁻¹

Question 26

How would one solve this problem?

Answer 26

• Draw a Hess cycle with CH₄ (g) at the top left, its elements in standard states (C (graphite) and H₂ (g)) at the bottom, and its atomised elements (C (g) and H (g)) at the top right.
• Connect the elements in their standard states to methane with an arrow labelled with the standard enthalpy of formation of CH₄ (–74.8 kJ mol⁻¹).
• Connect the elements in their standard states to the atomised elements with arrows labelled using their respective enthalpies of atomisation:
  – C (graphite) → C (g): +717.7 kJ mol⁻¹
  – 2H₂ (g) → 4H (g): 4 × 218 = +872.0 kJ mol⁻¹
• Connect the atomised elements (C (g) + 4H (g)) to CH₄ (g) with an arrow representing 4 × bond enthalpy (C–H) — this is the unknown to be calculated.
• Apply Hess’s Law:
  Total atomisation energy = (717.7 + 872.0) - (–74.8) = 4 × bond enthalpy
  1589.7 + 74.8 = 4 × bond enthalpy
  1664.5 = 4 × bond enthalpy
• Divide by 4 to find the average C–H bond enthalpy:
  Bond enthalpy = 1664.5 ÷ 4 = +416 kJ mol⁻¹