ย 

Combining energies for OCR A-level Chemistry

Combining energies

This page covers the following topics:

1. Bond energies
2. Hessโ€™s law

The breaking of bonds is endothermic, while the formation of bonds is exothermic. Combining these two processes can yield both endothermic and exothermic reactions depending on whether more energy is required to break the bonds or is given when new ones are created.

The energy transferred in a chemical reaction can be calculated by subtracting total bond enthalpies of products from the total bond enthalpies of reactants. If multiple same bonds are broken or formed, multiples of the bond energies are used.

Mean bond enthalpy is and average of bond energies across various molecules. Mean bond enthalpy can be calculated by combining an energy change in a reaction and other bond energies, essentially working backwards to how energy change of a reaction is found. Mean bond enthalpies do not always match to a bond energy of an atom in a specific molecule or structure since factors such as bond strength and interactions with other particles can influence the individual values.

Bond energies

Hessโ€™s law states that the overall energy change is the same regardless of the reactions used to get products from reactants. Thus, multiple enthalpies of reactions can be combined to work out the enthalpy of another reaction if adding up the equations gives the required equation. If an equation is opposite to the one required, it can be inverted, and an opposite value of the enthalpy change is used. If multiples of a reaction equation are required, multiples of the enthalpy change are used. To visualise Hessโ€™s law calculations, enthalpy cycle can be used which is a diagram that shows equations that can be combined to get a new one in order to find the enthalpy change of it.

For calculations that only use the enthalpies of formation of compounds involved in a reaction, the enthalpy change can be found by subtracting formation enthalpies of reactants from the formation enthalpies of products. For calculations that only use the enthalpies of combustion of compounds involved in a reaction, the enthalpy change can be found by subtracting combustion enthalpies of products from the combustion enthalpies of reactants.

Hessโ€™s law

1

Find the enthalpy change of the reaction provided by using the enthalpies of formation provided.

Cโ‚‚Hโ‚‚ + 2Hโ‚‚ โ†’ Cโ‚‚Hโ‚†

ฮ”Hf(Cโ‚‚Hโ‚‚) = โˆ’227 kJ/mol
ฮ”Hf(Cโ‚‚Hโ‚†) = โˆ’85 kJ/mol

For calculations that only use the enthalpies of formation of compounds involved in a reaction, the enthalpy change can be found by subtracting formation enthalpies of reactants from the formation enthalpies of products. Hydrogen formation enthalpy is not included in the calculations as it is a pure compound.

ฮ”H = ฮ”Hf(Cโ‚‚Hโ‚†) โˆ’ ฮ”Hf(Cโ‚‚Hโ‚‚)
ฮ”H = โˆ’85 โˆ’ (โˆ’227)
ฮ”H = 142 kJ/mol

142 kJ/mol

Find the enthalpy change of the reaction provided by using the enthalpies of formation provided.

Cโ‚‚Hโ‚‚ + 2Hโ‚‚ โ†’ Cโ‚‚Hโ‚†

ฮ”Hf(Cโ‚‚Hโ‚‚) = โˆ’227 kJ/mol
ฮ”Hf(Cโ‚‚Hโ‚†) = โˆ’85 kJ/mol

2

Li has drawn a diagram that shows the bonds of both reactants and products of a reaction. Calculate the molar change of energy of the reaction by using the mean bond enthalpies provided below.

ฮ”H(Hโˆ’H) = 436 kJ/mol
ฮ”H(O=O) = 495 kJ/mol
ฮ”H(Nโ‰กN) = 941 kJ/mol
ฮ”H(Nโˆ’H) = 391 kJ/mol
ฮ”H(Oโˆ’H) = 463 kJ/mol

The energy transferred in a chemical reaction can be calculated by subtracting total bond enthalpies of products from the total bond enthalpies of reactants. If multiple same bonds are broken or formed, multiples of the bond energies are used.

In this case two Hโˆ’H, one O=O bonds are broken and four Oโˆ’H bonds are made.
ฮ”H = 436 ร— 2 + 495 โˆ’ 463 ร— 4
ฮ”H = 872 + 495 โˆ’ 1852
ฮ”H = โˆ’485 kJ/mol

โˆ’485 kJ/mol

Li has drawn a diagram that shows the bonds of both reactants and products of a reaction. Calculate the molar change of energy of the reaction by using the mean bond enthalpies provided below.

ฮ”H(Hโˆ’H) = 436 kJ/mol
ฮ”H(O=O) = 495 kJ/mol
ฮ”H(Nโ‰กN) = 941 kJ/mol
ฮ”H(Nโˆ’H) = 391 kJ/mol
ฮ”H(Oโˆ’H) = 463 kJ/mol

3

How many bonds are broken during the reaction provided?

CHโ‚„ + 2Oโ‚‚ โ†’ COโ‚‚ + 2Hโ‚‚O

CHโ‚„ has 4 Cโˆ’H bonds and 2Oโ‚‚ has 2 O=O bonds; in total 4 + 2 = 6.

6

How many bonds are broken during the reaction provided?

CHโ‚„ + 2Oโ‚‚ โ†’ COโ‚‚ + 2Hโ‚‚O

4

Luca has drawn an enthalpy cycle for several reactions with ethane. Find ฮ”H if ฮ”Hโ‚ = 272 kJ/mol and ฮ”Hโ‚‚ = โˆ’3392 kJ/mol.

Hessโ€™s law states that the overall energy change is the same regardless of the reactions used to get products from reactants. Thus, multiple enthalpies of reactions can be combined to work out the enthalpy of another reaction if adding up the equations gives the required equation. If an equation is opposite to the one required, it can be inverted, and an opposite value of the enthalpy change is used.

ฮ”H = ฮ”Hโ‚ + ฮ”Hโ‚‚
ฮ”H = 272 โˆ’ 3392
ฮ”H = โˆ’3120 kJ/mol

โˆ’3120 kJ/mol

Luca has drawn an enthalpy cycle for several reactions with ethane. Find ฮ”H if ฮ”Hโ‚ = 272 kJ/mol and ฮ”Hโ‚‚ = โˆ’3392 kJ/mol.

5

Bethany has drawn an enthalpy cycle for several reactions with carbon compounds. Find ฮ”Hโ‚ if ฮ”H = โˆ’791 kJ/mol and ฮ”Hโ‚‚ = โˆ’566 kJ/mol.

Hessโ€™s law states that the overall energy change is the same regardless of the reactions used to get products from reactants. Thus, multiple enthalpies of reactions can be combined to work out the enthalpy of another reaction if adding up the equations gives the required equation. If an equation is opposite to the one required, it can be inverted, and an opposite value of the enthalpy change is used.

ฮ”H = ฮ”Hโ‚ + ฮ”Hโ‚‚
โˆ’791 = ฮ”Hโ‚ โˆ’ 566
ฮ”Hโ‚ = โˆ’791 + 566
ฮ”Hโ‚ = โˆ’225 kJ/mol

โˆ’225 kJ/mol

Bethany has drawn an enthalpy cycle for several reactions with carbon compounds. Find ฮ”Hโ‚ if ฮ”H = โˆ’791 kJ/mol and ฮ”Hโ‚‚ = โˆ’566 kJ/mol.

End of page

ย