# 1.5.4 Enthalpy & Bond Energies

### Enthalpy & Bond Energies

• During a reaction, enthalpy changes take place because bonds are being broken and formed
• Energy (in the form of heat) is needed to overcome attractive forces between atoms
• Bond breaking is therefore endothermic
• Energy is released from the reaction to the surroundings (in the form of heat) when new bonds are formed
• Bond forming is therefore exothermic

To break bonds energy is required from the surroundings and to make new bonds energy is released from the reaction to the surroundings

• If more energy is required to break bonds than energy is released when new bonds are formed, the reaction is endothermic
• If more energy is released when new bonds are formed than energy is required to break bonds, the reaction is exothermic
• In reality, only some bonds in the reactants are broken and then new ones are formed

### Enthalpy Calculations using Experimental Results

#### Measuring enthalpy changes

• Calorimetry is a technique used to measure changes in enthalpy of chemical reactions
• A calorimeter can be made up of a polystyrene drinking cup, a vacuum flask or metal can
• The energy needed to increase the temperature of 1 g of a substance by 1 oC is called the specific heat capacity (c) of the liquid
• The specific heat capacity of water is 4.18 J g-1 oC-1

#### Calculating enthalpy changes

• The energy transferred as heat can be calculated by:

ΔH = -m × c × ΔT

ΔH = the heat transferred, J

m = the mass of water, g

c = the specific heat capacity, J g-1 oC-1

ΔT = the temperature change, oC

• When there is a rise in temperature the value for ΔH becomes negative suggesting that the reaction is exothermic
• When the temperature falls the value for ΔH becomes positive suggesting that the reaction is endothermic

#### Worked example: Specific heat capacity calculations

• Step 1: Use the equation

ΔH = -m × c × ΔT

m (of water) = 500 g

c (of water) = 4.18 J g-1 K-1

ΔT (of water) = 68 oC – 25 oC

= 43 oC

= 43 K

The change in temperature in oC is equal to the change in temperature in K

• Step 2: Calculate the enthalpy change

Δ= -500 × 4.18 × 43

= -89 870 J

• Step 3:Calculate the total enthalpy change
This (-89 870 J) is only 30% of the total enthalpy change when methane is burnt

Total enthalpy change x 0.3 = -89 870 J

= -299 567 J

• Step 4: This is the enthalpy change when 2.50 g of methane is burnt

INSERT EQUATION

= 0.16 mol

So, when 0.16 moles of methane is burnt, the enthalpy change is -299 567 J

• Step 5 – Calculate the enthalpy change when 1 mole of methane is burnt

INSERT EQUATION

= -1 872 294 J

= -1.87 x 103 kJ mol-1

#### Exam Tip

Aqueous solutions of acid, alkalis and salts are assumed to be largely water so you can just use the m and c values of water when calculating the energy transferred.

And remember: the ΔT is the same in oC and K!

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