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Enthalpy of Neutralisation or Heat of Neutralization Chemistry Tutorial

Key Concepts

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Polstyrene Cup Calorimeter for Heat of Neutralisation Experiment

Expanded polystyrene (polystyrene foam or styrofoam™) cups are often used as take-away coffee cups because the expanded polystyrene is a good insulator, that is, your coffee stays hot but you don't burn your fingers holding the cup!
This same insulating property can be exploited to make a reasonably good calorimeter (a device used to measure energy, or heat, change during a chemical reaction).
A schematic diagram of a simple polystyrene foam cup calorimeter is shown below:

A known amount of a reactant, such as a dilute solution of a base, is placed in the polystyrene cup (insulated vessel in the diagram).

The following data for this solution are recorded:
concentration: c1 = ? mol L-1
volume = V1 mL

A small hole is placed in the polystyrene lid to allow a thermometer to be pushed through. The fit must be snug enough to hold the thermometer in place, suspended off the bottom of the cup and immersed in the reactant.
It is assumed no heat will be lost through the lid or the hole in the lid.

The intial temperature of the reactant is measured and recorded.

Ti = ? °C

A known amount of the second reactant, for example a dilute solution of acid, is added to the solution in the cup.

The following data for this second solution are recorded:
concentration: c2 = ? mol L-1
volume = V2 mL

The thermometer is also used to stir the solution while the reaction is taking place.
The temperature of the solution in the cup will rise.
The maximum temperature reached is recorded as the final temperature.

Tf = ? °C

For the calculation of heat of neutralization (enthalpy of neutralisation) we first need to determine:

Now we can calculate the heat released by the neutralisation reaction, q,

q = m × Cg × ΔT

In order to determine the molar heat of neutralization (molar enthalpy of neutralisation), we need to determine how many moles of water, n(H2O(l)), have been formed as a result of the reaction:

H+(aq) + OH-(aq) → H2O(l)

So, using the stoichiometric ratio (mole ratio), we can see that:

n(H+(aq)) = n(OH-(aq)) n(H2O(l))

Now we can calculate the energy released per mole of water, or the molar enthaply of neutralisation (molar heat of neutralization), ΔHneut:

ΔHneut = -q ÷ n(H2O(l))

Note that neuralisation is an exothermic reaction, it releases heat, so ΔHneut must be negative.

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Heat of Neutralisation: Strong Monoprotic Acid + Strong Monobasic Base

HCl(aq) is a strong monoprotic acid, it completely dissociates (ionises) in water to produce hydrogen ions (H+(aq)) and chloride ions (Cl-(aq)):

HCl → H+(aq) + Cl-(aq)

n(HCl(aq)) : n(H+(aq)) is 1:1 (that is, monoprotic)

NaOH(aq) is a strong monobasic base, it completely dissociates (ionises) in water to produce sodium ions (Na+(aq)) and hydroxide ions (OH-(aq)):

NaOH → Na+(aq) + OH-(aq)

n(NaOH(aq)) : n(OH-(aq)) is 1:1 (that is, monobasic)

A neutralization reaction occurs when HCl(aq) is added to NaOH(aq)

HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)

and heat energy is given off (the reaction is said to be exothermic)

In an experiment to determine the molar enthalpy of neutralisation, 50.0 mL of 1.0 mol L-1 NaOH(aq) is placed in the styrofoam cup.

The temperature of the NaOH(aq) is recorded.

1.0 mol L-1 HCl(aq) at the same temperature is added 10.0 mL at a time.

The reaction mixture is stirred between each addition.

The maximum temperature the solution reached is then recorded.

The results of the experiment are shown in the table below:

total volume HCl(aq)
added (mL)
0 10 20 30 40 50 60 70
temperature in
calorimeter (°C)
18.0 20.2 21.8 22.9 23.8 24.6 24.0 23.6

and the results have been plotted on the graph shown below:

temperature
°C
HCl(aq) added to NaOH(aq)

total volume of
HCl(aq) added (mL)

Initially, the temperature of the reaction mixture in the calorimeter (styrofoam cup) increases as HCl(aq) is added.

Energy (heat) is being produced by the reaction.
The reaction is exothermic.

Maximum temperature reached is 24.6°C when 50.0 mL of HCl(aq) had been added.

When 50.0 mL of the acid has been added, all the base has been neutralised.

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

moles HCl(aq) added = moles of NaOH(aq) present in the calorimeter

Adding more acid doesn't increase the temperature in the calorimeter any further3.

We can calculate the molar heat of neutralisation for the reaction if we assume:

Calculating the molar enthalpy of neutralisation using the data from the experiment:

  1. Extract the data needed to calculate the molar heat of neutralisation for this reaction:
    V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL
    V(HCl) = volume of HCl(aq) added to achieve neutralisation = 50.0 mL
    c(NaOH) = concentration of NaOH(aq) = 1.0 mol L-1
    c(HCl) = concentration of HCl(aq) = 1.0 mol L-1
    Ti = initial temperature of solutions before additions = 18.0°C
    Tf = final temperature of solution at neutralisation = 24.6°C
    d = density of solutions = 1 g mL-1 (assumed)
    Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed)
    q = heat liberated during neutralisation reaction = ? J
  2. Check the units for consistency and convert if necessary:
    Convert volume of solutions (mL) to mass (g):
        density × volume = mass
    since density = 1 g mL-1:
        1 × volume (mL) = mass (g)
    mass(NaOH) = 50.0 g
    mass(HCl) = 50.0 g
  3. Calculate the heat produced during the neutralisation reaction:
    heat produced = total mass × specific heat capacity × change in temperature
    q = mtotal × Cg × ΔT

    mtotal = mass(NaOH) + mass(HCl) = 50.0 + 50.0 = 100.0 g
    Cg = 4.18 J°C-1g-1
    ΔT = Tf - Ti = 24.6 - 18.0 = 6.6°C

    q = 100.0 × 4.18 × 6.6 = 2758.8 J

  4. Calculate the moles of water produced:
    OH-(aq) + H+(aq) → H2O(l)
    1 mol OH-(aq) + 1 mol H+(aq) → 1 mol H2O
    moles(H2O) = moles(OH-(aq))
    moles(OH-(aq)) concentration (mol L-1) × volume (L)
        = 1.0 × 50.0/1000 = 0.050 mol
    moles of water produced = 0.050 mol
  5. Calculate the heat liberated per mole of water produced, ΔHneut :
    ΔHneut will be negative because the reaction is exothermic
    ΔHneut = heat liberated per mole of water
        = -1 × q ÷ moles of water
    ΔHneut = -1 × 2758.8 ÷ 0.050
        = -55176.0 J mol-1

    We can convert J to kJ by dividing by 1000:
    ΔHneut = -55176.0 J mol-1 ÷ 1000 J/kJ = 55.2 kJ mol-1

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Enthalpy of Neutralization: Strong Diprotic Acid and Strong Monobasic Base

The experiment described above is repeated using 50.0 mL of 1.0 mol L-1 sodium hydroxide, a strong monobasic base, and 1.0 mol L-1 sulfuric acid, a strong diprotic acid, instead of 1.0 mol L-1 hydrochloric acid, a strong monoprotic acid.

When plotted on a graph as shown below, the second experiment's results look different when compared to the first experiment's results:

temperature
°C
Graph

total volume of acid added (mL)

Initially, the temperature of the reaction mixture in both experiments increases as acid is added.
Energy (heat) is being produced by the reaction. The reaction is exothermic.

Maximum temperature reached for the reaction with H2SO4(aq) is higher than the maximum temperature reached for the reaction with HCl(aq).

Volume of H2SO(aq) added to reach the maximum temperature is less than the volume of HCl(aq) needed to reach maximum temperature.

For the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
    we found that ΔH = -55.2 kJ mol-1 (per mole of water formed)
    So, H+(aq) + OH- → H2O(l)   ΔHneut = -55.2 kJ mol-1

For the reaction H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
    2H+(aq) + 2OH-(aq) → 2H2O(l)
    so, H+(aq) + OH-(aq) → H2O(l)
    We predict that ΔHneut = -55.2 kJ mol-1

We can use the results of the second experiment to calculate the value for the molar heat of neutralisation (ΔHneut), and see if they agree with our prediction.

Calculate the molar heat of neutralisation for the reaction:

H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)

  1. Extract the data needed to calculate the molar heat of neutralisation for this reaction:
    V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL
    V(H2SO4) = volume of H2SO4(aq) added to achieve neutralisation = 25.0 mL
    c(NaOH) = concentration of NaOH(aq) = 1.0 mol L-1
    c(H2SO4) = concentration of H2SO4(aq) = 1.0 mol L-1
    Ti = initial temperature of solutions before additions = 18.0°C
    Tf = final temperature of solution at neutralisation = 26.9°C
    d = density of solutions = 1 g mL-1 (assumed)
    Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed)
    q = heat liberated during neutralisation reaction = ? J
  2. Check the units for consistency and convert if necessary:
    Convert volume of solutions (mL) to mass (g):
        density × volume = mass
    since density = 1 g mL-1:
        1 × volume (mL) = mass (g)
    mass(NaOH) = 50.0 g
    mass(H2SO4) = 25.0 g
  3. Calculate the heat produced during the neutralisation reaction:
    heat produced = total mass × specific heat capacity × change in temperature
    q = mtotal × Cg × ΔT

    mtotal = mass(NaOH) + mass(H2SO4() = 50.0 + 25.0 = 75.0 g
    Cg = 4.18 J°C-1g-1
    ΔT = Tf - Ti = 26.9 - 18.0 = 8.9°C

    q = 75.0 × 4.18 × 8.9 = 2790.2 J

  4. Calculate the moles of water produced:

    OH-(aq) + H+(aq) → H2O(l)

    1 mol OH-(aq) + 1 mol H+(aq) → 1 mol H2O

    moles(H2O) = moles(OH-(aq))
    moles(OH-(aq)) = concentration (mol L-1) × volume (L)
        = 1.0 × 50.0/1000
        = 0.050 mol
    moles of water produced = 0.050 mol

  5. Calculate the heat liberated per mole of water produced, ΔHneut :
    ΔHneut will be negative because the reaction is exothermic
    ΔHneut = heat liberated per mole of water
       = -1 × q ÷ moles of water
    ΔHneut = -1 × 2790.2 ÷ 0.050
        = -55803 J mol-1 (of water produced)
    We can divide J by 1000 to convert this enthalpy change to kJ per mole:
    ΔHneut = 55803 J mol-1 ÷ 1000 J/kJ
        = 55.8 kJ mol-1 (of water produced)
  6. Compare the molar heat of neutralisation for:

    HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔHneut = -55.2 kJ mol-1 (of water)
    H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) ΔHneut = -55.8 kJ mol-1 (of water)

    There is close agreement between the two values for molar heat of neutralisation, so we can generalise and say that the molar heat of neutralisation for the reaction between a strong acid and a strong base is a constant.

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Molar Enthalpy of Neutralisation: Weak Acid + Strong Base

The experiment described above is repeated using 50.0 mL of 1.0 mol L-1 sodium hydroxide, a strong monobasic base, and 1.0 mol L-1 hydrogren cyanide (HCN), a weak monoprotic acid (Ka ≈ 6 × 10-10), instead of 1.0 mol L-1 hydrochloric acid, a strong monoprotic acid.

The results of these experiments is shown in the graph below:

temperature
°C
Graph

total volume of acid added (mL)

Initially, the temperature of the reaction mixture in both experiments increases as acid is added.
Energy (heat) is being produced by the reaction. The reaction is exothermic.

Maximum temperature reached for the reaction with HCN(aq) is much less than the maximum temperature reached for the reaction with HCl(aq).

Volume of HCN(aq) added to reach the maximum temperature is the same as the volume of HCl(aq) needed to reach maximum temperature (both volumes are 50.0 mL).

For the reaction HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
    we found that ΔH = -55.2 kJ mol-1 (per mole of water formed)
    So, H+(aq) + OH- → H2O(l)   ΔHneut = -55.2 kJ mol-1

For the reaction HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l)
    Will ΔH for this reaction be the same (-55.2 kJ mol-1) ?

We can use the results of the HCN experiment to calculate the value for the molar heat of neutralisation (ΔHneut), and see.

  1. Extract the data needed to calculate the molar heat of neutralisation for this reaction:
    V(NaOH) = volume of NaOH(aq) in the calorimeter = 50.0 mL
    V(HCN) = volume of HCN(aq) added to achieve neutralisation = 50.0 mL
    c(NaOH) = concentration of NaOH(aq) = 1.0 mol L-1
    c(HCN) = concentration of HCN(aq) = 1.0 mol L-1
    Ti = initial temperature of solutions before additions = 18.0°C
    Tf = final temperature of solution at neutralisation = 19.2°C
    d = density of solutions = 1 g mL-1 (assumed)
    Cg = specific heat capacity of solutions = 4.18 J°C-1g-1 (assumed)
    q = heat liberated during neutralisation reaction = ? J
  2. Check the units for consistency and convert if necessary:
    Convert volume of solutions (mL) to mass (g):
        density × volume = mass
    since density = 1 g mL-1 (assumed):
        1 × volume (mL) = mass (g)
    mass(NaOH) = 50.0 g
    mass(HCN) = 50.0 g
  3. Calculate the heat produced during the neutralisation reaction:
    heat produced = total mass × specific heat capacity × change in temperature
    q = mtotal × Cg × ΔT

    mtotal = mass(NaOH) + mass(HCN) = 50.0 + 50.0 = 100.0 g
    Cg = 4.18 J°C-1g-1
    ΔT = Tf - Ti = 19.2 - 18.0 = 1.2°C

    q = 100.0 × 4.18 × 1.2 = 501.6 J

  4. Calculate the moles of water produced:
    NaOH(aq) + HCN(aq) → NaCN(aq) + H2O(l)
    moles(H2O) = moles(NaOH)
    moles(NaOH(aq)) = concentration (mol L-1) × volume (L)
        = 1.0 × 50.0/1000 = 0.050 mol
    moles of water produced = 0.050 mol
  5. Calculate the heat liberated per mole of water produced, ΔHneut :
    ΔHneut will be negative because the reaction is exothermic
    ΔHneut = heat liberated per mole of water
        = -1 × q ÷ moles of water
    ΔHneut = -1 × 501.6 ÷ 0.050
        = -10032 J mol-1

    We can convert J to kJ by dividing by 1000:
    ΔHneut = -10032 J mol-1 ÷ 1000 J/kJ
        = -10.0 kJ mol-1

  6. Compare the molar heat of neutralisation for the neuralisation of NaOH(aq) by both HCl(aq) and HCN(aq):

    HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ΔHneut = -55.2 kJ mol-1 (of water)
    HCN(aq) + NaOH(aq) → NaCN(aq) + H2O(l) ΔHneut = -10.0 kJ mol-1 (of water)

    The heat released per mole of water for the hydrogen cyanide reaction is much less than the heat released per mole of water for the hydrochloric acid reaction.

The difference in molar heats of neutralisation is due to the type of reaction taking place:

The heat liberated per mole when a weak acid neutralises a strong base is less than the amount of heat liberated per mole when a strong acid neutralises a strong base.

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Footnotes

1. A better method for measuring heat of neutralization is to use an adiabatic calorimeter fitted with an electrical heater. A desciption of this type of calorimeter can be found in the calorimetry tutorial.

2. You will find slightly different values quoted for molar heat of neutralisation mostly because the neutralisation reaction is dependent on the temperature at which the reaction occurs. In general the values you see quoted will be between 55 kJ mol-1 and 58 kJ mol-1 and refer to reactions that take place at ambient temperatures in a laboratory.

3. The temperature is likely to decrease as more acid is added because the heat that was generated by the completed reaction is being dissipated in a greater mass of solution.