Enthalpy of Neutralisation or Heat of Neutralization Chemistry Tutorial

⚛ Neutralisation, or neutralization, is the name given to the reaction that occurs between an Arrhenius acid and an Arrhenius base. ⁽¹⁾

H⁺_(aq) + OH^-_(aq) → H₂O_(l)

⚛ When an acid is added to an aqueous solution of base, the temperature of the solution increases.

Or, if a base is added to an aqueous solution of an acid, the temperature of the solution increases.

⚛ Energy (heat) is produced when an acid reacts with a base in a neutralisation reaction.

· Neutralisation reactions are exothermic.

· ΔH for a neutralisation reaction is negative.

⚛ Molar enthalpy of neutralisation (molar heat of neutralization) is the energy liberated per mole of water formed during a neutralisation reaction.

· Δ_neutH (or ΔH_neut) is the symbol given to the molar enthalpy of neutralisation. ⁽²⁾

· Δ_neutH is usually given in units of kJ mol^-1
(kJ of energy released per mole of water produced)

⚛ Enthalpy of neutralisation can be determined in the school laboratory using a styrofoam™ cup solution calorimter⁽³⁾. Calculating the molar enthalpy of neutralisation from experimental results is a 3 step process:

Step 1: Calculate the heat evolved: q = m × C_g × ΔT
m = total mass of reaction mixture
C_g = specific heat capacity of solution
ΔT = change in temperature of solution

Step 2: Calculate the enthalpy change for the reaction: ΔH = −q

Step 3: Calculate the molar enthalpy of neutralisation: Δ_neutH = ΔH ÷ n(H₂O_(l))

⚛ Molar enthalpy of neutralisation for reactions between dilute aqueous solutions of strong acid and strong base is always the same⁽⁴⁾, that is,

Δ_neutH = -55.90 kJ mol ^-1

because no bonds need to be broken, and because making the H-O bonds in H₂O releases energy
(breaking bonds is an endothermic process, making bonds is an exothermic process)

⚛ Less than 55.90 kJ mol^-1 of energy is released when:

(a) a weak acid neutralises a strong base

(b) a strong acid neutralises a weak base

(c) a weak acid neutralises a weak base

because some of the energy is consumed in the process of breaking weak acid bonds and/or weak base bonds.

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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 aqueous 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.

initial temperature = T_i = ? °C

A known amount of the second reactant, for example a dilute solution of acid, at the same temperature 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.

final temperature = T_f = ? °C

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

(i) total mass, m, of the solution in the cup

First assume additivity of volumes so that total volume of solution, V_f, is the sum of the volume of the two reactants:

V_f = V₁ + V₂

Then assume that, because all of the solutions are dilute aqueous solutions, the density of each solution and hence the density of the final solution, d, is the same as water which we will assume is 1.00 g mL^-1

density = mass (g) ÷ volume (mL)
Substitute the density of water into the equation:
1.00 (g/mL) = mass (g) ÷ V_f (mL)
Multiplying both sides of the equation by V_f (mL)
V_f (mL) × 1.00 (g/mL) = V_f (mL) × mass (g) ÷ V_f (mL)
V_f × 1.00 = mass (g) = m g

(ii) specific heat capacity, C_g, of the solution

Assume that, because all of the solutions are dilute aqueous solutions, the specific heat capacity of each solution and hence the specific heat capacity of the final solution, C_g, is the same as water which we will assume is 4.18 J g^-1 °C^-1
C_g = 4.18 J g^-1 °C^-1

(iii) change in temperature, ΔT, as a result of the neutralisation reaction:

ΔT = (T_f − T_i) °C

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

q = m × C_g × ΔT

In order to determine the molar enthalpy of neutralisation (molar heat of neutralization), 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), Δ_neutH

Remember, the reaction is exothermic so the sign of ΔH will be negative!

Δ_neutH = −q ÷ n(H₂O_(l))

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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_(aq) → 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_(aq) → 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 1.00 mol L^-1 NaOH_(aq) is placed in the styrofoam cup.

The temperature of the NaOH_(aq) is recorded.

1.00 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:

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

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 further⁽⁵⁾.

We can calculate the molar enthalpy of neutralisation (molar heat of neutralization) for the reaction if we assume that the:

(i) density of each dilute aqueous solution is the same as water, 1 g mL^-1 at 25°C
so, the mass of solution in grams = volume of solution in mL

(ii) heat capacity of each solution is the same as for water, C_g = 4.18 J°C^-1g^-1

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

Step 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) = 1.00 mol L^-1
c_(HCl) = concentration of HCl_(aq) = 1.00 mol L^-1
T_i = initial temperature of solutions before additions = 18.0°C
T_f = final temperature of solution at neutralisation = 24.6°C
d = density of solutions = 1 g mL^-1 (assumed)
C_g = specific heat capacity of solutions = 4.18 J°C^-1g^-1 (assumed)
q = heat liberated during neutralisation reaction = ? J

Step 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

Step 3: Calculate the heat produced during the neutralisation reaction:

heat produced = total mass × specific heat capacity × change in temperature
q = m_total × C_g × ΔT

m_total = mass_(NaOH) + mass_(HCl) = 50.0 + 50.0 = 100.0 g
C_g = 4.18 J°C^-1g^-1
ΔT = T_f - T_i = 24.6 - 18.0 = 6.6°C

q = 100.0 g × 4.18 J g^-1 °C^-1 × 6.6 °C = 2758.8 J

Step 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 mol L^-1 × 50.0 mL/1000 mL/L = 0.050 mol
moles of water produced = 0.050 mol

Step 5: Calculate the heat liberated per mole of water produced, Δ_neutH :

Δ_neutH will be negative because the reaction is exothermic
Δ_neutH = heat liberated per mole of water
    = -1 × q ÷ moles of water
Δ_neutH = -1 × 2758.8 J ÷ 0.050 mol
    = -55176.0 J mol^-1

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

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The experiment described above is repeated using 50.0 mL of 1.00 mol L^-1 sodium hydroxide, a strong monobasic base, and 1.00 mol L^-1 sulfuric acid, a strong diprotic acid, instead of 1.00 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:

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 H₂SO_4(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 Δ_neutH = -55.2 kJ mol^-1 (per mole of water formed)
    So, H⁺_(aq) + OH^- → H₂O_(l)   Δ_neutH = -55.2 kJ mol^-1

For the reaction H₂SO_4(aq) + 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)
    2H⁺_(aq) + 2OH^-_(aq) → 2H₂O_(l)
    so, H⁺_(aq) + OH^-_(aq) → H₂O_(l)
    We predict that Δ_neutH = -55.2 kJ mol^-1

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

Calculate the molar heat of neutralisation for the reaction from the results of the experiment shown in the graph above:

H₂SO_4(aq) + 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

Step 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 H₂SO_4(aq) added to achieve neutralisation = 25.0 mL
c_(NaOH) = concentration of NaOH_(aq) = 1.00 mol L^-1
c_(H2SO4) = concentration of H₂SO_4(aq) = 1.00 mol L^-1
T_i = initial temperature of solutions before additions = 18.0°C
T_f = final temperature of solution at neutralisation = 26.9°C
d = density of solutions = 1 g mL^-1 (assumed)
C_g = specific heat capacity of solutions = 4.18 J°C^-1g^-1 (assumed)
q = heat liberated during neutralisation reaction = ? J

Step 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

Step 3: Calculate the heat produced during the neutralisation reaction:

heat produced = total mass × specific heat capacity × change in temperature
q = m_total × C_g × ΔT

m_total = mass_(NaOH) + mass_(H2SO4() = 50.0 + 25.0 = 75.0 g
C_g = 4.18 J°C^-1g^-1
ΔT = T_f - T_i = 26.9 - 18.0 = 8.9°C

q = 75.0 g × 4.18 J g^-1 °C × 8.9 °C = 2790.2 J

Step 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

Step 5: Calculate the heat liberated per mole of water produced, Δ_neutH :

Δ_neutH will be negative because the reaction is exothermic
Δ_neutH = heat liberated per mole of water
   = -1 × q ÷ moles of water
Δ_neutH = -1 × 2790.2 J ÷ 0.050 mol
    = -55803 J mol^-1 (of water produced)
We can divide J by 1000 to convert this enthalpy change to kJ per mole:
Δ_neutH = -55803 J mol^-1 ÷ 1000 J/kJ
    = -55.8 kJ mol^-1 (of water produced)

Now compare the molar enthalpy of neutralisation (molar heat of neutralization) for:

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

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The experiment described above is repeated using 50.0 mL of 1.00 mol L^-1 sodium hydroxide, a strong monobasic base, and 1.00 mol L^-1 aqueous hydrogren cyanide (HCN_(aq), hydrocyanic acid or prussic acid), a weak monoprotic acid (K_a ≈ 6 × 10^-10), instead of 1.00 mol L^-1 hydrochloric acid, a strong monoprotic acid.

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

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 Δ_neutH = -55.2 kJ mol^-1 (per mole of water formed)
So, H⁺_(aq) + OH^- → H₂O_(l)   Δ_neutH = -55.2 kJ mol^-1

For the reaction HCN_(aq) + NaOH_(aq) → NaCN_(aq) + H₂O_(l)
Will Δ_neutH for this reaction be the same (-55.2 kJ mol^-1) ?

We can use the results of the HCN experiment recorded in the graph above to calculate the value for the molar enthalpy of neutralisation (Δ_neutH), and see.

Step 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) = 1.00 mol L^-1
c_(HCN) = concentration of HCN_(aq) = 1.00 mol L^-1
T_i = initial temperature of solutions before additions = 18.0°C
T_f = final temperature of solution at neutralisation = 19.2°C
d = density of solutions = 1 g mL^-1 (assumed)
C_g = specific heat capacity of solutions = 4.18 J°C^-1g^-1 (assumed)
q = heat liberated during neutralisation reaction = ? J

Step 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

Step 3: Calculate the heat produced during the neutralisation reaction:

heat produced = total mass × specific heat capacity × change in temperature
q = m_total × C_g × ΔT

m_total = mass_(NaOH) + mass_(HCN) = 50.0 + 50.0 = 100.0 g
C_g = 4.18 J°C^-1g^-1
ΔT = T_f - T_i = 19.2 - 18.0 = 1.2°C

q = 100.0 g × 4.18 J g^-1 °C^-1 × 1.2 °C = 501.6 J

Step 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 mol L^-1 × 50.0 mL/1000 mL/L = 0.050 mol
moles of water produced = 0.050 mol

Step 5: Calculate the heat liberated per mole of water produced, Δ_neutH :

Δ_neutH will be negative because the reaction is exothermic
Δ_neutH = heat liberated per mole of water
    = -1 × q ÷ moles of water
Δ_neutH = -1 × 501.6 ÷ 0.050
    = -10032 J mol^-1

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

Compare the molar heat of neutralisation for the neuralisation of NaOH_(aq) by both HCl_(aq) and HCN_(aq):

The heat released per mole of water for the HCN_(aq) reaction is much less than the heat released per mole of water for the HCl_(aq) reaction.

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

(i) Strong Acid + Strong Base Reaction:

⚛ Strong base, NaOH, fully dissociates in water. The reacting species is OH^-_(aq)

⚛ Strong acid, HCl, fully dissociates in water. The reacting species is H⁺_(aq)

⚛ The reaction is therefore an Arrhenius neutralisation reaction:

H⁺_(aq) + OH^-_(aq) → H₂O_(l)

⚛ No bonds need to broken in the strong acid or strong base, no energy is lost in breaking bonds.

⚛ Energy is released when the H-O bonds form in the H₂O product.

(ii) Weak Acid + Strong Base

⚛ Strong base, NaOH, fully dissociates in water. The reacting species is OH^-_(aq)

⚛ Weak acid, HCN_(aq), only partially dissociates in water.
    Most of the acid species in solution are undissociated HCN molecules.

⚛ The reaction is therefore a Brønsted-Lowry proton transfer reaction:

HCN_(aq) + OH^-_(aq) → CN^-_(aq) + H₂O_(l)

⚛ In order for this reaction to occur, H-C bonds in HCN molecules must be broken before H₂O molecules can be produced.
    Breaking covalent bonds is an endothermic reaction, it absorbs energy.

⚛ So, even though we might expect the same amount of energy to be produced in both the HCl_(aq) and HCN_(aq) reactions because both reactions are producing the same number of moles of H₂O, we see that energy will be consumed in breaking bonds in HCN so the amount of energy produced overall will be less than that for the HCl + NaOH reaction.

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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Solving the Problem using the StoPGoPS model for problem solving:

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25.00 mL of 0.255 mol L^-1 HCl_(aq) is neutralised by exactly 19.73 mL of NaOH_(aq). If the initial temperature of the reactants was 18.78 °C, calculate the final temperature of the mixture.

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