Heat of Reaction

 

Types of heat of reactions
Heat of precipitation
Heat of displacement
Heat of neutralisation
Heat of combustion
 
  • The heat of reaction can be determined through experiments by establishing the temperature change when the reaction occurs.
  • The value of temperature change obtained is used to calculate heat of reaction.
 
General steps to calculate heat of reactions
Step 1 Determine the number of moles of the reactants and products formed, n mole
Step 2 Calculate the heat change in the reaction: \(Q = mc\theta\)
Step 3 Calculate the heat change for 1 mole of reactant or 1 mole of product formed in proportion
Step 4 State the heat of reaction, \(\Delta \text{H}\) with +/- signs and the correct units: \(\Delta \text{H}=+/- \text{kJ mol}^{-1}\)

 

Formula to calculate heat change

\(Q=mc\theta \\ \,\\\text{Where by,} \\ \,\\ \begin{aligned} m &= \text{Mass of reaction mixture}\\ c &= \text{Specific heat capacity}\\ \theta &= \text{Change of temperature}\\ \end{aligned}\)

 
  • The number of moles, n is according to the types of heat of reactions as follows:
    • Mole of precipitate formed for heat of precipitation.
    • Mole of metal displaced for heat of displacement.
    • Mole of water formed for heat of neutralisation.
    • Mole of fuel burnt for heat of combustion.
  • Most chemical reactions carried out to determine the heat of reaction involve aqueous solutions. 
  • Some assumptions are made during calculations:
    • The density of any aqueous solution is the same as the density of water, \(1 \text{ g cm}^{-3}\).
    • The specific heat capacity of any aqueous solution is equal to the specific heat capacity of water, which is \(4.2 \text{ J g}^{ -1 \,\text{ o}} \text{C}^{-1}\).
    • No heat is lost to the surroundings.
    • No heat is absorbed by the apparatus of the experiment.
 

Heat of Precipitation

 
Definition of heat of precipitation
The heat change when 1 mole of precipitate is formed from their ions in an aqueous solution

 

Example

Lead(II) nitrate, \(Pb(NO_3)_2\) and sodium sulphate, \(Na_2SO_4\) of certain volume and concentration, is mixed.

\(10\,000\,J\) of heat is released.

The change in temperature is 7°C.

What is the volume of the lead(II) nitrate, \(Pb(NO_3)_2\) solution?

[Specific heat capacity = \(\text{4.2 J g}^{-1}\,^\circ \text{C}^{-1}\); Density of solution = \(\text{1.0 g cm}^{-3}\)]

Answer: \(\text{340 cm}^3\)

\(\begin{aligned} m&=\dfrac{heat}{c \theta}\\ &=\dfrac{10 000 J}{\text{4.2 J g}^{-1}\,^\circ \text{C}^{-1} \times 7^\circ \text{C}}\\ &=\text{340 g} \end{aligned}\)

 

\(Density = \dfrac{Mass}{Volume}\\ \,\\ \begin{aligned} Volume&=\dfrac{Mass}{Density}\\ &=\dfrac{\text{340 g} }{\text{1.0 g cm}^{-3}}\\ &=\text{340 cm}^3 \end{aligned}\)

 

Heat of Displacement

 
Definition of heat of displacement
The heat change when 1 mole of a metal is displaced from its salt solution by a more electropositive metal
 
  • The more electropositive metal can displace a less electropositive metal from its salt solution.
 
Example

The ionic equation shows the displacement of lead from its salt solution.

\(Al(s)+Pb^{2+}(aq)\rightarrow Al^{3+}(aq)+Pb(s)\\ \Delta H=-150.7\,kJ\,mol^{-1}\)

The reaction displaced 20.7 g of lead solid.

Calculate the amount of heat released from the reaction.

[Relative atomic mass: \(Pb=207\)]

Answer: \(\text{15 070 J}\)

\(\text{Number of mole Pb}\\ =\dfrac{\text{Mass of Pb}}{\text{Molar mass of Pb}}\\ \,\\=\dfrac{20.7\,g}{207\,g\,mol^{-1}}\\ \,\\=0.1\,mol\)

 

\(\begin{aligned} \dfrac{\text{Heat released}}{0.1\,mol}&=\dfrac{150\,700\,J}{1\,mol} \end{aligned}\)

\(\text{Heat released}\\ =\dfrac{150\,700\,J}{1\,mol} \times 0.1\,mol\\ \,\\=\text{15 070 J}\)

 

Heat of Neutralisation

 
Definition of heat of neutralisation
The heat change when 1 mole of water is formed from the reaction between an acid and an alkali
 

Factors affecting heat of neutralisation

 

Strength of acid and base

  • The strength of the acid and base affects the amount of heat released from the reaction.
  • The heat released for the neutralisation of strong acid and strong base is the highest, in comparison to the neutralisation of weak acid and weak base.
  • Neutralisation reactions can occur between acids and alkalis of different strengths as follows:
    • strong acid and strong alkali
    • weak acid and strong alkali
    • strong acid and weak alkali
    • weak acid and weak alkali
 
  • The theoretical value of the heat of neutralisation between a strong acid and a strong alkali is \(-57 \text{ kJ mol}^{−1}\).
 
Example Heat of neutralisation, \(\Delta \text{H}\) (\(\text{ kJ mol}^{−1}\))
Strong acid + Strong alkali → Salt + Water -57
Weak acid + Strong alkali → Salt + Water -55
Strong acid + Weak alkali → Salt + Water -52
Weak acid + Weak alkali → Salt + Water -50
 
  • Note that there is an influence of acid and alkali strengths over the heat of neutralisation.
  • The value of the heat of neutralisation is lower when weak acids or weak alkalis are used.
 
Explanation on the lower value of the heat of neutralisation when weak acids or weak alkalis are used
Weak acids or weak alkalis ionise partially in water and some remain as molecules.
\(\downarrow\)
Some of the heat released during neutralisation is absorbed and used to completely ionise the weak acid or weak alkali in water.
\(\downarrow\)
Therefore, the heat released is lower.
 
  • The heat of neutralisation between weak acids and weak alkalis is the lowest:
    • More heat energy is needed to completely ionise both weak acids and weak alkalis.
    • Therefore, hydrogen ions, \( H^+\) and hydroxide ions, \(OH^−\) produced can completely react to form 1 mole of water.
 

Basicity of acids

  • Complete neutralisation between a strong diprotic acid and a strong alkali produces twice the quantity of heat compared to a strong monoprotic acid.
 
Comparison
Strong monoprotic acid + Strong alkali Strong diprotic acid + Strong alkali
\(HCl + NaOH \rightarrow NaCl + H_2O\\ \Delta H = -57 \text{ kJ mol}^{-1}\) \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\\ \Delta H = -114 \text{ kJ mol}^{-1}\)
 
  • One mole of a strong diprotic acid such as sulphuric acid, \(H_2SO_4\) ionises in water to produce two moles of hydrogen ions, \( H^+\).
  • Two moles of hydrogen ions, \( H^+\) will produce two moles of water, \(H_2O\) when they react with two moles of hydroxide ions, \(OH^−\).
  • 114 kJ heat is released because two moles of water are formed.
  • The heat of neutralisation of sulphuric acid, \(H_2SO_4\) with sodium hydroxide, \(NaOH\) solution is still the same, which is -57 kJ, because the meaning of the heat of neutralisation is the heat released for the formation of one mole of water.
  • Using monoprotic acids such as hydrochloric acid, \(HCl\) or nitric acid, \(HNO_3\) with a strong alkali, such as sodium hydroxide, \(NaOH\) or potassium hydroxide, \(KOH\), will produce one mole of water.
 
Example

Which of the following would have the highest amount of heat released from the reaction?

(A)   \(CH_3COOH(aq) +NaOH(aq)\)
     
(B)   \(HCl (aq)+ KOH(aq)\)
     
(C)   \(HCl (aq)+ NH_4\,^+(aq)+OH\,^-(aq)\)
     
(D)   \(CH_3COOH(aq) +NH_4\,^+(aq)+OH\,^-(aq)\)

Answer: B

Hydrochloric acid, \(HCl\) is a strong acid and potassium hydroxide, \(KOH\) is a strong base.

 

Heat of Combustion

 
Definition of heat of combustion
The heat change when 1 mole of a substance is completely burnt in excess oxygen, \(O_2\)

 

Example

When \(\text{5 g}\) of fuel \(X\) is burnt in excess oxygen, it raised the temperature of \(\text{250 cm}^3\) water by \(\text{40 }^OC\).

Calculate the heat of combustion of the reaction.

[Specific heat capacity of water= \(4.2 \, J \, g^{-1}\,^OC^{-1}\), Density of water= \(1.0 \,g\,cm^{-3}\); Molar mass \(X= \text{20 g mol}^{-1}\)]

Answer: \(\text{168 kJ mol}^{-1}\)

\(\text{Heat change}\\ =mc\theta\\ = 250\,g \times 4.2\,J\,g^{-1}\,^OC^{-1} \times 40^OC\\ =42\,000\,J.\)

 

\(\text{Number of mole X}\\ =\dfrac{\text{Mass}}{\text{Molar mass}}\\ =\dfrac{\text{5 g}}{\text{20 g mol}^{-1}}\\ =\text{ 0.25 mol.}\)

 

\( \text{Heat of combustion}\\=\dfrac{\text{Heat change}}{\text{Number of mole}}\\ =\dfrac{\text{42 000 J}}{\text{0.25 mol}}\\ = \text{168 000 J mol}^{-1}\\ =\text{168 kJ mol}^{-1} \)

 
  • Alcohol molecules consist of carbon atom, C, hydrogen atom, H and oxygen atom, O.
  • Complete combustion of alcohols produces carbon dioxide, \(CO_2\) and water, \(H_2O\).
  • Combustion of alcohol also releases energy, which is an exothermic reaction.
  • As the number of carbon atoms per alcohol molecule increases, the combustion of alcohol produces more carbon dioxide and water molecules.
  • Therefore, more heat is released. 
  • The increment in the heat of combustion between successive members of alcohols is almost the same.
  • This is because each member differs from the following member with one \(CH_2\) group.

Heat of Reaction

 

Types of heat of reactions
Heat of precipitation
Heat of displacement
Heat of neutralisation
Heat of combustion
 
  • The heat of reaction can be determined through experiments by establishing the temperature change when the reaction occurs.
  • The value of temperature change obtained is used to calculate heat of reaction.
 
General steps to calculate heat of reactions
Step 1 Determine the number of moles of the reactants and products formed, n mole
Step 2 Calculate the heat change in the reaction: \(Q = mc\theta\)
Step 3 Calculate the heat change for 1 mole of reactant or 1 mole of product formed in proportion
Step 4 State the heat of reaction, \(\Delta \text{H}\) with +/- signs and the correct units: \(\Delta \text{H}=+/- \text{kJ mol}^{-1}\)

 

Formula to calculate heat change

\(Q=mc\theta \\ \,\\\text{Where by,} \\ \,\\ \begin{aligned} m &= \text{Mass of reaction mixture}\\ c &= \text{Specific heat capacity}\\ \theta &= \text{Change of temperature}\\ \end{aligned}\)

 
  • The number of moles, n is according to the types of heat of reactions as follows:
    • Mole of precipitate formed for heat of precipitation.
    • Mole of metal displaced for heat of displacement.
    • Mole of water formed for heat of neutralisation.
    • Mole of fuel burnt for heat of combustion.
  • Most chemical reactions carried out to determine the heat of reaction involve aqueous solutions. 
  • Some assumptions are made during calculations:
    • The density of any aqueous solution is the same as the density of water, \(1 \text{ g cm}^{-3}\).
    • The specific heat capacity of any aqueous solution is equal to the specific heat capacity of water, which is \(4.2 \text{ J g}^{ -1 \,\text{ o}} \text{C}^{-1}\).
    • No heat is lost to the surroundings.
    • No heat is absorbed by the apparatus of the experiment.
 

Heat of Precipitation

 
Definition of heat of precipitation
The heat change when 1 mole of precipitate is formed from their ions in an aqueous solution

 

Example

Lead(II) nitrate, \(Pb(NO_3)_2\) and sodium sulphate, \(Na_2SO_4\) of certain volume and concentration, is mixed.

\(10\,000\,J\) of heat is released.

The change in temperature is 7°C.

What is the volume of the lead(II) nitrate, \(Pb(NO_3)_2\) solution?

[Specific heat capacity = \(\text{4.2 J g}^{-1}\,^\circ \text{C}^{-1}\); Density of solution = \(\text{1.0 g cm}^{-3}\)]

Answer: \(\text{340 cm}^3\)

\(\begin{aligned} m&=\dfrac{heat}{c \theta}\\ &=\dfrac{10 000 J}{\text{4.2 J g}^{-1}\,^\circ \text{C}^{-1} \times 7^\circ \text{C}}\\ &=\text{340 g} \end{aligned}\)

 

\(Density = \dfrac{Mass}{Volume}\\ \,\\ \begin{aligned} Volume&=\dfrac{Mass}{Density}\\ &=\dfrac{\text{340 g} }{\text{1.0 g cm}^{-3}}\\ &=\text{340 cm}^3 \end{aligned}\)

 

Heat of Displacement

 
Definition of heat of displacement
The heat change when 1 mole of a metal is displaced from its salt solution by a more electropositive metal
 
  • The more electropositive metal can displace a less electropositive metal from its salt solution.
 
Example

The ionic equation shows the displacement of lead from its salt solution.

\(Al(s)+Pb^{2+}(aq)\rightarrow Al^{3+}(aq)+Pb(s)\\ \Delta H=-150.7\,kJ\,mol^{-1}\)

The reaction displaced 20.7 g of lead solid.

Calculate the amount of heat released from the reaction.

[Relative atomic mass: \(Pb=207\)]

Answer: \(\text{15 070 J}\)

\(\text{Number of mole Pb}\\ =\dfrac{\text{Mass of Pb}}{\text{Molar mass of Pb}}\\ \,\\=\dfrac{20.7\,g}{207\,g\,mol^{-1}}\\ \,\\=0.1\,mol\)

 

\(\begin{aligned} \dfrac{\text{Heat released}}{0.1\,mol}&=\dfrac{150\,700\,J}{1\,mol} \end{aligned}\)

\(\text{Heat released}\\ =\dfrac{150\,700\,J}{1\,mol} \times 0.1\,mol\\ \,\\=\text{15 070 J}\)

 

Heat of Neutralisation

 
Definition of heat of neutralisation
The heat change when 1 mole of water is formed from the reaction between an acid and an alkali
 

Factors affecting heat of neutralisation

 

Strength of acid and base

  • The strength of the acid and base affects the amount of heat released from the reaction.
  • The heat released for the neutralisation of strong acid and strong base is the highest, in comparison to the neutralisation of weak acid and weak base.
  • Neutralisation reactions can occur between acids and alkalis of different strengths as follows:
    • strong acid and strong alkali
    • weak acid and strong alkali
    • strong acid and weak alkali
    • weak acid and weak alkali
 
  • The theoretical value of the heat of neutralisation between a strong acid and a strong alkali is \(-57 \text{ kJ mol}^{−1}\).
 
Example Heat of neutralisation, \(\Delta \text{H}\) (\(\text{ kJ mol}^{−1}\))
Strong acid + Strong alkali → Salt + Water -57
Weak acid + Strong alkali → Salt + Water -55
Strong acid + Weak alkali → Salt + Water -52
Weak acid + Weak alkali → Salt + Water -50
 
  • Note that there is an influence of acid and alkali strengths over the heat of neutralisation.
  • The value of the heat of neutralisation is lower when weak acids or weak alkalis are used.
 
Explanation on the lower value of the heat of neutralisation when weak acids or weak alkalis are used
Weak acids or weak alkalis ionise partially in water and some remain as molecules.
\(\downarrow\)
Some of the heat released during neutralisation is absorbed and used to completely ionise the weak acid or weak alkali in water.
\(\downarrow\)
Therefore, the heat released is lower.
 
  • The heat of neutralisation between weak acids and weak alkalis is the lowest:
    • More heat energy is needed to completely ionise both weak acids and weak alkalis.
    • Therefore, hydrogen ions, \( H^+\) and hydroxide ions, \(OH^−\) produced can completely react to form 1 mole of water.
 

Basicity of acids

  • Complete neutralisation between a strong diprotic acid and a strong alkali produces twice the quantity of heat compared to a strong monoprotic acid.
 
Comparison
Strong monoprotic acid + Strong alkali Strong diprotic acid + Strong alkali
\(HCl + NaOH \rightarrow NaCl + H_2O\\ \Delta H = -57 \text{ kJ mol}^{-1}\) \(H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O\\ \Delta H = -114 \text{ kJ mol}^{-1}\)
 
  • One mole of a strong diprotic acid such as sulphuric acid, \(H_2SO_4\) ionises in water to produce two moles of hydrogen ions, \( H^+\).
  • Two moles of hydrogen ions, \( H^+\) will produce two moles of water, \(H_2O\) when they react with two moles of hydroxide ions, \(OH^−\).
  • 114 kJ heat is released because two moles of water are formed.
  • The heat of neutralisation of sulphuric acid, \(H_2SO_4\) with sodium hydroxide, \(NaOH\) solution is still the same, which is -57 kJ, because the meaning of the heat of neutralisation is the heat released for the formation of one mole of water.
  • Using monoprotic acids such as hydrochloric acid, \(HCl\) or nitric acid, \(HNO_3\) with a strong alkali, such as sodium hydroxide, \(NaOH\) or potassium hydroxide, \(KOH\), will produce one mole of water.
 
Example

Which of the following would have the highest amount of heat released from the reaction?

(A)   \(CH_3COOH(aq) +NaOH(aq)\)
     
(B)   \(HCl (aq)+ KOH(aq)\)
     
(C)   \(HCl (aq)+ NH_4\,^+(aq)+OH\,^-(aq)\)
     
(D)   \(CH_3COOH(aq) +NH_4\,^+(aq)+OH\,^-(aq)\)

Answer: B

Hydrochloric acid, \(HCl\) is a strong acid and potassium hydroxide, \(KOH\) is a strong base.

 

Heat of Combustion

 
Definition of heat of combustion
The heat change when 1 mole of a substance is completely burnt in excess oxygen, \(O_2\)

 

Example

When \(\text{5 g}\) of fuel \(X\) is burnt in excess oxygen, it raised the temperature of \(\text{250 cm}^3\) water by \(\text{40 }^OC\).

Calculate the heat of combustion of the reaction.

[Specific heat capacity of water= \(4.2 \, J \, g^{-1}\,^OC^{-1}\), Density of water= \(1.0 \,g\,cm^{-3}\); Molar mass \(X= \text{20 g mol}^{-1}\)]

Answer: \(\text{168 kJ mol}^{-1}\)

\(\text{Heat change}\\ =mc\theta\\ = 250\,g \times 4.2\,J\,g^{-1}\,^OC^{-1} \times 40^OC\\ =42\,000\,J.\)

 

\(\text{Number of mole X}\\ =\dfrac{\text{Mass}}{\text{Molar mass}}\\ =\dfrac{\text{5 g}}{\text{20 g mol}^{-1}}\\ =\text{ 0.25 mol.}\)

 

\( \text{Heat of combustion}\\=\dfrac{\text{Heat change}}{\text{Number of mole}}\\ =\dfrac{\text{42 000 J}}{\text{0.25 mol}}\\ = \text{168 000 J mol}^{-1}\\ =\text{168 kJ mol}^{-1} \)

 
  • Alcohol molecules consist of carbon atom, C, hydrogen atom, H and oxygen atom, O.
  • Complete combustion of alcohols produces carbon dioxide, \(CO_2\) and water, \(H_2O\).
  • Combustion of alcohol also releases energy, which is an exothermic reaction.
  • As the number of carbon atoms per alcohol molecule increases, the combustion of alcohol produces more carbon dioxide and water molecules.
  • Therefore, more heat is released. 
  • The increment in the heat of combustion between successive members of alcohols is almost the same.
  • This is because each member differs from the following member with one \(CH_2\) group.