• 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.

Calculate the heat change in the reaction: 𝑄=𝑚𝑐𝜃 \(Q = mc\theta\).

• 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.

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

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

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

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

• 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.

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

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

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

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

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

• 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.
  • Wtrong 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 \(-57 \text{ kJ mol}^{−1}\) seperti di bawah:

• 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.

• 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.

• Peneutralan lengkap antara asid diprotik kuat dan alkali kuat menghasilkan dua kali ganda kuantiti haba berbanding asid monoprotik kuat.

• 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.

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

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.

• 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.

Calculate the heat change in the reaction: 𝑄=𝑚𝑐𝜃 \(Q = mc\theta\).

• 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.

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

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

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

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

• 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.

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

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

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

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

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

• 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.
  • Wtrong 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 \(-57 \text{ kJ mol}^{−1}\) seperti di bawah:

• 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.

• 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.

• Peneutralan lengkap antara asid diprotik kuat dan alkali kuat menghasilkan dua kali ganda kuantiti haba berbanding asid monoprotik kuat.

• 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.

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

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.