4.4 Heat of Neutralisation

Heat of neutralisation
■ Neutralisation reaction

	►	The reaction between acid and alkali to produce salt and water.
	►	The hydrogen ion, H+ from the acid will combine with the hydroxide ion, OH− from the alkali to form H2O.
	►	Examples of neutralisation reaction: ○ HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ○ 2HNO3(aq) + Mg(OH)2 → Mg(NO3)2(aq) + 2H2O(l) ○ CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)

■ Heat of neutralisation

►

The heat that is released when 1 mol of hydrogen ion, H+ neutralises 1 mol of hydroxide ion, OH−, to form 1 mol of water.

■ The amount of heat released during neutralization reaction depends on:

	►	Concentration of the acid or alkali
	►	Quantity of the acid and alkali
	►	Basicity of the acid

■ Concentration of the acid or alkali

►

The heat released in the reaction of a strong acid with a strong alkali is higher than the heat released in the reaction of a weak acid with a strong alkali. ○ Strong acid + Strong alkali 1 mole of HCl will react with 1 mole of NaOH to produce 57.0kJ of heat. ○ Weak acid + Strong alkali 1 mole of CH3COOH will react with 1 mole of NaOH to produce 55.0kJ of heat.

■ Quantity of the acid and alkali

►

Quantity of heat released in the neutralization reaction is directly proportional to the quantity of acid and alkali used. ○ 1 mole of HCl + 1 mole of NaOH Heat released = 57.0kJ ○ 2 mole of HCl + 2 mole of NaOH Heat released = 2 x 57.0kJ = 114kJ ○ 3 mole of HCl + 3 mole of NaOH Heat released = 3 x 57.0kJ = 171kJ

■ Basicity of the acid

►

Neutralisation of a diprotic acid will produce more heat energy than a monoprotic acid. ○ Monoprotic acid + alkali 1 mole of HCl will react with 1 mole of NaOH to produce 57.0kJ of heat. ○ Diprotic acid + alkali1 mole of H2SO4 will react with 2 mole of NaOH to produce 114 kJ of heat.

■ This video contains information regarding heat of neutralisation

■ This video contains calculation guidelines on heat of neutralisation.

Laboratory Activity 4.4.1 : Heat of Neutralisation between a Strong Acid and a Strong Alkali

✍ Worked-example 4.4(a) When 50cm3 of 0.5mol dm-3 NaOH is added to 50cm3 of 0.5mol dm-3 HCl acid in a plastic bottle, the temperature rise by 3.27°C. Find the heat of reaction for the reaction in kJ mol-1. [Specific heat capacity = 4.2Jg-1 °C-1, density of a solution of 1g cm-3] Solution: Step 1 Write chemical equation H+(aq) + OH−(aq) → H2O(aq) Step 2 Calculate heat change = mc θ = (50 + 50) X 4.2 X 3.27 = 1373.4J Step 3 Calculate the number of moles Number of moles of H+ = $\frac{\text{MV}}{1000}$ = $\frac{50\mathrm{\times }0.5}{1000}$ = 0.025mol Number of moles of OH− = $\frac{\text{MV}}{1000}$ = $\frac{50\mathrm{\times }0.5}{1000}$ = 0.025mol Step 4 Calculate the heat of reaction by linking the number moles of reactants with the heat change which occurs 0.025 mol of ion H+ react with 0.025 mol of ion OH− to produce 1373.4 J Thus, 1 mol of ion H+ react with 1 mol of ion OH− to produce = $\frac{1373.4}{0.025}$ = 54936J = 54.963kJ Heat of neutralisation of sodium hydroxide and hydrochloric acid, ∆H = -54.963kJ mol-1

✍ Worked-example 4.4(b) Conducts the following experiment to determine the heat of neutralization. Solution: (a) Step 1 Write chemical equation H2SO4(aq) + 2NaOH(s) → Na2SO4(aq) + 2H2O(l) Step 2 Calculate heat change = mc θ = (100 + 50) X 4.2 X (34 – 30) = 2520J Step 3 Calculate the number of moles Number of moles of H2SO4 = $\frac{\text{MV}}{1000}$ = $\frac{100\mathrm{\times }2}{1000}$ = 0.2mol Number of moles of NaOH = $\frac{\text{MV}}{1000}$ = $\frac{50\mathrm{\times }1}{1000}$ = 0.05mol Step 4 Calculate the heat of reaction by linking the number moles of reactants with the heat change which occurs 0.05 mol of NaOH react with 0.025 mol of H2SO4 to produce 2520 J Thus, 1 mol of NaOH react with 1 mol of H2SO4 to produce = $\frac{2520}{0.05}$ = 50400J = 50.4kJ Heat of neutralisation of sodium hydroxide and sulphuric acid, ∆H = -50.4kJ mol-1 (b) HNO3(aq) + NaOH(s) → NaNO3(aq) + H2O(l) Number of moles of HNO3 = $\frac{\text{MV}}{1000}$ = $\frac{100\mathrm{\times }2}{1000}$ = 0.2mol 0.05 mol of NaOH react with 0.05 mol of HNO3 to produce 2520 J. There is an excess of HNO3 that is not involved in the reaction. The volume used for HNO3 is same with the volume used for H2SO4, thus the temperature change for both experiments are same, that is 4°C.

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