Heat of Neutralization

  1. Heat of neutralisation is the heat energy evolved when an acid reacts with a base, per mole of the acid or base. All measurements are made under standard state conditions.
  2. An example of the enthalpy change of neutralisation is the heat change obtained in the reaction between sodium hydroxide and hydrochloric acid. The equation for the reaction is

NaOH(aq) + HC1(aq) → NaCl(aq) + H2O(1); ∆H = -57 kJ mol-1

Examples of Neutralization Reaction

Equation
Half Equation
HCl + NaOH → NaCl + H2O H+(aq) + OH-(aq) → H2O(l)
H2SO4 + 2KOH → K2SO4 + 2H2O H+(aq) + OH-(aq) → H2O(l)
CH3COOH + NaOH → CH3COONa +H2O H+(aq) + OH-(aq) → H2O(l)
2HNO3 + Ba(OH)2 → Ba(NO3)2 + 2H2O H+(aq) + OH-(aq) → H2O(l)

  1. The heat of neutralisation of a strong acid with a strong alkali is almost the same for all acids and alkalis. This is because the same reaction always takes places. The reaction is H+(aq) + OH-(aq) → H2O(l)
  2. Heat change of neutralization reaction is affected by 3 factors:
    1. Quantity of acid and alkali
    2. Basicity of the acid and alkali
    3. Strength of acid and alkali


Example:
NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l)
∆H = -57 kJ mol-1

An experiment is carried out by adding 25cm³ of sodium hydroxide 0.5 mol/dm³ into 25 cm³ of dilute nitric acid 0.5 mol/dm³. Calculate the temperature change of the mixture. [Specific heat capacity of the solution = = 4.2 Jg-1°C-1, density of the solution =1 g/cm³ ]

Answer:
Number of mole of NaOH,
n= MV 1000 n= (0.5)(25) 1000 =0.0125mol
Number of mole of HNO3,
n= MV 1000 n= (0.5)(25) 1000 =0.0125mol
Number of mole of water produced = 0.0125mol

Amount of heat released, Q = 0.0125 x 57,000J = 712.5J

Mass of the solution, m = 25 + 25 = 50 cm³
Specific heat capacity of the solution = 4.2 Jg-1°C-1

Q = mcθ
712.5 = 50(4.2)θ
θ = 3.4°


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