The heat absorbed or evolved during thermo chemical reaction is called heat of reaction. Exothermic and endothermic reactions can easily be detected by touching the vessel before and after chemical reaction. The increases in temperature indicate that reaction is exothermic and decrease in temperature indicates that the reaction is endothermic. The accurate values of ΔH can be determined by using calorimeter. The simple type of calorimeter is an insulated container fitted with thermometer and a stirrer. Known amounts of reactants are placed in calorimeter, when reaction proceeds the heat energy evolved or absorbed will either cool or warm the system. ΔH for reaction may be calculated by determining the difference in temperature, mass of reactants and specific heat of reaction mixture.

Heat of Neutralization:

The reaction between an acid and base to form a salt and water is called neutralization reaction. Neutralization reaction is an example of exothermic reaction. The amount of heat released during a neutralization reaction in which 1 mole of water is formed is called as the heat of neutralization or the amount of heat released when 1 mole of hydrogen ions (H+) from an acid reacts with 1 mole of hydroxide ions (OH-) from base to form salt and one mole of water is called heat of neutralization. Procedure:

Take 50 cm3 of molar NaOH solution and note its temperature (t1) and 50 cm3 of molar HCl solution and note its temperature (t2). The two temperatures will be usually same, but they need not. Pour the HCl solution in 250 mls breaker (Calorimeter) and then add quickly NaOH solution the solution being stirred all the time, and note down the highest temperature reached during the reaction. At the end weigh the calorimeter with salt solution. Heat of neutralization is calculated by the following formula.

ΔH = m x S x (t2 - t1) or      ΔH = m x S x Δt

Observation and Calculations

(1)  Mass of calorimeter along with stirrer                 = w1   = 50g

(2)  Mass of calorimeter along with stirrer + salt sol.   = w2   = 150g

(3)   Mass of salt solution (w1 w2)                        = m    = 100g

(4)   Specific heat of salt solution                             = s     = 4.25J/g/oC

(5)  Initial temperature of reactants t1                      = t2    = toC   = 20oC

(6)  Final highest temperature                                 = t3oC = 20oC

(7)   Increase in temperature i.e. t3oC - t                 = Δt    = 6.8oC


     Heat of neutralization is given by Δ H = ms Δ t

ΔH = 100g x 4.25J/1gx1OCx6.8OC = 2856J

The value obtained i.e. 2856J is for 50 mls of solution, it must be multiplied by 20 to give the amount of heat evolved, when 1 mole of NaOH is neutralized by HQ

2856 J x 20 = 57120 J = 57.12 K.J/mol

Result: The heat of neutralization of NaOH by HCl is ΔH = -57.12 KJ/mol.

Note: The heat of neutralization for any strong acid with strong base is approximately same.

For example:

NaOH (aq) + HCl (aq) ------à KOH (aq) + HNO3 (aq)     ΔH: 57.3 KJ/mol. 57.3 KJ/mol.

NaCl (aq) + H2O (1) AH ------à KNO3 (aqJ+H2O (1)      Δ H: 57.3 KJ/mol. 57.3 KJ/mol.

2NaOH (aq) + H2SO4 (aq) ------àNa2SO4 (aq) + 2H2O (1) Δ H = - (2x57.3) KJ/mol.

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