Key Concepts
- Heat of Reaction is the heat liberated or absorbed when a chemical reaction takes place.
An exothermic reaction liberates heat, temperature of the reaction mixture increases.
An endothermic reaction absorbs heat, temperature of the reaction mixture decreases.
- The units of heat of reaction are kJ mol-1 for a specified reactant or product.
- The heat of reaction for a neutralisation reaction is known as the heat of neutralisation.
- The heat of reaction for a solute dissolving in a solvent is known as the heat of solution.
- The heat of reaction for a precipitation reaction is known as the heat of precipitation.
- Heat of reaction can be measured experimentally.
Measuring Heat of Reaction Experimentally
- A known quantity of reactant is placed in a well insulated vessel (eg, a styrofoam cup)
- The initial temperature of this reactant is recorded, Ti.
- A known quantity of the second reactant is added, the vessel is sealed with a lid and the reaction mixture stirred.
- The final temperature of the reaction mixture is recorded, Tf.
- The the heat released or absorbed, in joules, for the reaction is calculated:
q = mass x specific heat capacity x change in temperature
- The enthalpy change in kJ per mole of a given reactant for the reaction is calculated:
 H = q/1000 ÷ moles
exothermic reactions: H is negative
endothermic reactions: H is positive
Common assumptions for reaction mixtures made up of aqueous solutions:
- density of aqueous solution assumed to be the same as for water, 1g/mL
eg, 100mL of solution is said to have a mass of 100g
- additivity of volumes of reactants is assumed
eg, 100mL of reactant a + 200mL of reactant b = 300mL of reaction mixture
- specific heat capacity of the reaction mixture assumed to be the same as water,
ie, specific heat capacity = 4.184 JK-1g-1
- Heat is not lost to, or absorbed by, the surroundings.
Typically, the calculation for heat released or absorbed for the reaction of aqueous solutions is:
q = (mass in grams reactant a + mass in grams reactant b) x 4.184 x (Tf - Ti)
Heat of Reaction in kJ/mol = q/1000 ÷ moles of reactant
For a reaction that liberates heat, an exothermic reaction, H is negative.
For a reaction that absorbs heat, an endothermic reaction, H is positive.
Heat of Solution Example
In an experiment, 1.2g of sodium hydroxide pellets, NaOH(s), were dissolved in 100mL of water at 25oC.
The temperature of the water rose to 27.5oC.
Calculate the enthalpy change (heat of solution) for the reaction.
- Calculate the heat released, in J, by the reaction:
q = mass(water) x specific heat capacity x change in temperature
q = mH2O(l) x cg x (Tf - Ti)
q = 100 x 4.184 x (27.5 - 25) = 1046 J
- Calculate the moles of solute (NaOH(s)):
n = mass ÷ MM
n (NaOH) = 1.2 ÷ (22.99 + 16.00 + 1.008) = 0.030 mol
- Calculate the enthalpy change, H, in kJ/mol of solute:
H = -q/1000 ÷ n(solute) = -1046/1000 ÷ 0.030 = -35 kJ/mol
H is negative because the reaction is exothermic (energy is released).
Heat of Neutralisation Example
In an experiment to determine H for the neutralisation reaction:
NaOH(aq) + HCl(aq) -----> NaCl(aq) + H2O(l)
The following results were obtained:
| Mass of 100mL of 0.50M HCl (ma) |
= 100g |
| Mass of 100mL of 0.50M NaOH (mb) |
= 100g |
| Initial Temperature (Ti) |
= 20.1oC |
| Final Temperature (Tf) |
= 23.4oC |
| Specific heat capacity of solutions (cg) |
= 4.184 JK-1g-1 |
Calculate H in kJ/mol for the reaction.
- Calculate the heat released, in J, by the neutralisation reaction:
q = mass x specific heat capacity x change in temperature
q = (ma + mb) x cg x (Tf - Ti)
q = (100 + 100) x 4.184 x (23.4 - 20.1) = 200 x 4.184 x 3.3 = 2761.44 J
- Calculate the moles of reactants:
n = M x V
n (NaOH) = 0.50M x (100 x 10-3)L = 0.05 mol
n (HCl) = 0.50M x (100 x 10-3)L = 0.05 mol
NaOH(aq) and HCl(aq) are in 1:1 mole ratio which is an exact stoichiometric ratio as shown by the neutralisation equation.
- Calculate the heat of reaction, H, in kJ/mol
Since the reactants are present in a 1:1 stoichiometric ratio, H will be calculated as H per mole of NaOH or H per mole of HCl
H = -q/1000 ÷ n = -2761.44/1000 ÷ 0.05 = -55.2 kJ/mol
H is negative because the reaction is exothermic.
Heat of Precipitation Example
50.mL of 0.20M lead (II) nitrate solution, Pb(NO3)2(aq), at 19.6oC was added to 30mL of a solution containing excess potassium iodide, KI(aq) also at 19.6oC.
The solutions reacted to form a yellow lead (II) iodide precipitate, PbI2(s), and the temperature of the reaction mixture increased to 22.2oC.
Calculate the enthalpy change per mole of lead (II) nitrate for the reaction.
- Calculate the heat released, in J, by the precipitation reaction:
q = mass x specific heat capacity x change in temperature
q = [mPb(NO3)2(aq) + mKI(aq)] x cg x (Tf - Ti)
q = [50 + 30] x 4.184 x (22.2 - 19.6) = 870.27 J
- Calculate the moles of reactant specified, n[Pb(NO3)2(aq)]:
n = M x V
n[Pb(NO3)2(aq)] = 0.20 x 50 x 10-3 = 0.010 mol
- Calculate the heat of precipitation, H, in kJ/mol of Pb(NO3)2(aq):
H = -q/1000 ÷ n
H = -0.870/1000 ÷ 0.010 = -87 kJ mol-1
H is negative because the reaction is exothermic.
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