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Heat (enthalpy) of Reaction

Key Concepts

  • Enthalpy of Reaction (Heat of Reaction) is the heat liberated or the heat 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 enthalpy of reaction, or heat of reaction, are kJ mol-1 for a specified reactant or product.

  • The enthalpy (heat) of reaction for a neutralisation reaction is known as the enthalpy (heat) of neutralisation.

  • The enthalpy (heat) of reaction for a solute dissolving in a solvent is known as the enthalpy (heat) of solution.

  • The enthalpy (heat) of reaction for a precipitation reaction is known as the enthalpy (heat) of precipitation.

  • Enthalpy (heat) of reaction can be measured experimentally.

Measuring Heat of Reaction Experimentally

  1. A known quantity of reactant is placed in a well insulated vessel (eg, a polystyrene foam cup, that is, a styrofoam™ cup)

  2. The initial temperature of this reactant is recorded, Ti.

  3. A known quantity of the second reactant is added, the vessel is sealed with a lid and the reaction mixture stirred.

  4. The final temperature of the reaction mixture is recorded, Tf.

  5. The the heat released or absorbed (the heat change) q, in joules (J), for the reaction is calculated:
    heat change = mass × specific heat capacity × temperature change

    q = m × cg × ΔT

  6. The enthalpy change, ΔH, in kJ per mole of a given reactant for the reaction is calculated:

    ΔH = heat change/1000 ÷ moles

    ΔH = q/1000 ÷ n

    exothermic reactions: ΔH is negative

    endothermic reactions: ΔH is positive

Common assumptions for reaction mixtures made up of aqueous solutions:

  • density of aqueous solution assumed to be the same as for water, 1 g mL-1 at 25°C
    eg, 100 mL of solution is said to have a mass of 100 g

  • additivity of volumes of reactants in solution is assumed
    eg, 100 mL of "reactant a(aq)" + 200 mL of "reactant b(aq)" = 300 mL of "aqueous solution"

  • 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, q, for the reaction of aqueous solutions is measured in units of joules (J):

q = (mass in grams of "reactant a" + mass in grams of "reactant b") x 4.184 x (Tfinal - Tinitial)

Enthalpy (heat) of Reaction in kJ mol-1 = q/1000 ÷ moles of reactant

For a reaction that liberates heat, an exothermic reaction, ΔH is negative.

For a reaction that absorbs heat, an endothermic reaction, ΔH is positive.

Heat of Solution Example

In an experiment, 1.2 g of sodium hydroxide pellets, NaOH(s), were dissolved in 100 mL of water at 25oC.

The temperature of the water rose to 27.5oC.

Calculate the enthalpy change (heat of solution) for the reaction in kJ mol-1 of solute.

  1. Calculate the heat released, q, in joules (J), by the reaction:

    q = mass(water) x specific heat capacity(water) x change in temperature(solution)
    q = mH2O(l) x cg x (Tf - Ti)
    q = 100 x 4.184 x (27.5 - 25) = 1046 J

  2. Calculate the moles of solute (NaOH(s)):

    moles = mass ÷ molar mass

    moles (NaOH) = 1.2 ÷ (22.99 + 16.00 + 1.008) = 0.030 mol

  3. Calculate the enthalpy change, ΔH, in kJ mol-1 of solute:

    ΔH = -q/1000 ÷ n(solute) = -1046/1000 ÷ 0.030 = -35 kJ mol-1
    ΔH is negative because the reaction is exothermic (energy is released causing the temperature of the solution to increase).

For a more detailed tutorial go to Heat of Solution tutorial

Heat of Neutralisation Example

In an experiment to determine ΔH for the neutralisation reaction:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)

The following results were obtained:

Mass of 100 mL of 0.50 mol L-1 HCl (ma) = 100 g
Mass of 100 mL of 0.50 mol L-1 NaOH (mb) = 100 g
Initial Temperature (Ti) = 20.1oC
Final Temperature (Tf) = 23.4oC
Specific heat capacity of solutions (cg) = 4.184 JK-1g-1

Calculate ΔH in kJ mol-1 of water formed for the reaction.

  1. Calculate the heat released, q, in Joules (J) by the neutralisation reaction:

    q = mass(reaction mixture) x specific heat capacity(water) x change in temperature(solution)
    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

  2. Calculate the moles of reactants:

    moles = molarity x volume

    moles (NaOH) = 0.50 mol L-1 x (100 x 10-3) L = 0.05 mol
    moles (HCl) = 0.50 mol L-1 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.
    0.05 mole of NaOH(aq) reacts with 0.05 mole HCl(aq) to produce 0.05 mole of water.

  3. Calculate the enthalpy (heat) of reaction, ΔH, in kJ mol-1

    Since the reactants are present in a 1:1 stoichiometric ratio, 0.05 mole of NaOH(aq) reacts with 0.05 mole HCl(aq) to produce 0.05 mole of water, moles (n) = 0.05 mol

    ΔH = -q/1000 ÷ n = -2761.44/1000 ÷ 0.05 = -55.2 kJ mol-1
    ΔH is negative because the reaction is exothermic.

For a more detailed tutorial go to Heat of Neutralisation tutorial

Heat of Precipitation Example

50 mL of 0.20 mol L-1 lead(II) nitrate solution, Pb(NO3)2(aq), at 19.6oC was added to 30 mL 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.

Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq)

Calculate the enthalpy change per mole of lead(II) iodide precipitated for the reaction.

  1. Calculate the heat released, q, in Joules (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

  2. Calculate the moles of species specified, n(PbI2(s)):

    From the equation, 1 mole Pb(NO3)2(aq) reacts with excess KI(aq) to produce 1 mol PbI2(s)
    moles Pb(NO3)2(aq) = moles PbI2(s)

    n(Pb(NO3)2(aq)) = n(PbI2(s))

    n(Pb(NO3)2(aq)) = molarity x volume
    n = moles(Pb(NO3)2(aq)) = moles(PbI2(s))= 0.20 x 50 x 10-3 = 0.010 mol

  3. Calculate the enthalpy (heat) of precipitation, ΔH, in kJ mol-1 of PbI2(s):

    ΔH = -q/1000 ÷ n
    ΔH = -0.870/1000 ÷ 0.010 = -87 kJ mol-1
    ΔH is negative because the reaction is exothermic (energy was released causing the temperature to increase).


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