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Introductory Chemistry

Energy Change and Reactions

Measuring Energy Changes

The enthalpy change is the energy that would be exchangeed with the surroundings if the reaction occured in such a way that the temperature of the reaction mixture before and after the reaction were the same. The reaction mixture is referred to as the system.

The enthalpy change of a reaction is the energy exchanged with the surroundings at constant pressure. When a reaction releases energy, the reaction is said to be exothermic and delta H (change in heat) is negative, i.e. −∆H. When a reaction absorbs energy, the reaction is said to be endothermic where delta H is positive, i.e. +∆H.

- The reaction between copper(ii) sulphate and zinc

Cu²⁺ (aq) + Zn (s) → Cu (s) + Zn²⁺ (aq)

• Measure 25.0cm³ of 2.0M copper(ii) sulphate into a well insulated container and measure temperature.
• Add 0.01 mol (an excess) of zinc powder, Zn. (6.5g).
• Stir gently and continuously and note the highest temperature reached.
• Work out the temperature change to the nearest 0.1°C.

- The reaction between citric acid and sodium hydrocarbonate

C₆H₈O₇ (aq) + 3HCO₃^- (s) → C₆H₅O₇^3- (aq) + 3CO₂ (g) + 3H₂O (l)

• Same as the first reaction, using
  • 25cm³ of 1.0M citric acid
  • 0.1 mol (an excess) of sodium hydrocarbonate, NaHCO₃ (8.4g)

The first reaction will be exothermic, the second reaction will be endothermic. The thermometer should range from −10 to 50°C. A well insulated container can be a polystyrene cup with a foil lid.

We can calculate the enthalpy change by using the following relationship:

Enthalpy change (J) = total mass of solution (g) × specific heat capicity (4.18 J) × temperature change (ºC or K)

E = mc∆t

Where the specific heat capacity of water is when you raise the temperature of 1cm³ of water by 1ºC, it will take 4.18 Joules of energy. We take this to be the specific heat capacity in all of these calculations, and we take 1cm³ of solution to equal 1g.

The standard enthalpy change of a reaction, ∆H^Θ, is the enthalpy change of a reaction under standard conditions. Standard conditions (Θ) are:

• Physical state of reagents: Normal state (solid, liquid, or gas) at 298K (25ºC) and 1 atmosphere.
• Solutions: 1.0M
• Experimental conditions: Energy exchange with the surroundings is at 298K and 1 atmosphere pressure.

The standard enthalpy change of formation of a compound, ∆H_f^Θ, is the enthalpy change that takes place when one mole of the compound is formed from its elements under standard conditions. Because this is strictly one mole, when balancing equations, we can use fractions. e.g.

• 2Fe (s) + 1½O₂ (g) → Fe₂O₃ (s)
  • ∆H_f^Θ [Fe₂O₃ (s)] = −824.2 kJmol^-1.
• ½H₂ (g) + ½Cl₂ (g) → HCl (g)
  • ∆H_f^Θ [HCl (g)] = −92.3 kJmol^-1.

Worked Example

Addition of an excess of magnesium to 100cm³ of 2.00 mol dm^-3 CuSO₄(aq) raised the temperature from 20.0ºC to 65.0ºC. Find the enthalpy change for the reaction:

Mg(s) + CuSO₄(aq) → MgSO₄ + Cu(s)

Take the specific heat capacity to be 4.18Jg^-1K^-1.

Find the heat energy change:

• 100cm³ of solution has a mass of 100g.
• Temperature change, ∆t = 65.0 − 20.0 = +45.0ºC
• Heat energy loss from reactants, E = mc∆t = 100 × 4.18 × 45.0 = −18810 J

Work out the number of moles that reacted:

• No. moles = 2.00 × 100 / 1000 = 0.200 mol

Scale up the quantities to those in the equation:

• Mg(s) + CuSO₄(aq) → MgSO₄ + Cu(s)
• 1 mol + 1 mol → 1 mol + 1 mol
• ∴ 0.200 mol (1/5th mol) of CuSO₄ has ∆H of −18810 J
• ∴ 1 mol of CuSO₄ has ∆H of 5 × -18810 = −94050 J
• ∆H = −94.1 kJmol^-1 (3sf)

Hess' Law

The First Law of Thermodynamics states that all energy is conserved. Hess' Law states that the total enthalpy change accompanying a chemical change is independent of the route by which a chemical reaction takes place.

The great value of Hess' Law is that it can be used to calculate enthalpy changes that cannot be determined by experiment. An example of a Hess Cycle is shown in Fig. 1.

• (1) is ∆H_f^Θ[reactants].
• (2) is ∆H^Θ[reaction].
• (3) is ∆H_f^Θ[products].

Some of these values will be given to you in the exam question, and you may be asked to work out the other value using the relationship:

∆H^Θ[reaction] = ∆H_f^Θ[products] − ∆H_f^Θ[reactants]

(2) = (3) − (1)

Worked Example

Sodium hydrocarbonate, NaHCO₃, decomposes on heating. The decomposition products are sodium carbonate, carbon dioxide and water. The equation for the reation is:

2NaHCO₃ (s) → Na₂CO₃ (s) + CO₂ (g) + H₂0 (l)

The standard enthalpy changes of formation at 298 K of the four compounds are listed below:

a) Draw a Hess Cycle to show this data (shown in Fig. 2).

b) Use your completed Hess Cycle to calculate the standard enthalpy change (in kJ) accompanying the thermal decomposition of 2 moles of sodium hydrocarbonate. (Remember to include the appropriate sign in your answer).

Notice that in the Hess cycle, I have multiplied the enthalpy change of the NaHCO₃ by 2. This is because there are two moles of it being formed. We can also use fractions in the elements section of the Hess cycle if needed. Notice also that state symbols must be included.

• b) (1) + (2) = (3)
• −1902 + (2) = −1811 kJmol^-1
• ∆H^Θ(2) = + 91 kJmol^-1