4.2 Heat of Precipitation

Heat of reaction
■ Heat of reaction

	►	The change in heat content (either heat released or absorbed) when one mole of reactants react completely or when one mole of the reaction products formed.
	►	The symbol for heat of reaction is ∆H. ○ ∆H = Hproducts – Hreactants ◉ ∆H is negative (–) for exothermic reaction. ◉ ∆H is positive (+) for endothermic reaction.
	►	The unit for heat of reaction is joule (J).

■ Thermochemical equation

	►	Chemical equation together with the heat of reaction.
	►	Example: Pb(NO3)2(aq) + K2SO4 → PbSO4(s) + 2KNO3(aq), ∆H = -50kJ mol-1 ○ The thermochemical equation above shows that: ◉ 50kJ of heat energy is released when 1 mol of lead(II) sulphate, PbSO4 is formed. ◉ The heat of precipitation of PbSO4, ∆H is -50kJ mol-1.

■ Calculation of heat change

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Heat change in a chemical reaction can be determined by using the following formula. ○ Heat change = solution mass X specific heat capacity X temperature change = mcθ ◉ Solution mass (ignore the mass of chemical in solid form) ◉ Specific heat capacity : the quantity of heat required to raise the temperature of 1g of any substance by 1°C. For example, the specific heat capacity of water is C = 4.2Jg-1 °C-1 or 4200Jg-1 °C-1 or 4.2kJ kg-1 °C-1. ◉ Temperature change, θ = final temperature(T2) – initial temperature(T1)

■ Types of heat of reaction

	►	Heat of precipitation: Pb(NO3)2(aq) + K2SO4 → PbSO4(s) + 2KNO3(aq), ∆H = -50kJ mol-1
	►	Heat of displacement: Mg(s) + FeSO4(aq) → MgSO4(aq) + Fe(s), ∆H = -80kJ mol-1
	►	Heat of neutralisation: HNO3(aq) + KOH(aq) → HNO3(aq) + H2O(l), ∆H = -57.3kJ mol-1
	►	Heat of combustion: ${\text{CH}}_3\text{OH}(\text{l})+\frac{3}{2}{\text{O}}_2\to {\text{CO}}_2(\text{g})+2{\text{H}}_2\text{O}(\text{l})$ , ∆H = -592.3kJ mol-1

■ This video contains information about heat of reaction.

Calculation involving heat of reaction
■ Heat of reaction calculation guidelines

	►	Heat change in a chemical reaction ○ Volume of solution and temperature change are given. ◉ Heat change = solution mass X specific heat capacity X temperature change = mcθ ○ Thermochemical equation (∆H) is given. ◉ Heat change = Number of moles X ∆H ◉ Number of moles can be calculated from the following equations: If the mass of substances is given, number of moles = $\frac{\text{mass}}{\text{molar mass}}$ If the volume and concentration of the solution is given, number of moles = $\frac{\text{MV}}{1000}$
	►	Temperature change in a chemical reaction ○ Thermochemical equation (∆H) is given. θ = $\frac{\text{heat change}}{\text{mc}}$
	►	Heat of reaction ○ Step 1: Write chemical equation ○ Step 2: Calculate heat change ○ Step 3: Calculate the number of moles ○ Step 4: Calculate the heat of reaction by linking the number moles of reactants with the heat change which occurs

■ This video contains guidelines on the calculation of heat of reaction.

✍ Worked-example 4.2(a) Calculate the heat change that occurs when 50cm3 of dilute sulphuric acid, H2SO4, solution is mixed with 40cm3 potassium hydroxide solution, KOH. In this reaction, the temperature of the mixture increases from 29°C to 42°C. Solution: The mass of the solution, m = 50 + 40 = 90g Temperature change, θ = 42°C – 29°C = 13°C Heat change = mc θ = 90 X 4.2 X 13 = 4914J

✍ Worked-example 4.2(b) When 2g of magnesium powder is added to 50cm3 of ferum(II) chloride solution,the temperature increases from 28°C to 40°C. What is the heat change of the reaction? Solution: The mass of the solution, m = 50g (ignored the mass of chemical in solid form) Temperature change, θ = 40°C – 28°C = 12°C Heat change = mc θ = 50 X 4.2 X 12 = 2520J

✍ Worked-example 4.2(c) Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), ∆H = - 210kJ mol-1 What is the mass of zinc, that is needed to release 21 kJ of heat for the reaction? [Relative atomic mass, Zn, 65] Solution: Number of moles = $\frac{\text{heat change}}{\text{∆H}}$ = $\frac{21}{210}$ = 0.1mol Thus, the mass of zinc = 0.1 X 65 = 6.5g

✍ Worked-example 4.2(d) An experiment is carried out by mixing excess zinc powder into 50cm3 of 1mol dm-3 copper(II) sulphate solution. Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), ∆H = - 210kJ mol-1 What is the expected temperature change in this experiment? [Specific heat capacity = 4.2Jg-1 °C-1, density of a solution of 1g cm-3] Solution: Number of moles = $\frac{\text{MV}}{1000}$ = $\frac{1\mathrm{\times }50}{1000}$ = 0.05mol Heat change = number of moles X ∆H = 0.05 X 210kJ = 10.5kJ = 10500J θ = $\frac{\text{heat change}}{\text{mc}}$ = $\frac{10500}{50\mathrm{\times }4.2}$ = 50°C

Heat of precipitation
■ Heat of precipitation

	►	The heat change that occurs when 1 mole of precipitate is formed from the reaction solution at standard conditions.
	►	After a reaction, the solid that appears in the solution is called a precipitate.
	►	Example of a precipitation reaction ○ Pb(NO3)2(aq) + K2SO4(aq) → PbSO4(s) + 2KNO3(aq) ○ 2AgNO3(aq) + MgCl2(aq) → 2AgCl(s) + Mg(NO3)(aq)

■ This video contains information regarding heat of precipitation

Laboratory Activity 4.2.1 : Heat of Precipitation of Silver Chloride

Calculation involving the heat of precipitation

✍ Worked-example 4.2(e) An experiment is carried out to determine the heat of precipitation of barium sulphate. In this reaction, 25cm3 of 1.0mol dm-3 barium chloride, is poured into a polystyrene cup and the initial temperature of solution is recorded. 25cm3 of 1.0mol dm-3 of sodium sulphate solution is poured into the same polystyrene cup. The resulting solution mixture is stirred and the highest temperature is recorded. The recorded temperatures are shown below. Initial temperature = 29°C Highest temperature reached by the solution = 34°C Calculate the heat of precipitation of barium sulfate and draw an energy level diagram for the reaction in this experiment. Solution: Step 1 ○ BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq) ○ Ba2+(aq) + SO42-(aq) → BaSO4(s) Step 2 Calculate heat change = mc θ = (25 + 25) X 4.2 X (34 – 29) = 1050J Step 3 Calculate the number of moles barium chloride $\frac{\text{MV}}{1000}$ $\frac{25\mathrm{\times }1}{1000}$ = 0.025mol Calculate the number of moles sodium sulphate $\frac{\text{MV}}{1000}$ $\frac{25\mathrm{\times }1}{1000}$ = 0.025mol Step 4 Calculate the heat of reaction by linking the number moles of reactants with the heat change which occurs: 0.025mol of barium chloride react with 0.025mol of sodium sulphate to produce 1050J Thus, 1 mol of barium chloride react with 1 mol of sodium sulphate to produce = $\frac{1050}{0.025}$ = 42000J = 42kJ Heat of precipitation of barium sulphate, ∆H = -42kJ mol-1

✍ Worked-example 4.2(f) Solid silver bromide is formed when silver ions combine with bromide ions, as shown by the following thermochemical equation: Ag+(aq) + Br−(aq) → AgBr(s), ∆H = -50kJ mol-1 In an experiment, 50cm3 of 0.5mol dm-3 silver nitrate solution is added to 100cm3 of 2.0mol dm-3 sodium bromide. (a) Write a balanced chemical equation for the reaction of silver nitrate with sodium bromide. (b) Calculate the number of moles of silver bromide which precipitated (c) Calculate the change in temperature that occurs in this reaction. (d) Sketch the energy level diagram for the precipitation of silver bromide. [Specific heat capacity = 4.2Jg-1 °C-1, density of a solution of 1g cm-3] Solution: (a) AgNO3(aq) + NaBr(aq) → AgBr(s) + NaNO3(aq) (b) Number of moles of silver nitrate $\frac{\text{MV}}{1000}$ $\frac{50\mathrm{\times }0.5}{1000}$ = 0.025mol Number of moles of sodium bromide $\frac{\text{MV}}{1000}$ $\frac{100\mathrm{\times }2}{1000}$ = 0.2mol 0.025 mol of silver nitrate react with excess of sodium bromide to produce 0.025 mol of argentum bromide. (c) Heat change = number of moles X ∆H = 0.025 X 50kJ = 1.25kJ θ = $\frac{\text{heat change}}{\mathrm{mc}}$ = $\frac{1250}{\mathrm{(50\; +\; 100)}\mathrm{\times }4.2}$ = 2°C (d) Energy level diagram for the precipitation of silver bromide

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