2 heat of precipitation

2 HEAT OF PRICIPITATION

Thermochemistry Heat of precipitation

- precipitate is unsoluble salt - precipitate must be prepared through double bond decomposition or precipitation method Do you still remember what is meant by double bond decomposition? [please refer to salts notes]

General equation double bond decomposition/precipitation;

Ionic equation for precipitation reaction.

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The heat of precipitation is the heat change when one mole of a precipitate is formed from

their ions in aqueous solution

- two aqueous solution of a substances/soluble salt was mix together. - one of the solution contain cation, while another one contain anion for insoluble salt/precipitate that need to be prepared. - both of the aqueous solution exchanging their ions to produce 2 substances, which is unsoluble salt (precipitate) and one soluble salt

MX (aq) + NY (aq) → MY (s) + NX (aq)

M+ (aq) + Y- (aq) → MY (s)


Salt Solubility in waterLi+, Na+, K+, NH4

+ All salt dissolve in waterNitrate, NO3

- All nitrate salt dissolve in water

Chloride

All chloride salt dissolve dissolve in water except; PbCl2 - lead(II) chloride (dissolve in hot water)AgCl - argentums/silver chlorideHgCl - hydroargentum chloride, mercury chloride

Sulphate All sulphate salt dissolve in water except; PbSO4 , BaSO4 , CaSO4

Carbonate All carbonate salt not dissolve in water except; Li2CO3 Na2CO3 , K2CO3 , (NH4)2CO3

Oxide All oxide not dissolve in water except; Na2O , K2O , CaO

HydroxideAll hydroxide not dissolve in water except; NaOH, KOH, Ca(OH)2 , Ba(OH)2

Formula to determine the heat change;

Heat released/absorbed, H = mcӨ [unit = J or kJ]

Symbol Description Unitm mass of solution 1cm3 = 1 gc specific heat capacity of solution 4.2 J g-1 oC-1

Ө temperature change oC

2Method to determine the heat of precipitation


To determine precipitation heat of silver chloride, AgCl

In this experiment you must have the following data;

●

○

Thermometer

Polystyrene cup

Procedure 25 cm3 sodium chloride solution 0.5 mol dm-3 is measured with measuring cylinder 50ml, and poured into polystyrene cup, record the temperature with termometer (0-110)oC. 25 cm3 silver nitrate solution 0.5 mol dm-3 is measured with measuring cylinder 50ml, and poured into another polystyrene cup, record the temperature with termometer (0-110)oC. Sodium chloride solution is added to silver nitrate solution quickly. The reacting mixture is stirred using thermometer.Highest temperature obtained is recorded. Repeat all the step by using different substance.

Precaution steps;Use polystyrene cup. (polystyrene cup is insulator, to avoid loss of heat)Stir the mixture.

25 cm3 sodium chloride solution 0.5 mol dm-3

●

Thermometer

Polystyrene cup

25 cm3 silver nitrate solution 0.5 mol dm-3

○

3


Data tabulationInitial temperature of sodium chloride NaCl /oC

x oC

Initial temperature of silver nitrate, AgNO3 /oC

y oC

Average initial temperature for both solution

Highest temperature for the solution z oC

Temperature change z – (x + y) oC = Ө oC 2

Chemical equation for the reaction;

AgNO3 (aq) + NaCl (aq) → AgCl (s) + NaNO3 (aq)

Ionic equation for the reaction;

Ag+ (aq) + Cl- (aq) AgCl (s)

Calculation of heat of precipitation for AgCl;1. Calculate the number of mole of precipitate formed

No. of mol NaCl = = = 0.0125 mol

No. of mol AgNO3 = = = 0.0125 mol

FBCE;

4

(x + y) oC 2

MV 1000

0.5 X 25 1000

MV 1000

0.5 X 25 1000

Formula of mole


No. of mol AgCl = 0.0125 mol

2. Calculate the heat released/given out [From the experiment]Total volume of the mixture = 25 cm3 AgNO3 + 25 cm3 NaCl

= 50 cm3

Mass of solution = 50 g

Temperature change = Ө oC

Heat given out, H = = 50 × 4.2 × Ө J

= kJ

Therefore, heat given out during the experiment is kJ

3. Calculate the heat of precipitation

0.0125 mol of AgCl produces kJ

Therefore;1 mol of AgCl produces = kJ mol-1

= × kJ mol-1

∆H = kJ mol-1

Thus;The heat of precipitation of silver chloride, AgCl;

5

50 × 4.2 × Ө 1000

1 .0.0125

m = mass of solution ( 1cm3 = 1 g)c = specific heat capacity of solution (4.2 J g-1 oC-1)Ө = temperature change ( oC)

Heat given out No. of mole

50 × 4.2 × Ө 100050 × 4.2 × Ө 12.5

50 × 4.2 × Ө 1000

mcӨ

1st formula

2nd formula

50 × 4.2 × Ө 1000


∆H = – kJ mol-1

Example 1: Precipitation for lead(II) sulphate

Chemical equation;

Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3

∆H = -50 kJmol-1

Ionic equation;

Pb2+ + SO42- → PbSO4 ∆H = -50 kJmol-1

50 kJ heat released when 1 mol of lead(II) ions react with 1 mol of sulphate ions to form 1 mol precipitate of lead(II) sulphate.

The heat of precipitation for PbSO4 = – 50 kJmol-1

Example 2: Precipitation for silver chloride

Chemical equation ;

Energy

Pb2+ + SO42-

PbSO4

∆ H = - 50 kJmol-1

Energy level diagram

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50 × 4.2 × Ө 12.5


AgNO3 + KCl → AgCl + KNO3

∆H = – 65.5 kJmol-1 Ionic equation;

Ag+ + Cl- → AgCl ∆H = – 65.5 kJmol-1

65.5 kJ heat released when 1 mol of silver ions react with 1 mol of chloride ions to form 1 mol precipitate of silver chloride.

The heat of precipitation for AgCl = – 65.5 kJmol-1

Example 3: Precipitation for copper(II) hydroxide

Energy

Ag+ + Cl-

AgCl

∆ H = – 65.5 kJmol-1


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Chemical equation;

CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4

∆H = -60 kJmol-1

Ionic equation;

Cu2+ + 2OH- → Cu(OH)2 ∆H = -60 kJmol-1

60 kJ heat released when 1 mol of copper(II) ion react with 2 mol of hydroxide ion to form 1 mol precipitate of copper(II) hydroxide.

The heat of precipitation for Cu(OH)2 = – 60 kJmol-1

Calculation for heat of precipitate Example 1When 50 cm3 calcium nitrate solution, Ca(NO3)2 2 mol dm-3

Energy

Cu2+ + 2OH-

Cu(OH)2

∆ H = – 60 kJmol-1


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is added to 50 cm3 sodium carbonate solution, Na2CO3 2.0 mol dm-3, precipitate of calcium carbonate, CaCO3 is produce.Temperature of the mixture solution decrease 3.0 oC. Calculate the heat of precipitation of calcium carbonate.

[Specific heat capacity of solution: 4.2 J g-1 oC-1. Density of solution: 1 g cm-3]

Chemical equation;Ca(NO3)2 + Na2CO3 → CaCO3 + 2NaNO3

Ionic equation: Ca2+ + CO3

– → CaCO3

Step 1: Calculate the number of mole of precipitate formed

No. of mol Ca(NO3)2 = = =

No. of mol Na2CO3 = = =

FBCE;1 mol Ca2+ react with 1 mol CO3

2- , produce 1 mol calcium carbonatTherefore;0.1 mol Ca2+ react with 0.1 mol CO3

2- , produce 0.1 mol CaCO3

Thus;No. of mol CaCO3 = 0.1 mol

Step 2 : Calculate the total heat absorb in exp. [From the experiment]Total volume of the mixture = 50 cm3 Ca(NO3)2 + 50 cm3 Na2CO3

9

MV 1000

2 X 50 1000

0.1 mol

MV 1000

2 X 50 1000

0.1 mol


= 100 cm3 Mass of solution = 100 g

Temperature change = 3 oC

Heat absorbs, H = = 100 × 4.2 × 3 J

= 1260 JTherefore, heat given out during the experiment is 1.26 kJ

Step 3 : Calculate the heat of precipitstion 0.1 mol CaCO3 absorb at 1260 J of heat.Therefore; 1 mol CaCO3

absorb heat = J mol-1

= 12600 J mol-1

= 12.6 kJ mol-1

ThusThe heat of precipitation CaCO3, ∆H = + 12.6 kJ mol -1

Step 4 : Draw energy level diagram

Example 2:Thermochemical equation for precipitate of magnesium carbonate, Mg(NO3)2.

Energy

Ca2+ + CO32-

CaCO3

∆ H = +12.6 kJ mol-1

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1 kJ = 1000 J

1260 0.1

sign (+) must write

m = mass of solution ( 1cm3 = 1 g)c = specific heat capacity of solution (4.2 J g-1 oC-1)Ө = temperature change ( oC)

mcӨ

1st formula


Mg(NO3)2 (ak) + Na2CO3 → MgCO3 (p) + 2NaNO3 (ak)

∆H = +25 kJmol-1

Calculate the changes of temperature when 50 cm3 magnesium nitrate solution , Mg(NO3)2 2.0 mol dm-3 is added to 50 cm3 sodium carbonate solution, Na2CO3 2.0 mol dm-3/ Chemical equation has been given the question.

Ionic equation: Mg2+ + CO32- MgCO3

Step 1: Calculate the number of mole of precipitate formed

No. of mol Mg(NO3)2 = = =

No. of mol Na2CO3 = = =

FBCE;1 mol Mg2+ react with 1 mol CO3

2- , to produce 0.1 mol MgCO3

Therefore;0.1 mol Mg2+ react with 0.1 mol CO3

2- , to produce 0.1 mol MgCO3

Thus;No of mole MgCO3 = 0.1 mol

Step 2 : Calculate the total heat absorb in exp.

Heat absorb = mcӨ

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MV 1000

2.0 X 50 1000

0.1 mol

MV 1000

2.0 X 50 1000

0.1 mol

m = solution mass ( 1cm3 = 1 g)c = specific heat capacity (4.2 J g-1 oC-1)Ø = temperature


(endothermic) = (50 + 50) x 4.2 x Ө J = 100 x 4.2 x Ө J = 420 Ө J

[from chemical equation, (∆H = +25 kJ)]The heat of precipitation for magnesium carbonate is +25 kJ,

Therefore;1 mol precipitate of MgCO3 absorbs heat 25 kJ,

Thus;0.1 mol precipitate MgCO3 absorb heat,

= 0.1 x 25 kJ,= 2.5 kJ= 2.5 x 1000 J= 2500 J

Thus;The heat absorbs in experiment is 2500 J

Step 3 : determine the value of Ө, temperature change in exp. In experiment,

Total heat absorb = mcӨ 420 Ø J = 2500 J

Ø =

Ø = 5.95 oC

The reaction is endothermic, temperature decrease 5.95 oCEnergy level diagram of precipitate of MgCO3

12

1 kJ = 1000 J

Change to ‘J’, we want substitute it into formula

2500 J 420 J


3. thermochemical equation for precipitate of copper(II)hydroxide, Cu(OH)2 given below,

Cu2+ (ak) + 2OH- (ak) → Cu(OH)2 (p) ∆H = -60 kJ

How many its volume solution for copper(II)sulphate, CuSO4 1.0 mol dm-3 that need to mixture with 50 cm-3 sodium hydroxide solution, NaOH 2.0 mol dm-3 to increase temperature of the solution mixture to 6.3oC?

Step 1: write the chemical equation for this reactionChemical equation has been state at the question.

Step 2 : calculate no. of mol for the substance

No. of mol CuSO4/Cu2+ = = =

No. of mol NaOH /OH- = = =

Step 3 : ratio for no. of mol

Energy

Mg2+ + CO32-

MgCO3

∆ H = +25 kJ mol-1

13

MV 1000

1.0 X V 1000

0.001 V mol

MV 1000

2.0 X 50 1000

0.1 mol


FBCE;2 mol NaOH /OH- react with 1 mol CuSO4/Cu2+ , to produce1 mol Cu(OH)2, copper(II) hydroxide,

so ;0.1 mol NaOH /OH- react with mol CuSO4/Cu2+, to produce mol Cu(OH)2.

So ;0.1 mol NaOH /OH- react with 0.05 mol CuSO4/Cu2+, to produce 0.05 mol Cu(OH)2.

* carefull to calculate value of ‘V’, we must include the value of temperature changes, Ø (6.3 oC) from the question that give above.

Step 4 : calculate the total heat release/absorb in this exp.Total heat released = mcØ (t/b exothermic) = (50 + V) x 4.2 x 6.3 J

= (50 + V) x 26.46 J = (50 + V) 26.46 J

(V not yet known)

[from chemical equation], (∆H = -60 kJmol-1 )]

Heat of precipitate for copper(II)hydroxide is -60 kJ, (from question)

1 mol precipitate of Cu(OH)2 releaseing heat 60 kJ, So ;0.05 mol precipitate of Cu(OH)2 releasing heat,

= 0.05 x 60 kJ,= 3 kJ

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Ø = 6.3 oC (substitute in equation)

1 kJ = 1000 J

Change to ‘J’, we want substitute it into equation

0.1 2 0.1

2

Don’t calculate the volume of CuSO4 from this value


= 3 x 1000 J= 3000 J

So ;Total heat released by 0.05 mol precipitate of Cu(OH)2 in experiment is 3000 J

Step 5 : determine the value of V, solution volume of CuSO 4, in exp

In experiment,Total heat released = mcØ J

(50 + V) 26.46 J = 3000 J

(50 + V) =

(50 + V) = 113.38 cm3

V = (113.38 – 50) cm3

V = 63.38 cm3

Energy level diagram for precipitate of Cu(OH)2

4. when solution of 500 cm3 M2+ 2.0 mol dm-3 is mixture with 500 cm3 solution of ion Cl- , heat of solution increase to Ø oC.

energy

Cu2+ (ak) + 2OH- (ak)

Cu(OH)2 (p)

∆ H = -60 kJ mol-1

15

3000 J 26.46 J


Calculate the heat of precipitate for the reacted ion M2+ and ion Cl- to precipitate the formula substance MCl2.

Step 1: write the chemical equationIonic equation

M2+ (ak) + 2Cl- (ak) → MCl2 (p)


No. of mol M2+ = = =

No. of mol Cl- = = =

Step 3 : ratio for no. of molDPKYS;1 mol M2+ react with 2 mol Cl- to produce 1 mol precipitate MCl2

so;1 mol M2+ reacted with 2 mol Cl- to produce1 mol precipitate MCl2

Step 4 : calculate the total heat release/absorb in this exp.Total heat released = mcØ (t/b exothermic) = (500 + 500) x 4.2 x Ø J

= 1000 x 4.2 x Ø J = 4.2 x Ø kJ

So ;

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MV 1000

2.0 X 500 1000

1 mol

MV 1000

M X 500 1000

0.5 M mol

1 kJ = 1000 J

Don’t use this no. of mol because the actual no. of mol we don’t know yet


Total heat released by 1 mol precipitate of MCl2 in experiment is 4.2 x Ø kJ,

Step 5 determine the heat of precipitate of MCl2

Because the total heat released by 1 mol precipitate ofMCl2 in this experiment is 2 x Ø kJ ,

So ;Heat of precipitate for MCl2 is -4.2 x Ø kJ mol-1

∆H = -4.2 x Ø kJ mol-1

energy level diagram for heat of precipitate of MCl2

5. In one experiment to determine the heat of precipitate between ion M2+ and ion SO4

2- , found that when 250 cm3 of M2+ 2 mol dm3

solution is mix with 250 cm3 solution of ion SO4

2- , temperature of the solution increase at 30 oC. Calculate heat of precipitation for the reaction between ion M2+ and ion SO4

2-

Energy

M2+ (ak) + Cl- (ak)

MCl2 (p)

∆ H = -4.2 x Ø kJ mol-1

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Step 1: write the chemical equation for this reactionIonic equation

M2+ (ak) + SO42- (ak) → MSO4 (p)


No. of mol M2+ = = =

No. of mol SO42- = = =

Step 3 : ratio for no. of molDPKYS;1 mol of M2+ react with 1 mol SO4

2- to produce1 mol precipitate of MSO4

So ;0.5 mol M2+ react with 0.5 mol SO4

2- to produce 0.5 mol precipitate of MSO4

Step 4 : calculate the total heat release/absorb in this exp.Total heat release = mcØ (t/b exothermic) = (250 + 250) x 4.2 x 30 J

= 63000 J = 63 kJ

So ;Total heat release by 0.5 mol precipitate of MSO4 in experiment is 63 kJ

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MV 1000

2.0 X 250 1000

0.5 mol

MV 1000

M X 250 1000

0.25 M mol

1 kJ = 1000 J

Don’t use this no. of mol because the actual no. of mol we don’t know yet


Step 5 determine the heat of precipitate of MSO4

In experiment;0.5 mol precipitate of MSO4 release heat 63 kJ

So ;1 mol precipitate of MSO4 releasing heat ;

=

= 126 kJ mol-1 So ;Heat of precipitation of MSO4 is -126 kJ mol-1

∆H = -126 kJ mol-1

Energy level diagram for heat of precipitation of MSO4

6. mixture between 75 cm3 hydrochloric acid solution 0.15 mol dm-3

Energy

M2+ (ak) + SO42- (ak)

MSO4 (p)

∆ H = -126 kJ mol-1

19

63 kJ 0.5


and 75 cm3 silver nitrate solution 0.15 mol dm-3 has increase the temperature at 1.9 oC.

i. how much the quantity of heat of energy that release in this experiment?

ii. heat of precipitate for this reaction iii. draw energy level diagram for this reaction.

Step 1: write the chemical equation for this reactionChemical equation;

AgNO3 (ak) + HCl (ak) → AgCl (p) + HNO3 (ak)

Ionic equation;Ag+ (ak) + Cl- (ak) → AgCl (p)


No. of mol HCl = = =

No. of mol AgNO3 = = =

Step 3 : ratio for no. of molDPKYS;1 mol HCl react with 1 mol AgNO3 to produce1 mol precipitate of AgCl

So ;0.01125 mol HCl react with 0.01125 mol AgNO3 to produce 0.01125 mol precipitate of AgCl

Step 4 : calculate the total heat release/absorb in this exp.20

MV 1000

0.15 X 75 1000

0.01125 mol

MV 1000

0.15 X 75 1000

0.01125 mol


Total heat release = mcØ (t/b exothermic) = (75 + 75) x 4.2 x 1.9 J

= 1197 J = 1.197 kJ

So ;Total heat release by 0.01125 mol precipitate of AgCl in experiment is 1.197 kJ

Step 5 determine the heat of precipitate of MSO4

In experiment;0.01125 mol precipitate of releasing heat at 1.197 kJ

So ;1 mol precipitate of AgCl releasing heat;

=

= 106.4 kJ mol-1 So ;Heat of precipitate of AgCl is -106.4 kJ mol-1

∆H = -106.4 kJ mol-1

Energy level diagram for heat of precipitation of AgCl

Kamal Ariffin B Saaim

Energy

Ag+ (ak) + Cl- (ak)

AgCl (p)

∆ H = -106.4 kJ mol-1

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1 kJ = 1000 J

1.197 kJ mol-1 0.01125

2 heat of precipitation

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