3 heat of diplacement

3 HEAT OF DIPLACEMENT

Thermochemistry

Heat of Displacement

What is meant by ‘displacement’ reaction?

Example 1;

Zn + Cu2+ Zn2+ + Cu ∆H= –210 kJmol-1

Energy

Zn + Cu2+

Zn2+ + Cu

When 1 mole Cu is displaced by Zn, 210 kJ heat energy is released/given out.

∆H= –210 kJmol-1

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The Heat of Diplacement is the heat change when one mole of metal is diplaced from its solution

by a more electropositive metal.

Metal that more electropositive will displacemetal that is less electropositive from its salt solution.


Example 2;

Zn + Pb2+ Zn2+ + Pb ∆H= –112 kJmol-1

Example 3;

Mg + Fe2+ Mg2+ + Fe ∆H= –80 kJmol-1

Energy

Zn + Pb2+

Zn2+ + Pb

When 1 mole Pb is displaced by Zn, 112 kJ heat energy is released/given out.

∆H= –112 kJmol-1

Energy

Mg + Fe2+

Mg2+ + Fe

When 1 mole Fe is displaced by Mg, 80 kJ heat energy is released/given out.

∆H= –80 kJmol-1

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To determine the heat of diplacement of copper by zinc

Data tabulation

Procedure; 50 cm3 copper(II) sulphate solution of 0.1 mol dm-3 is measured with measuring cylinder 50ml and poured into polystyrene cup, record the temperature with

termometer (0-110)oC. 1.0 g metal powder is weighed by using electronic balance and quickly added into the polystyrene cup that contain copper(II) sulphate solution. The mixture is stirred using the thermometer. The highest/maximum temperature of heat is recorded. Repeat the step by using different substance. [if necessary]

●

○

Thermometer

Polystyrene cup

50.0 cm3 copper(II) sulphate solution 0.1 mol dm-3

Beaker that contain1 g zinc powder(excess)

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Method to determine heat of displacement


Initial temperature of copper(II) sulphate , CuSO4 /oC

y

Highest/maximum temperature for the solution /oC

z

Temperature change /oC (z – y) = Ө[Note : mass of zinc is used in excess]

Chemical equation for the reactionZn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)

Ionic EquationZn + Cu2+ → Cu + Zn2+

Calculation the heat of displacement of copper by zinc;1. Calculate the number of mole of Cu formed

No of moles of Zn = . mass . = 1.0 = 0.015 mol molar mass 65

No. of moles CuSO4 = = = 0.005 mol

(No. of moles of CuSO4) 0.005 < 0.015 (No of moles of Zinc)[Important notes: calculation MUST based on CuSO4 solution because quantity of zinc used in excess]

FBCE;1 mole of CuSO4 produces 1 mole of CuTherefore;0.005 mole of CuSO4 produces 0.005 mole of CuThus;No. of mole of Cu formed = 0.005 mol

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M V 1000

0.1 X 50 1000


2. Calculate the heat released/given out

Heat release/given out = mcӨ(in exp.)

= 50 x 4.2 x Ө J

= kJ

3. Calculate the heat of diplacement for 1 mole copper by zinc

0.005 mole of copper diplaced produced kJ

Therefore; When 1 mole of copper is diplaced by zinc, the heat released is

= kJ mol-1

= X kJ mol-1

= kJ mol-1

Thus; The heat of diplacement of copper by zinc

∆H = – kJ mol-1

Calculation heat of displacement5

50 x 4.2 x Ө 1000

50 x 4.2 x Ө 1000

1 .0.005

50 x 4.2 x Ө 5

50 x 4.2 x Ө 1000 . 0.005

50 x 4.2 x Ө 1000

50 x 4.2 x Ө 5

Volume of solution


Question 1Excess iron powder is added into 50 cm3 copper(II) chloride solution, CuCl2 1.0 mol dm-3, brown solid is formed and blue solution change to green. Iron has displace copper from its salt solution. The following data is get from above experiment.

Initial temperature for copper(II)chloride solution = 28.0 oCHighest temperature for mixture solution = 57.0 oC

Calculate heat changes when 1 mol of copper is displace by iron.

Solution

chemical equation;Fe (s) + CuCl2 (aq) → FeCl2 (aq) + Cu (s)

Ionic equation; Fe (s) + Cu2+ (aq) → Cu (s) + Fe+2 (aq)

Step 1 : Calculate the number of mole of Cu formed

No. of mol CuCl2 = = =

No. of mol Fe = (no need to calculate because is in excess)

FBCE;No. of mole of Cu formed = 0.05 mol

Step 2 : Calculate the heat released/given outHeat released/given out = mcӨ (exothermic) = 50 x 4.2 x Ө J

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MV 1000

1 X 50 1000

0.05 mol


Temperature change, Ө = (highest temperature– initial temperature) = (57.0 – 28.0) oC

= 29.0 oC

Total heat release = mcӨ = 6090 J

= 6.09 kJ

Step 3 : determine the heat of displacement of Cu by Fe Displacement of 0.05 mol Cu releasing 6.09 kJ of heat.

Therefore;When 1 mole of copper is diplaced by zinc, the heat released is

= kJ mol-1

= 121.8 kJ mol-1 Thus;The heat of displacement of Cu by iron;

∆H = –121.8 kJmol-1

Draw energy level diagram

Question2Study the following equation; Fe (s) + CuSO4 (aq) → Cu (s) + FeSO4 (aq)

Energy

Fe + Cu2+

Fe2+ + Cu

∆ H = -121.8 kJ mol-1

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(determine the changes of temperature)

1 kJ = 1000 J

6.09 0.05


∆H = -250 kJ mol-1

If excessive iron powder is add into 100 cm3 copper(II) sulphate solution 0.25 mol dm-3, calculate

i. Heat releasedii. Temperature riseiii. Mass of copper that is displaceiv. Mass of salt that formed if it crystalizev. Draw energy level diagram

If magnesium powder is use to replace iron powder, is it the energy that release is more higher, same or lower. [Ar = Cu, 64; Fe, 56; S, 32; O, 16]

Solution

Chemical equation;Fe (p) + CuSO4 (ak) → FeSO4 (ak) + Cu (p)

Ionic equation; Fe (p) + Cu2+ (ak) → Fe+2 (ak) + Cu (p)

i : calculate no. of mole of Cu formed

No. of mol CuSO4 = = =

No. of mol Fe = (no need to calculate because excessive)

FBCE;1 mol CuSO4 produced 1 mol Cu

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MV 1000

0.25 X 100 1000

0.025 mol


Therefore;0.025 mol CuSO4 produce 0.025 mol CuThus;No. of mole of Cu formed = 0.025 mol

calculate the heat released by 0.025 mol of Cu in the exp.[from question: ∆H = -250 kJ mol-1] That’s mean;When 1 mole Cu is displaced by Fe, 250 kJ heat energy is released/given out.

Therefore;0.025 mol Cu is diplaced by Fe, heat released is;

= 0.025 × 250 kJ= 6.25 kJ = 6250 J

Thus;Heat released/given out = 6.25 kJ or 6250 J

ii: determine the temperature changes during the reaction[Heat released/given out = 6250 J]

Heat released/given out = mcӨ 6250 = 100 x 4.2 x Ө

Ө = 14.9 oCTherefore;The increases in temperature = 14.9 oC

iii: determine the mass of copper that is diplace[from the previous information above]No. of mole of Cu formed = 0.025 mol

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Change to unit of kJ, because we want to find Ө


No. of mol Cu = mass of Cu Ar Cu Mass of Cu = 0.025 × 64 = 1.6 g

iv: determine the mass of salt formedFBCE;1 mol CuSO4 produced 1 mol FeSO4

Therefore;0.025 mol CuSO4 produce 0.025 mol FeSO4

Thus;No. of mole of FeSO4 formed = 0.025 mol

No. of mol FeSO4 = mass of FeSO4

Mr FeSO4

Mass of FeSO4 = 0.025 × [56 + 32 + 4(16)] = 3.8 g

v : draw energy level diagram

Question 3100 cm3 copper(II) nitrate solution 0.2 mol dm-3 is poured into plastic container. Temperature is recorded. Then excess magnesium powder

Energy

Fe + Cu2+

Fe2+ + Cu

∆ H = -250 kJ mol-1

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is added to the solution. The mixture is stirred and the temperature rises is recorded. The temperature shows an increases of 5 oC.

(a) What is the colour of the solution in the plastic container;i. before magnesium powder is place?ii. after magnesium powder is place?

(b) Write the total ionic equation of the reaction.(c) How many mole of copper(II) nitrate reacts?(d) Calculate the heat releases in this experiment.(e) Calculate the heat energy release when one mol of copper is formed.(f) What is the heat of displacement of copper? (g) Draw energy level diagram for this experiment.(h) Why magnesium used is in form of fine powder not granulated?(i) i. if potassium hydroxide solution is mix with the solution in plastic beaker in the end of the experiment, what can you observe? ii. write the chemical equation for the reaction in (i)(i). iii. write the ionic equation for the reaction in (i)(i).

SOLUTION (a) i. blue ii. colourless

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(b) Mg (aq) + Cu2+ (aq) → Mg2+ (aq) + Cu (s)

(c) no. of mol Cu(NO3)2 = =

=

(d) Heat release = mcӨ (during reaction) = 100 x 4.2 x 5.0 J

= 2100 J= 2.1 kJ

(e) FBCE; 1 mol CuSO4 produced 1 mol Cu Therefore; 0.02 mol CuSO4 produce 0.02 mol Cu Thus; No. of mole of Cu formed = 0.02 mol When 0.02 mol of Cu diplaced by Mg, 2.1 kJ of heat released. Therefore; 1 mol of Cu diplaced by Mg will releases heat;

=

= 105 kJ mol-1

(f) The heat of displacement of copper = -105 kJ mol-1 Heat change, ∆H = -105 kJ mol-1

(g) Energy level diagram

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MV 1000

0.2 x 100 1000

0.02 mol

2.1 kJ mol-1

0.02


(h) To increase the total surface area per volume for magnesium, thus it will increase the rate of reaction.

(i) i. white precipitate formed ii. Mg(NO3)2 (aq) + 2KOH (aq) → Mg(OH)2 (s) + 2KNO3 (aq)

iii. Mg2+ (aq) + 2OH- (aq) → Mg(OH)2 (s)

Learning task: pg 158 no. 1 & 2Effective Practise: pg. 158 no. 3

Kamal Ariffin B SaaimSMKDBLhttp://kemhawk.webs.com/

Energy Mg (s) + Cu2+ (aq)

Mg2+ (aq) + Cu (s)

∆ H = - 105 kJ mol-1

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3 heat of diplacement

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