Wednesday, May 18, 2011

Form 5 Chapter 4 Thermochemistry

Describe an experiment to determine the heat of precipitation of silver chloride


Diagram:






Apparatus:
Measuring cylinders, thermometer and plastic cup

Materials:
0.5 mol dm-3 silver nitrate solution, 0.5 mol dm-3 sodium chloride solution


Procedure:
1.      25cm3 of 0.5 mol dm-3 sodium chloride solution is measured and poured into a clean and dry plastic cup using a measuring cylinder.
2.      The initial temperature of sodium chloride solution is measured and then recorded.
3.      Using another measuring cylinder, 25 cm3 of 0.5 mol dm-3 silver nitrate solution is measured. The initial temperature of the silver nitrate solution is measured after 5 minutes and recorded.
4.      The silver nitrate solution is poured quickly and carefully into the sodium chloride solution.
5.      The mixture is stirred with a thermometer throughout the experiment and the highest temperature obtained is recorded.
Result:


Mass, m
c
q
= (25 + 25) g = 50 g
4.2
Initial temperature of sodium chloride solution     = T1oC
Initial temperature of the silver nitrate            = T2 oC
Highest temperature reached                                            = T4 oC
=> Average initial temperature = (T1+T2)/2                       = T3 oC
è Increase in temperature,  q = (T4 - T3) oC         



Calculation:


No of moles of sodium chloride used

No of moles of silver nitrate used
= MV/1000 = 0.0125 mole

= MV/1000= 0.0125 mole

Chemical equation: AgNO3 (aq) + NaCl (aq) à AgCl (s) + NaNO3 (aq)
Ionic equation:              Ag + (aq)                              +              Cl- (aq) à           AgCl (s)
                                                             1 mole                                  1 mole                  1 mole


Heat released =              mc
q
                                             =              (25 + 25 ) x 4.2 x (T4 - T3)
                                             =              x J
0.0125 mole of AgCl precipitated releases x J of heat energy.
Therefore, 1 mole of AgCl precipitated will released
Heat of precipitation, DH = x/0.0125 J = y J

Thermochemical equation:
AgNO3 (aq) + NaCl (aq) à AgCl (s) + NaNO3 (aq)                             DH = - y/1000 kJmol-1

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