Treatment of results
Calculate the # of moles of:
i) calcium carbonate used
Molecular mass of CaC03 ii) calcium Oxide used
Mass=1g RMM of CaO=40+16 gmol-1
Mole=Mass/RMM
1g/56gmol-1
Moles of CaO= 0.01786mol
1*CA +1*c + 3*0
=1*40+1*12+3*16
=40+12+48
=100gmol-1
No of moles=1g/100=
Mole =given/RMM =0.01M
Calculate the enthalpy change of the reaction involving
i) calcium oxide
Enthalpy change of CaO=
Mass=50 g(since 1g=1cm3)),c=4.18Jg-1°C-1(s.h.c. Of water), ∆T
∆T=30.0-25.0 =5.0°c
Enthalpy Change=mc∆T =50*4.18*5.0= -1,045J
Calcium Carbonate mass= 50 g(since 1g=1cm3),c=4.18Jg-1°C(s.h.c),
∆T=26.5-25.0=1.5°c
Enthalpy change= mc∆T
=50*4.18*1.5= -313.5J
Given that the heat capacity of the solution is 4.18 J g-1 °C-1 and that 1 cm3 of aqueous solution has a mass of 1 g
Using your answers to questions 1 and 2, determine the enthalpy change of reaction for
i) one mole of calcium oxide
Enthalpy change of CaO=-1,045J,Moles of CaO=0.01786mol
Enthalpy change per mole=Enthalpy change/mole
=- 1.045J/0.01786mol
=-58,510.6Jmol. -59KJmol
Calcium carbonate of CaCO= -313.5J Moles of CaCO= 0.01 mol
Enthalpy change per mole =Enthalpy change/mole= -313.5/0.01 mol= -31,350 Jmol-1 /-31kJmol-1
Draw an energy cycle for the enthalpy change of the reaction CaO(s) + CO2(g) → CaCO3(s)
using the TWO BALANCED chemical equations for the reactions you have conducted.
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