ANSWERS
      

Lesson 1:1

Chemical kinetics studies the rate of progress of a chemical reaction. Thermodynamics studies the reaction only when it has achieved equilibrium.      Back

Lesson 2:1

Reaction: Fe + 2HCl   �   FeCl2 + H2. That is one mole of hydrogen gas is released.
One mole of hydrogen occupies 22.4 dm� at SATP.
So w = - pΔV = 101.3 x 10� Jm‾� x 22.4 x 10‾� m� = - 2.27 kJ.      Back

Lesson 2:2

18g of water is equivalent to one mole. So the heat energy given to the system at constant temperature q = ΔH = 40.6 kJ mol‾�.
ΔU = q - pΔV = q - Δ(nRT) = q - RT
     = 40.6 kJ mol‾� - 8.314 JK‾� mol‾� x 373.15 K = 37.5 kJ mol‾�.      Back

Lesson 2:3

Since there is no interaction between the molecules in an ideal gas there will be no change in the internal energy with the change in volume at constant temperature. So πT = 0. This will be proved mathematically very, very much later.      Back

Lesson 2:4

For a perfect gas πT = 0 and {∂(pV)/∂T}p,m = R.     So Cp,m + Cv,m = R.      Back

Lesson 2:5

49.6 JK‾� mol‾�      Back

Lesson 3:1

Lesson 3:2

ΔfHΘ{propane(g)} = -103.8 x 103 J mol‾�     Back

Lesson 3:3

The standard enthalpy of hydrogenation of the compounds in its natural states at SATP to its products in its most stable state at SATP?      Back

Lesson 3:4

Plot Cp versus temperature, you will get a fairly linear plot for the region 300 - 700 K. Compute the triangular area under plot from 300 - 700 K. This gives the value of   ∫Cp dT.

Compounds n-HexanePropanePropeneNeo-hexane
∫Cp dT / kJ mol‾� 2613.8 10.928
ΔfH�(700 K) / kJ mol‾� −141−90.231.3 −158
Back

Lesson 3:5

Thermochemical process;

[ 2 C(s)→C(g) + 3 H2(g) � 2 H(g) ]   →   [ C-C(g) ]   �   [ C-C(g) + 6 H(g) ]   �   [CH3CH3(g)]

  2 C(s) �2 C(g)  2 x 716.7 kJ
 3 H2(g) �6 H(g)  6 x 218.0 kJ
 C(g) + C(g) �C−C(g)   − 331.0 kJ
 6 H(g) + C−C(g) �CH3CH3(g)   − 6 x 416.0 kJ
SUM: 2 C(s) + 3 H2(g) �CH3CH3(g)   − 85.6 kJ mol‾�

The exact value is ΔfHΘ{ethane (g)} = − 84.5 kJ mol‾�      Back

Lesson 4:1

Plot Cp/Temp versus temperature, you will get a fairly linear plot for the region 300 - 700 K. Compute the triangular area under plot from 300 - 700 K. This gives the value of   ∫(Cp/T) dT.

Compounds n-HexanePropanePropeneNeo-hexane
∫(Cp/T) dT / J mol‾� 178 914
S�(700 K) / J K‾� mol‾� 406278276 373
Back

Lesson 4:2

S(10K) = {S(0) + 1.92} JK‾� mol‾�      Back

Lesson 4:3

  • ΔfG = ΔfH   +   TΔfS. So ΔfS = −80.84 J mol‾�.
  • C + 2 H2 � CH4. ΔfS = S(CH4) − S(C) − 2 x S(H2) = − 80.84 J mol‾�.      Back

    • Lesson 5:1

    dU = TdS   −   pdV ; for expansion work only
    Consider U = f(S,V), we can write; dU = ( ∂U

    ∂S    		 )V dS + ( ∂U

    ∂V    		 )S dV Back

    Lesson 5:2

    ΔG˚ = ΔH˚ − TΔS˚     where
    ΔH˚ = ∑ ΔfH˚(products) − ∑ ΔfH˚(reactants) and ΔS˚ = ∑S˚(products) − ∑ S˚(reactants)
    ΔfH˚ (400) = ΔfH˚ (298) + Cp ∫dT ; assuming that Cp is constant for the short temperature range from 298 K to 400 K. Note ΔfH˚ (400) for Pb = 0; since it is an element.
    S˚ (400) = S˚(298) + Cp ∫dT/T ; assuming that Cp is constant for the short temperature range from 298 K to 400 K

    298 K: ΔG˚ = − 68.26 kJ mol‾� ; K = 9.22 x 10�� and;
    400 K: ΔG˚ = − 68.97 kJ mol‾� ; K = 1.02 x 109

    298 K: ΔG˚ = 68.26 kJ mol‾� ; K = 1.08 x 10‾��. That is most of the compound exist as Pb and CO2. In other words the reaction will not happen.          Back
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