The standard free energy change for the following reaction is Delta G degree = -

Question

The standard free energy change for the following reaction is Delta G degree = -225.8 kJ at 25 degree C: 2 KClO3 (s) doubleheadarrow 2 KCl(s) + 3 O2(g) (a) Calculate the standard molar entropy S degree for O2 at 25 degree C given the following data - S degree (KClO3) = 143 J.K^-1.mol^-1 S degree (KCl) = 82.6 J.K^-1.mol^-1 Delta H degree f (KClO3) =-397.7 kJ.mol^-1 Delta H degree f (KCI) = -436.7 kJ.mol^-1 b) The Second Law of Thermodynamics states that "the total entropy of a system and its surroundings always increases in a spontaneous process". Show that the reaction above supports the Second Law. c) Calculate the pressure of O2 at equilibrium, giving your answer in the correct units.

Explanation / Answer

a.) Calculate Delta  H0reax using the given data:

Delta  H0reax = Delta  H0f products- Delta  H0f reactants

Delta  H0reax = (2 X (-436.7 kJ/mol)) + (3 X 0 kJ/mol) - (2 X (-397.7 kJ/mol))

Delta  H0reax = -873.4 kJ/mol + 0 - (-795.4 kJ/mol)

Delta  H0reax = -78 kJ/mol

Delta  H0reax = -78 X 103 J/mol

The molar entropy for the reaction is given by:

Delta G0reax = Delta H0reax - T Delta S0reax

-225.8 X 103 J = -78 X 103 J/mol - (298 X Delta S0reax)

-147800 = -298 X Delta S0reax

495.973 J.K-1mol-1 = Delta S0reax

Use the value of Delta S0reax to calculate standard molar entropy of O2.

Delta S0reax = Delta S0products - Delta S0reactants

495.973 J.K-1mol-1 = (2 X 82.6 J.K-1mol-1) + (3 X O2 J.K-1mol-1) - (2 X 143 J.K-1mol-1)

495.973 J.K-1mol-1 = 165.2 J.K-1mol-1 + (3 X O2 J.K-1mol-1) - 286 J.K-1mol-1

495.973 J.K-1mol-1 = -120.8 J.K-1mol-1 + (3 X O2 J.K-1mol-1)

616.773 J.K-1mol-1 = 3 X O2 J.K-1mol-1

205.591 J.K-1mol-1 = O2

Thus, the standard molar entropy for O2 at 250 C is 205.591 J.K-1mol-1.

b.) We know that, Delta Suniverse = Delta Ssystem + Delta Ssurroundings

Using Delta Hsystem and Delta Ssystem values from part a.),

Delta Ssurroundings = - Delta Hsystem / T

Delta Ssurroundings = - 78000 / 298

Delta Ssurroundings = -261.744 J.K-1.mol-1

Delta Suniverse = 495.973 + (-261.744)

Delta Suniverse = 234.228 J.K-1mol-1

Delta Suniverse here is a positive value i.e. > 0. Thus, we have proved the second law that the total entropy (Delta Suniverse) of a spontaneous process is always increasing.

A second and slightly less time consuming way to prove the second law is to look at the Delta G0reax value. It is negative. We know that when Delta G0 < 0, then the process is spontaneous. In this spontaneous process the Delta Ssystem value is however positive i.e. > 0. Thus, in a spontaneous process, the entropy has appeared to increase proving the second law of thermodynamics.

c.) Two substances out of three are solids and hence must not be taken into consideration. So, we only need to worry about O2. We know the formula,

Delta G0reax = -RT lnK

-225.8 X 103 J = - 8.314 J.K-1.mol-1 X 298 K X ln K

-225.8 X 103 J = -24770572 X ln K

ln K = 9.116 X 10-3

K = 1.009

K = p3 (O2). (Taking into consideration the stoichiometric coefficient of O2).

p = 1.003 bar.

Thus, the pressure of O2 at equilibrium is 1.003 bar.

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