Understand enthalpy, entropy, free energy concepts and relationships, and determ
ID: 997791 • Letter: U
Question
Understand enthalpy, entropy, free energy concepts and relationships, and determine and apply thermodynamic quantities to chemical reactions; describe and calculate dependence of chemical equilibria on H, S, and G values.
For the reaction
Ca(OH)2 (s) CaO (s) + H2O (g) the following data given at 25 0C
So J/mol K 63.24 26.8 188.84
Ho f kJ/mol -924.7 - 601.8 - 241.82
a) Calculate standard Gibbs free energy change at 25oC
Is this reaction spontaneous at this temperature? (2 %) ____________Why?_______________________________
If the reaction is non-spontaneous, would a change in temperature make it spontaneous? ______________How?____________________________
b) calculate the equilibrium constant for the reaction above at 25 oC
Explanation / Answer
Given reaction is
Ca(OH)2 (s) -------------> CaO (s) + H2O (g)
So (J/mol K) 63.24 26.8 188.84
Hof (kJ/mol) -924.7 - 601.8 - 241.82
a)
So= So (products) - So(reactants)
= So [CaO(s)]+ So[H2O(g) - So[Ca(OH)2(s)]
= [26.8 + 188.84 ] - [63.24]
= + 152.4 J/ (mol.K)
Horxn = Hof(products) - Hof(reactants)
= Hof [CaO(s)]+ Hof[H2O(g) - Hof[Ca(OH)2(s)]
= [-601.8-241.82 ] - [-924.7]
= + 81.08 kJ/mol
= + 81080 J/mol
We know that
Gorxn = Horxn - T Sorxn
= + 81080 J/mol - ( 298 K x 152.4 J/ (mol.K)
= + 35664 J/mol
Gorxn = + 35664 J/mol
Since Gorxn is positive, the reaction is non-spontaneous.
[ Spontaneous reactions have Gorxn = -ve ]
b)
Gorxn = + 35664 J/mol
Gorxn = -RTInK K = equilibrium constant
K = e^-(Gorxn/RT)
= e^-(35664 J/mol/8.314 J/K/mol x 298 K)
= 5.6 x 10-7
K = 5.6 x 10-7
Therefore,
Equilibrium constant for the given reaction at 25oC = 5.6 x 10-7
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