måndag 31 oktober 2011

Thermodynamics - calculations of ∆H°, ∆S° and K

Howdy folks!

Time for some thermodynamics. I'll go through strategies of solving common questions.

  1. How does one calculate H° for the following reaction :

CO2 (g, 30 mbar) + H2O H+ (aq, pH =8) + HCO-3(aq, c=010 mM)


First you have to find values corresponding to each of the species in the reaction. I don't know where you find yours, but I use SI Chemical data, table 5. Units: kJ/mol


Reactants:

CO2 (g, 30 mbar) = -394

H2O = -286

Products :

H+ (aq, pH =8) = 0

HCO-3(aq, c=010 mM) = -690


Once you've listed these values, use the formula : (product+product) – (reactant - reactant)

So, in this case : (0 +(-690 )) - (-394) - (-286) → -10 kJ/mol

H° = -10 kJ/mol


In order to find S°, use a similar strategy. First; find the S°f for the different spieces. Units : J/K/mol

Reactants:

CO2 (g, 30 mbar) = 214

H2O = 70

Products :

H+ (aq, pH =8) = 0

HCO-3(aq, c=010 mM) = 98

Again, use the formula : (product + product) – (reactant- reactant) = 0 + 98 – 214 – 70 = -186 J/K/mol.

S° = -186 J/K/mol.


Ok, moving on to how you can calculate K for such a reaction.

The equation is : G° = H° - T

We've just calculated the values corresponding to H° and S°, -10 kJ/mol & -186 J/K/mol, respectively.

T has to be tranformed into Kelvins. 25°C corresponds to 298°K . Furthermore, we have to transform -10 kJ/mol into J/mol; -10103 J/mol. Now, just insert your values into the equation

G° =- -10103 J/mol – 298 ∙ (-186 J/K/mol) = 45428 J/mol OR 45,4 kJ/mol.


K is related to ∆G° in this way : K = e(G°/RT) R denoting the gas constant; 8,314J/K/mol or 8314 kJ/K/mol. So, in this case : K = e- (45,4/8314∙298) gives 1,08-08


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Another one then...

Calculate ∆H° for the following reaction :

2 Fe2O3 (s) + 3 C (s) 4 Fe(s) + 3 CO2 (g)

2 Fe2O3 (s) = -824

Fe = 0

3 CO2 = 214

3 C = 0


(Product + product) – (reactant-reactant) = 0 + 3(-394) - 2(-824 – 0) = 466 kJ/mol


So H° = 466 kJ/mol


Calculate S° for the reaction 2 Fe2O3 (s) + 3 C (s) 4 Fe(s) + 3 CO2 (g)

2 Fe2O3 (s) = 87

Fe = 27

3 CO2 = 214

3 C = 6


Ja osv...


  1. 2011-02-24

Calculate H° and S° at 25°C for the following reaction :

C(s) + 2H2 (g)→ CH4

CH4 = -74

C = 0

H2 = 0


(product + product) – (reactants – reactants) = -74 – 0 - 2∙ 0 = -74 kJ/mol

Enthalpy : -74 kJ/mol


S° is calculated through S°f -values.

CH4 = 186

C = 6

H2 = 131


186 – 6- (2131) = -82 J/mol/K

Entrophy : -82 J/mol/K


Calculate K for the reaction!

G° = H° - T-74∙103 J/mol- (298)∙(-82 J)) = enligt mig : 74000- 298 x -82 = - 4,9862 ∙104

  • 4,9862 ∙104 /8,314∙298 = -20 → e20 = 4,8∙108

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