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 :

CO_{2 (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:

CO_{2 (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:

CO_{2 (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∆S°

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; -10∙10³ J/mol. Now, just insert your values into the equation

∆G° =- -10∙10³ 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 Fe₂O_{3 (s)} + 3 C _(s) → 4 Fe_(s) + 3 CO_{2 (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 Fe₂O_{3 (s)} + 3 C _(s) → 4 Fe_(s) + 3 CO_{2 (g)}

2 Fe2O3 (s) = 87

Fe = 27

3 CO₂ = 214

3 C = 6

Ja osv...

6. 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- (2∙131) = -82 J/mol/K

Entrophy : -82 J/mol/K

Calculate K for the reaction!

∆G° = ∆H° - T∆S° -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