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