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52.

Substance

Enthalpy of Combustion

 /\Ho kJ/mol

Absolute Entropy

So J/mol K

C(s)

- 393.5

5.74

H2(g)

- 285.8

130.6

C2H5OH(l)

- 1366.7

160.7

H2O(l)

----------

69.91


a.
   
Write a separate balanced equation for the combustion of each of the following: C(s), H2(g), and C2H5OH(l). Consider the only products to be CO2(g) and/or H2O(l).
   
C(s)  +  O2(g)  -->  CO2(g)                                              Combustion Reactions will have O2 as 1

2H2(g)  +  O2(g)  -->  2H2O(l)                                         of the reactants.

C2H5OH(l)  +  3O2(g)  -->  2CO2(g)  +  3H2O(l)

   
    
b. In principle, ethanol can be prepared by the following reaction.
2C(s) + 2H2(g) + H2O(l) ---> C2H5OH(l)
Calculate the standard enthalpy change, /\Ho, for the preparation of ethanol.
   
2C(s) + 2H2(g) + H2O(l) ---> C2H5OH(l)

1.  Steps must add up to the total reaction.
2.  Multiply
/\Ho by same value you multiply the coefficients of the step.
3.  Multiply by a negative if the reactants and products are reversed.

    

2C(s)  +  2O2(g)  -->  2CO2(g)                                              /\Ho = 2 mol x ( - 393.5 kJ/mol)

2H2(g)  +  O2(g)  -->  2H2O(l)                                              /\Ho = 1 mol x ( - 285.8 kJ/mol)

2CO2(g)  +  3H2O(l) -->  C2H5OH(l)  +  3O2(g)                 /\Ho = - 1 mol x ( - 1366.7 kJ/mol)
---------------------------------------------------------------               -------------------------------------------------
2C(s) + 2H2(g) + H2O(l) ---> C2H5OH(l)                           /\Ho = 293.9 kJ

    
    
c. Calculate the standard entropy change, /\So, for the reaction in part (b).
    
2C(s) + 2H2(g) + H2O(l) ---> C2H5OH(l)

/\Sorxn  = n Soproducts  + n Soreactants  
    

/\Sorxn  = 1 mol SoC2H5OH  -  2 mol SoC  - 2 mol SoH2 - 1 mol SoH2O  

/\Sorxn  = 1mol x 160.7J/molK - 2mol x 5.74J/molK - 2mol x 130.6J/molK - 1mol x 69.91J/mol K

/\Sorxn  = - 181.9 J/K

   
    
d. Calculate the value of the equilibrium constant at 25oC for the reaction represented by the equation in part (b).
   
/\Gorxn  =  /\Horxn  -  T /\Sorxn

/\Gorxn  =  293.9 kJ - 298 K x (- 0.1819 kJ/K) = 348.1 kJ

/\Gorxn  = - 2.303 RT log Keq
    

                         - /\Gorxn                                       - 348.1 kJ
Keq = antilog(--------------) = antilog( -------------------------------------------------------)
                        2.303 RT                     2.303 x 0.008314 kJ mol -1 K -1 x 298 K

    
Keq = 9.83 x 10 -62