C2H5OH(aq)+O2(aq) = CH3COOH(aq)+H2O(l) C2H5OH(aq) H = -288.3 KJ/mol. S= 148.5 J/

Question

C2H5OH(aq)+O2(aq) = CH3COOH(aq)+H2O(l)

C2H5OH(aq) H = -288.3 KJ/mol. S= 148.5 J/mol K

CH3COOH H = -485.8 KJ/mol. S= 178.7 J/mol K

1. Calculate H and S for this process.

2. Calculate G for this reaction at 35C. Is this rxn spontaneous at this temperature?

3. Calculate temperature at which this reaction will reach equilibrium. Will this reaction be spontaneous above or below the equilibrium temperature?

4. Calculate G at 25C for the formation of acetic acid from ethanol (see rxn above) when [CH3COOH] = .200M, (P)O2 = 1.13 atm, and [C2H5OH] = .125 atm.

Thanks for your help!

Explanation / Answer

Reaction: C2H5OH (aq) + O2 (g) = CH3COOH (aq) + H2O (l)

1) Hrxn = (n.H)products – (n.H)reactants = [(1 mole)*H (CH3COOH) + (1 mole)*H (H2O)] – [(1 mole)*H (C2H5OH) + (1 mole)*H (O2)]

= [(1 mole)*(-485.8 kJ/mol) + (1 mole)*(-285.82 kJ/mol)] – [(1mole)*(-288.3 kJ/mol) + (1 mole)*(0)] = [(-485.8) + (-285.82)] – [(-288.3)] kJ = -483.32 kJ (the enthalpy of formation of water is required and the value is taken from internet sources and the enthalpy of formation of a pure compound like O2 gas is zero). (ans)

Srxn = (n.S)products – (n.S)reactants = [(1 mole).S(CH3COOH) + (1 mole).S(H2O)] – [(1 mole).S(C2H5OH) + (1 mole).S(O2)] = [(1 mole)*(178.7 J/mol.K) + (1 mole)*(69.9 J/mol.K)] – [(1 mole)*(148.5 J/mol.K) + (1 mole)*(205.0 J/mol.K) = [(178.7 J/K) + (69.9 J/K)] – [(148.5 J/K) + (205.0 J/K)] = (248.6 J/K) – (353.5 J/K) = -104.9 J/K

(the values for entropy of formation of water and O2 are obtained from the internet) (ans)

Note that the values obtained from the internet are standard values at 25°C and we assume the enthalpy change and entropy change to remain constant.

2) Temperature = 35°C = (35 + 273)K = 308 K

Grxn = Hrxn – T.Srxn = (-483.32 kJ) – (308 K)*(-104.9 J/K) = (-483.32 kJ) – (-32.3092 kJ) = -451.0108 kJ -451.01 kJ (ans)

Since Grxn is negative, the reaction is spontaneous at 35°C.

3) At equilibrium Grxn is zero. Since Hrxn and Srxn are constants, we must have,

0 = (-485.32 kJ) – T*.(-104.9 J/K) where T* is the equilibrium temperature in K.

Therefore,

0 = (-485.32 kJ) + 104.9T* J/K

===> 485.32 kJ = 104.9T* J/K

===> T* = (485.32*1000 J)/(104.9 J/K) = 4626.50 K

The equilibrium temperature is 4626.50 K = (4626.50 – 273)°C = 4353.50°C (ans)

Since both Hrxn and Srxn are negative, Grxn will be negative and hence spontaneous at lower temperatures. As temperature is increased beyond equilibrium temperature, Grxn becomes positive and the reaction becomes non-spontaneous. The reaction will be spontaneous below the equilibrium temperature (ans).

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