Question

The U.S. requires automobile fuels to contain a renewable component. The fermentation of glucose from corn produces ethanol, which is added to gasoline to fulfill this requirement: C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g) Calculate ΔH o , ΔS o and ΔG o for the reaction at 25°C. Is the spontaneity of this reaction dependent on temperature? ΔH o rxn = kJ ΔS o rxn = J/K ΔG o rxn = kJ


The reaction is:


spontaneous at all temperatures

nonspontaneous at all temperatures

spontaneous only at high temperatures

spontaneous only at low temperatures

spontaneous only in the reverse direction

Answer 1

Answer:

ΔH° = -67.9 kJ

ΔS° = 536.7 J/K = 0.5367 kJ/K

ΔG° = -227.8 kJ

the reaction is spontaneous at all temperatures

Explanation:

Step 1: Data given

Temperature = 25.0 °C

ΔH°f(glucose) = - 1274.5 kJ/ mol

ΔH°f(C2H5OH) = -277.7 kJ/mol

ΔH°f(CO2) = -393.5 kJ/mol

S°(glucose) = 212.1 J/ K)

S°(C2H5OH) = 160.7 J/K

S°(CO2) = 213.7 J/K

Step 2: The balanced equation

C6H12O6(s) → 2 C2H5OH(l) + 2 CO2(g)

Step 3: Calculate ΔH°

ΔH° = 2*ΔH°f(C2H5OH) + 2ΔH°f(CO2) - ΔH°f(glucose)

ΔH° = 2*(-277.7 kJ) + 2*(-393.5 kJ) - (-1274.5 kJ)

ΔH° = -555.4 kJ + (-787 kJ) +1274.5 kJ)

ΔH° = -67.9 kJ

Step 4: Calculate ΔS°

ΔS° = 2*S°(C2H5OH)  + 2*S°(CO2) - S°(glucose)

ΔS° = 2*(160.7 J/K) + 2(213.7 J/K) - 212.1 J/K

ΔS° = 321.4 + 427.4 J/K - 212.1 J/K

ΔS° = 536.7 J/K = 0.5367 kJ/K

Step 5: Calculate ΔG°

ΔG° =ΔH° - T*ΔS°

ΔG° = -67.9 kJ - 298K * 0.5367 kJ/K

ΔG° = -227.8 kJ

Since ΔS° is positive and  ΔH° is negative, ΔG° will be negative

This means ΔG° is negative at all temperature.

A negative ΔG° means the reaction is spontaneous at all temperatures

Answer 2

a) ΔH° = -67.9 kJ

b) ΔS° = 536.7 J/K = 0.5367 kJ/K

c) ΔG° = -227.8 kJ

The reaction is spontaneous at all temperatures

What information do we have?

Temperature = 25.0 °C

ΔH°f(glucose) = - 1274.5 kJ/ mol

ΔH°f(C₂H₅OH) = -277.7 kJ/mol

ΔH°f(CO₂) = -393.5 kJ/mol

S°(glucose) = 212.1 J/ K

S°(C₂H₅OH) = 160.7 J/K

S°(CO₂) = 213.7 J/K

Balanced chemical equation:

$C_6H_{12}O_6(s) --- > 2 C_2H_5OH(l) + 2 CO_2(g)$

Calculation of ΔH°:

ΔH° = 2*ΔH°f(C₂H₅OH) + 2ΔH°f(CO₂) - ΔH°f(glucose)

ΔH° = 2*(-277.7 kJ) + 2*(-393.5 kJ) - (-1274.5 kJ)

ΔH° = -555.4 kJ + (-787 kJ) +1274.5 kJ)

ΔH° = -67.9 kJ

Calculation of ΔS°:

ΔS° = 2*S°(C₂H₅OH)  + 2*S°(CO₂) - S°(glucose)

ΔS° = 2*(160.7 J/K) + 2(213.7 J/K) - 212.1 J/K

ΔS° = 321.4 + 427.4 J/K - 212.1 J/K

ΔS° = 536.7 J/K = 0.5367 kJ/K

Calculation of  ΔG°:

ΔG° =ΔH° - T*ΔS°

ΔG° = -67.9 kJ - 298K * 0.5367 kJ/K

ΔG° = -227.8 kJ

Since, ΔS° is positive and ΔH° is negative, ΔG° will be negative,

ΔG° is negative at all temperatures.

A negative ΔG° means the reaction is spontaneous at all temperatures.

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