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
