for the equation C6H12O6(s) ---> 3C2H4(g)+3O2(g), find Delta S, Delta G, and the Temperature using the information given.

Question

C6​H12​O6​(s)⟶3C2​H4​(g)+3O2​(g)

Given S°= 212.1 J/molK, ΔG°= -910.56 kJ/mol and}ΔH°= -1274.5 kJ/mol for C₆H₁₂O₆ (s). You can look up the other values

 

Using the tabulated ΔG° values find, ΔS ,ΔG°, and the Temperature in C.

Answer 1

Look up ethylene and oxygen, then use products minus reactants for all three quantities. Standard table values:

C2H4(g): dHf = +52.4 kJ/mol, dGf = +68.4 kJ/mol, S = 219.3 J/mol-K
O2(g): dHf = 0, dGf = 0 (element in standard state), S = 205.0 J/mol-K

dH(rxn) = 3(52.4) + 3(0) - (-1274.5) = 157.2 + 1274.5 = +1431.7 kJ/mol

dG(rxn) = 3(68.4) + 3(0) - (-910.56) = 205.2 + 910.56 = +1115.8 kJ/mol

dS(rxn) = [3(219.3) + 3(205.0)] - 212.1 = 1272.9 - 212.1 = +1060.8 J/mol-K

Note O2 has dHf = dGf = 0 but its entropy is NOT zero - that trips people up constantly. Only formation energies are zero for elements.

Check your work before moving on: dG = dH - TdS at 298 K gives 1431.7 - 298(1.0608) = 1115.6 kJ. That matches the 1115.8 you got from the dGf table, so the three numbers are consistent. Always run that check.

For the temperature, set dG = 0, the crossover where the reaction turns spontaneous:

T = dH / dS = 1431.7 kJ / 1.0608 kJ/mol-K = 1350 K = 1077 C

Watch the units there - dS is in joules and dH is in kilojoules, so convert one before dividing. That is the single most common error on this problem.

Interpretation: dH is positive and dS is positive, so the reaction is non-spontaneous at room temperature and only becomes spontaneous above about 1077 C. That fits chemically - you are tearing a stable solid sugar apart into six moles of gas, which costs a lot of energy but buys a large entropy increase, so only high temperature makes it go.

Your numbers may differ by a few units depending on which table your textbook uses; the method is what matters.