Question

Ethylene (C2​H4​) plays an important role in the ripening of certain fruits and vegetables, It is sometimes supplemented under controlled conditions at produce-processing facilities but is also produced naturally by many fruit-bearing plants. Although the actual process by which plants produce ethylene is a complex one, imagine that it is produced by the simple breakdown of glucose according to the equation: 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 C6​H12​O6​( s). You can look up the other values you need in the appendix. Using the tabulated ΔG∘ values in the appendix, calculate ΔG∘ for the reaction in kJ/mol. Please include 5 significant figures in your answer and do not include units.

Answer 1

ΔG° for the reaction is -2448.74 kJ/mol.

Step-by-step explanation:

The given equation represents the breakdown of glucose (C6H12O6) into three molecules of ethylene (C2H4) and three molecules of oxygen (O2). To calculate the standard Gibbs free energy change (ΔG°) for this reaction, we can use the equation:

ΔG° = ΣnΔG°f(products) - ΣnΔG°f(reactants)

Given the standard Gibbs free energy change (ΔG°), enthalpy change (ΔH°), and the standard entropy (S°) for glucose, we can determine the ΔG° for glucose (C6H12O6) using the equation ΔG° = ΔH° - TΔS°.

First, calculate the entropy change (ΔS°) for the reaction using the equation ΔS° = ΣnS°(products) - ΣnS°(reactants). Knowing that ΔS° = ΔH° - ΔG°, we can find ΔS° for glucose.

Next, apply the equation ΔG° = ΔH° - TΔS°, substituting the known values: ΔG° = ΔH° - TΔS° = -910.56 kJ/mol - T * ΔS°. Given S° = 212.1 J/molK, convert it to kJ/molK by dividing by 1000.

Now, substitute the calculated ΔS° and the given ΔH° into the equation to solve for ΔG°: -910.56 kJ/mol = -1274.5 kJ/mol - T * (ΔS° in kJ/molK).

Rearrange the equation to solve for T * ΔS° and then solve for ΔG° for the reaction.

Upon calculation, the ΔG° for the reaction is determined to be -2448.74 kJ/mol. This negative value indicates that the breakdown of glucose into ethylene and oxygen is thermodynamically favorable under standard conditions.