Question

Find ΔG∘rxnΔGrxn∘ at 25.0 ∘C

I2(s) I2(g)
ΔHf,I2(g)∘=62.42kJ/mol,
S of formation of I2(s)=116.14J/(mol⋅K)
and s of formation of ∘I2(g)=260.69J/mol⋅

Answer 1

The standard free-energy change (ΔG°rxn) at 25.0 °C for the reaction I2(s) → I2(g) is 26.64 kJ/mol.

Step-by-step explanation:

To find the standard free-energy change (ΔG°rxn) at 25.0 °C, we can use the formula ΔG°rxn = ΔH°rxn - TΔS°rxn. First, we need to calculate the enthalpy change (ΔH°rxn) and entropy change (ΔS°rxn) for the reaction I2(s) → I2(g). Given the standard enthalpy of formation (ΔH°f) of I2(g) is 62.42 kJ/mol and the standard entropy of formation (S°f) of I2(s) is 116.14 J/(mol·K), we need to convert the units to match. One kilojoule (kJ) is equal to 1000 joules (J), so the enthalpy change is 62.42 kJ/mol and the entropy change is 0.11614 kJ/(mol·K). Now we can substitute these values into the formula. However, I noticed that there is a typo in the given equation (I2(g) → I2(g)). Let's assume that it was supposed to be I2(s) → I2(g).

Substituting the values into the formula, we have:

ΔG°rxn = ΔH°rxn - TΔS°rxn

= (62.42 kJ/mol) - (298 K)(0.11614 kJ/(mol·K))

= 26.64 kJ/mol

Therefore, the standard free-energy change (ΔG°rxn) at 25.0 °C for the reaction I2(s) → I2(g) is 26.64 kJ/mol.

Answer 2

The standard Gibbs free energy change (∆G°rxn) at 25.0°C for the reaction I2(s) → I2(g) is approximately 53.55 kJ/mol.

Explanation

Gibbs free energy (∆G°) relates to spontaneity and directionality of a reaction. For this process (I2(s) → I2(g)), it involves the phase change of solid iodine (I2) into gaseous iodine. To calculate ∆G°rxn, we use the equation ∆G°rxn = ∆H°rxn - T∆S°rxn, where ∆H°rxn is the standard enthalpy change and ∆S°rxn is the standard entropy change. Given the enthalpy of formation for gaseous iodine (∆Hf,I2(g)° = 62.42 kJ/mol) and the entropy values for both solid (∆S°f, I2(s) = 116.14 J/(mol⋅K)) and gaseous iodine (∆S°f, I2(g) = 260.69 J/(mol⋅K)), we can calculate ∆G°rxn.

First, convert the entropy values to kJ/mol⋅K by dividing by 1000: S°f, I2(s) = 0.11614 kJ/(mol⋅K) and S°f, I2(g) = 0.26069 kJ/(mol⋅K). Then, use the equation ∆G°rxn = ∆Hf,I2(g)° - T∆S°rxn, substituting the values to find ∆G°rxn at 25.0°C (298.15 K). Calculations yield ∆G°rxn = 62.42 kJ/mol - 298.15 K * (0.26069 kJ/(mol⋅K) - 0.11614 kJ/(mol⋅K)), resulting in approximately 53.55 kJ/mol as the standard Gibbs free energy change at 25.0°C for the given reaction. This positive value indicates a non-spontaneous process at standard conditions.