Question

For a given reaction, ΔH = -19.9 kJ/mol and ΔS = -55.5 J/Kmol.

The reaction will have ΔG = 0 at __________ K.

Assume that ΔH and ΔS do not vary with temperature.

Answer 1

The reaction will have ΔG = 0 at approximately 359.46 K.

Step-by-step explanation:

To determine the temperature at which the reaction will have a standard free energy change (ΔG) of 0, we can use the equation ΔG = ΔH - TΔS. By substituting the given values: ΔH = -19.9 kJ/mol and ΔS = -55.5 J/Kmol, we can rearrange the equation to solve for T.

ΔG = -19.9 kJ/mol - T(-55.5 J/Kmol) = 0

Simplifying the equation:

T = (-19.9 kJ/mol) / (-55.5 J/Kmol)

By converting units and calculating:

T ≈ 359.46 K

Therefore, the reaction will have ΔG = 0 at approximately 359.46 K.

Answer 2

359 K

Step-by-step explanation:

The Gibbs free energy can be calculated using the following expression.

ΔG = ΔH - T.ΔS

where,

ΔG is the Gibbs free energy

ΔH is the enthalpy of the reaction

T is the absolute temperature

ΔS is the entropy of the reaction.

When ΔG = 0,

ΔH - T.ΔS = 0

ΔH = T.ΔS

T = ΔH/ΔS = (-19.9 × 10³ J/mol) / (-55.5 J/K.mol) = 359 K