Which statement describes a reaction at 298 K if [tex]$\Delta H=31$ $kJ / mol , \Delta S=0.093 kJ /( mol \cdot K )$[/tex]?
A. Gibbs free energy is negative.
B. It is not spontaneous.
C. It is exothermic.
D. It is at equilibrium.
Answer (2)
Calculate the Gibbs free energy change using the formula: Δ G = Δ H − T Δ S .
Substitute the given values: Δ G = 31 − ( 298 × 0.093 ) = 31 − 27.714 = 3.286 kJ/mol .
Since 0"> Δ G > 0 , the reaction is non-spontaneous.
The correct answer is B. It is not spontaneous.
Explanation
Problem Analysis and Given Data We are given the enthalpy change ( Δ H = 31 kJ/mol ), the entropy change ( Δ S = 0.093 kJ/(mol ⋅ K) ), and the temperature ( T = 298 K ). We need to determine whether the reaction is spontaneous, non-spontaneous, exothermic, or at equilibrium. To do this, we will calculate the Gibbs free energy change ( Δ G ) using the formula Δ G = Δ H − T Δ S .
Calculating Gibbs Free Energy We will now calculate the Gibbs free energy change ( Δ G ) using the formula Δ G = Δ H − T Δ S . Substituting the given values, we have: Δ G = 31 kJ/mol − ( 298 K × 0.093 kJ/(mol ⋅ K) )
Calculating the Value Δ G = 31 − ( 298 × 0.093 ) = 31 − 27.714 = 3.286 kJ/mol The result of the calculation is Δ G = 3.286 kJ/mol .
Determining Spontaneity and Correct Answer Since 0"> Δ G > 0 , the reaction is non-spontaneous. Also, since 0"> Δ H > 0 , the reaction is endothermic. Therefore, the correct statement is that the reaction is not spontaneous.
Examples
The Gibbs free energy helps determine if a reaction will occur spontaneously at a given temperature. For example, in drug development, chemists use Gibbs free energy to predict whether a drug molecule will bind spontaneously to a target protein in the body. If the Gibbs free energy is negative, the binding is spontaneous, which is essential for the drug to be effective. This concept is also used in designing industrial chemical processes to ensure reactions proceed efficiently and economically.
The reaction at 298 K with Δ H = 31 kJ/mol and Δ S = 0.093 kJ/(mol ⋅ K) has a Gibbs free energy change of Δ G = 3.286 kJ/mol , indicating it is non-spontaneous. Therefore, the correct answer is B: It is not spontaneous.
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