Question

Consider the signs for ∆Hrxn and ∆Srxn for several different reactions. In which case is the reaction spontaneous at all temperatures?

a) ∆Hrxn < 0, ∆Srxn > 0
b) ∆Hrxn > 0, ∆Srxn < 0
c) ∆Hrxn < 0, ∆Srxn < 0
d) ∆Hrxn > 0, ∆Srxn > 0

Answer 1

A reaction is spontaneous at all temperatures if the enthalpy change (ΔHrxn) is negative and the entropy change (ΔSrxn) is positive. Hence, the answer is a) ΔHrxn < 0, ΔSrxn > 0, as this ensures that the Gibbs free energy (ΔG) is negative regardless of temperature.

Step-by-step explanation:

To determine the spontaneity of a reaction at all temperatures, we look at the signs of the enthalpy change (ΔHrxn) and the entropy change (ΔSrxn). According to Gibbs free energy equation (ΔG=ΔH-TΔS), a reaction will be spontaneous at all temperatures if ΔG is always negative.

For a reaction to be spontaneous at all temperatures, ΔH must be negative (exothermic) and ΔS must be positive, as a positive ΔS ensures that the -TΔS term will increasingly favor spontaneity at higher temperatures. This would lead to ΔG being negative at all temperatures due to the equation ΔG=ΔH-TΔS. Hence, the correct answer is a) ΔHrxn < 0, ΔSrxn > 0, which corresponds to both driving forces (enthalpy and entropy changes) being in favor of product formation.

Option b) ΔHrxn > 0, ΔSrxn < 0 indicates that the reaction is never spontaneous because it would require energy input and results in decrease in disorder. For the options c) ΔHrxn < 0, ΔSrxn < 0, and d) ΔHrxn > 0, ΔSrxn > 0, these reactions may be spontaneous under certain temperature conditions, but not at all temperatures.