Import Question JSON

$\text{Hint:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{To determine the temperature range at which the reaction is spontaneous, we need to analyze the Gibbs free energy equation:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{For a reaction to be spontaneous, }\Delta\text{G must be negative.}$ $\text{Given information:}$ $\text{- The reaction releases energy (q kJ/mol)}$ $\text{- The entropy change (}\Delta\text{S) is positive}$ $\text{Step 1: Identify the enthalpy change of the reaction.}$ $\text{Since the reaction releases energy (q kJ/mol), this indicates an exothermic reaction. For exothermic reactions, the enthalpy change is negative:}$ $\Delta\text{H} = -\text{q kJ/mol} < 0$ $\text{Step 2: Analyze the Gibbs free energy equation with the given parameters.}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{With }\Delta\text{H} < 0\text{ and }\Delta\text{S} > 0\text{:}$ $\text{Both terms in the equation (}\Delta\text{H and -T}\Delta\text{S) are negative, which means }\Delta\text{G will always be negative, regardless of the temperature value.}$ $\text{This can be explained as follows:}$ $\text{- The negative }\Delta\text{H contributes to making }\Delta\text{G negative}$ $\text{- Since }\Delta\text{S is positive, the term -T}\Delta\text{S is negative (as T is always positive in Kelvin)}$ $\text{- The sum of two negative terms is always negative}$ $\text{Step 3: Determine the temperature range for spontaneity.}$ $\text{Since }\Delta\text{G is negative at all temperatures, the reaction will be spontaneous at any temperature.}$ $\text{When both enthalpy and entropy changes favor the reaction (}\Delta\text{H} < 0\text{ and }\Delta\text{S} > 0\text{), we have what is sometimes called an "entropy-enthalpy driven reaction" or "win-win situation" in thermodynamic terms. Such reactions are spontaneous under all conditions.}$ $\text{Therefore, the reaction is possible at any temperature, making option 4 the correct answer.}$