Import Question JSON

$\text{Hint: For a spontaneous reaction, }\Delta\text{G must be -ve.}$ $\text{To determine when an exothermic reaction with negative entropy change is spontaneous, we need to analyze the Gibbs free energy equation:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{For a reaction to be spontaneous, }\Delta\text{G must be negative.}$ $\text{Given information:}$ $\text{- The reaction is exothermic, so }\Delta\text{H is negative}$ $\text{- The entropy change (}\Delta\text{S) is negative}$ $\text{Substituting into the Gibbs equation:}$ $\Delta\text{G} = \Delta\text{H} - \text{T}\Delta\text{S}$ $\text{Since }\Delta\text{S is negative, the term -T}\Delta\text{S will be positive (a negative multiplied by a negative is positive).}$ $\text{For the reaction to be spontaneous, we need:}$ $\Delta\text{G} < 0$ $\text{Therefore:}$ $\Delta\text{H} - \text{T}\Delta\text{S} < 0$ $\Delta\text{H} < \text{T}\Delta\text{S}$ $\text{However, since both }\Delta\text{H and }\Delta\text{S are negative, we need to be careful with the inequality. When we divide both sides by the negative }\Delta\text{S, the inequality flips:}$ $\frac{\Delta\text{H}}{\Delta\text{S}} > \text{T}$ $\text{This means T must be less than }\frac{\Delta\text{H}}{\Delta\text{S}}\text{. In other words, temperature must be low for the reaction to be spontaneous.}$ $\text{At high temperatures, the T}\Delta\text{S term would become more significant, potentially making }\Delta\text{G positive and the reaction non-spontaneous.}$ $\text{Therefore, the answer is option 4: Low temperature.}$