Consider the following spontaneous exothermic reaction :3X2(g) → 2X3(g)What are the sign of ΔH, ΔS, and ΔG for the reaction.+, +, ++, –, ––, +, ––, –, –

The sign of ΔH for the reaction can be determined by examining the stoichiometric coefficients of the reactants and products. In this case, since the reaction is exothermic, the sign of ΔH is negative (-).

The sign of ΔS for the reaction can be determined by considering the change in the number of moles of gas molecules. In this case, since the number of moles of gas molecules decreases from 3X2(g) to 2X3(g), the sign of ΔS is negative (-).

The sign of ΔG for the reaction can be determined using the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin. Since both ΔH and ΔS are negative (-), the sign of ΔG depends on the temperature. If the temperature is low, the magnitude of TΔS will be smaller than ΔH, resulting in a positive (+) ΔG. However, if the temperature is high, the magnitude of TΔS will be larger than ΔH, resulting in a negative (-) ΔG.