Given the equation for free energy, (delta)G = (delta)H-T(delta)S, we can determine that the reaction is nonspontaneous at all temperatures if H is positive and S is negative. This combination would always lead to a positive G value, meaning that free energy is required for the reaction to take place and it is therefore nonspontaneous.

A reaction is spontaneous if the Gibb's Free Energy of the reaction is negative.

If , the enthalpy, and , the entropy, are both negative, then the reaction will be spontaneous if and only if the magnitude of the enthalpy is greater than the magnitude of the entropy times the temperature.

A reaction is spontaneous if the change of Gibb's free energy (G) is less than zero. Recall that (G) is related to H and S by the equation below.

For this particular reaction, S is negative because the total number of moles of gas decreases from reactants to products. Since the reaction is spontaneous at lower temperatures, then G must be negative when T is small. Since the -TS term would be positive for all values of T, the only way G can be negative is if H is negative. At higher temperatures, the positve -TS term would outweigh the negative H term, resulting in a positive G and a non-spontaneous reaction.

A reaction is spontaneous when Gibb's free energy is negative. As a result, we need to determine the temperature range where Gibb's free energy is less than zero. Since we know the values for changes in enthalpy and entropy, we can plug them into the Gibb's free energy equation, and set it equal to zero.

437K is the temperature at which Gibb's free energy is zero. Since entropy is positive for this reaction, increasing the temperature will result in a more negative value for Gibb's free energy.

As a result, any temperature that is greater than 437K will make this reaction spontaneous.

For this question, we're told that a reaction is being run under standard conditions, and that the equilibrium constant for this reaction is much greater than 1. With this information in mind, we can find the correct answer without even having to resort to math. Since the equilibrium constant is greater than one, we know that the products of this reaction are favored over the reactants. And since the products are favored, this means that the reaction must be shifted to the right, in which case it is spontaneous. It's important to know that a spontaneous reaction will always have a negative change in Gibb's energy.

For completion's sake, however, we can show that the Gibb's free energy change is negative by utilizing the free energy change equation.

From this equation, we can see that if is greater than one (as it is in the reaction for this question) the natural logarithm of this value will also be positive. And since the ideal gas constant and the absolute temperature are also positive values, the product of all these values will be positive. But, there is a negative sign in front of these terms, hence making our final answer negative.

If Q is greater than K, the reaction has exceeded the equilibrium state. It will proceed nonspontaneously (since equilibrium has already been reached), and this must mean that the ΔG (gibbs free energy) must be positive, or greater than zero.

Based on Gibbs Free Energy, ∆G = ∆H – T∆S, we find that ∆H– will contribute to spontaneity (∆G<0). Since we are looking at temperatures, in order for ∆S to make an nonspontaneous reaction spontaneous at only high temperatures, ∆H will be positive, and ∆S will be positive.

Approximately 467 K. ∆G=∆H-T∆S and a rxn proceeds spontaneously when ∆G < 0
and is non-spontaneous when ∆G > 0. So if we set ∆G=0 and solve the equation for T, we will see that the crossover from spontaneous to non-spontaneous occurs when T=467K.

Change in free energy must always be negative for a spontaneous process. Additionally, Q must be less than K so that the reaction will proceed in the forward reaction, toward equilibrium.

A reaction being favorable or unfavorable is largely determined by the thermochemistry of the reaction. More specifically it is determined by the change in gibbs free energy which is determined by the change in enthalpy, the change in entropy, and the temperature of the reaction. Here we are told that the reaction has a favorable enthalpy change (- means energy is released). Qualitatively we can see that the reaction will have an unfavorable change in entropy because the product, being a solid, is more ordered than the reactants which are both solid and gaseous. Thus we can conclude that the reaction is favorable because of the favorable change in enthalpy, which helps to overcome the unfavorable change in entropy.