What will be the result of a reaction if its enthalpy change is positive, its entropy change is negative and Gibbs free energy is positive?

A. Never spontaneous
B. Nonspontaneous, if the temperature is low
C. Spontaneous, if the temperature is high
D. Always spontaneous

http://www.kentchemistry.com/links/Kinetics/Gibbs.htm

ΔG = ΔH - TΔS

Consider answer C, if Temp goes high, then the - TΔS goes positive, and delta G is even more positive.
Answer A looks inviting here.

Oh, I love talking about reactions! Especially the ones with positive enthalpy change, negative entropy change, and positive Gibbs free energy. It's like a recipe for disaster!

So, based on these factors, the reaction would be in the mood to party, but it wouldn't actually happen by itself. It's like inviting a bunch of rowdy guests to your house, but they won't show up unless you really crank up the temperature.

In other words, the correct answer is B. Nonspontaneous, if the temperature is low. So, if you want this reaction to happen, you better grab a thermometer and start cranking that temperature dial!

The correct answer is B. Nonspontaneous, if the temperature is low.

The positive enthalpy change indicates an endothermic reaction, meaning heat is absorbed from the surroundings. The negative entropy change indicates a decrease in disorder or randomness in the system. And the positive Gibbs free energy indicates that the reaction is not thermodynamically favorable.

For a reaction to be spontaneous, the overall change in Gibbs free energy (ΔG) must be negative. However, in this case, the positive ΔG indicates that the reaction is nonspontaneous. However, if the temperature is raised, the positive ΔG might become negative as the equation ΔG = ΔH - TΔS suggests, allowing the reaction to become spontaneous at higher temperatures.

To determine the spontaneity of a reaction given its enthalpy change (ΔH), entropy change (ΔS), and Gibbs free energy (ΔG), we can use the equation:

ΔG = ΔH - TΔS

where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, ΔS is the change in entropy, and T is the temperature in Kelvin.

In this case, the question states that the enthalpy change is positive (ΔH > 0), the entropy change is negative (ΔS < 0), and the Gibbs free energy is positive (ΔG > 0).

Plugging these values into the equation, we get:

ΔG = positive (ΔH) - T * negative (ΔS)

Since both ΔH and ΔS have opposite signs, the sign of ΔG will depend on the temperature (T).

Now, let's analyze the options:

A. Never spontaneous: This option can be ruled out because if the temperature is high enough, the negative TΔS term can overcome the positive ΔH term, resulting in a negative ΔG and making the reaction spontaneous.

B. Nonspontaneous, if the temperature is low: This option is consistent with the information given. At low temperatures, the negative TΔS term is more significant, making the positive ΔG term dominate and making the reaction nonspontaneous.

C. Spontaneous, if the temperature is high: This option is also consistent with the information given. At high temperatures, the negative TΔS term becomes less significant compared to the positive ΔH term, resulting in a negative ΔG and making the reaction spontaneous.

D. Always spontaneous: This option can be ruled out because it does not account for the temperature dependence. The spontaneity of the reaction depends on the temperature.

Therefore, the correct answer is option B. Nonspontaneous, if the temperature is low.