I have two questions about this problem, I understand how it's supposed to be done but I don't understand how my teacher came to this answer, so I was wondering if somebody could confirm or deny his answer here's the problem.
Below what temperature is: I2(g) --> I2(s)spontaneous?
When my teacher switches the equation around he ends up with this: T < -dela H/ delta S. Is that right? I would think that it would be positive
When you say "switched the equations around" are you referring to switching
I2(s) ==> I2(g)
Yes, when you mention "switching the equation around," I assume you mean reversing the chemical reaction. In this case, "I2(g) --> I2(s)" would become "I2(s) --> I2(g)."
To determine whether the reaction is spontaneous or not at a certain temperature, we can use Gibbs free energy (ΔG). The equation you mentioned, T < -ΔH/ΔS, is actually the condition for spontaneity at constant temperature and pressure. The correct equation should be 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.
By examining the sign of ΔG, we can determine whether the reaction is spontaneous. If ΔG is negative, the reaction is spontaneous, and if ΔG is positive, the reaction is non-spontaneous.
In the case of the reaction I2(g) --> I2(s), when the equation is reversed to I2(s) --> I2(g), the sign of the enthalpy change (ΔH) remains the same, but the sign of the entropy change (ΔS) is opposite. Therefore, to find the condition for spontaneity, we use ΔG = -ΔH + TΔS.
If we want the reaction to be spontaneous, ΔG should be negative. That means -ΔH + TΔS < 0. Rearranging the equation gives us T > ΔH/ΔS, not T < -ΔH/ΔS as your teacher stated.
So, the correct condition for spontaneity in this case is T > ΔH/ΔS, indicating that the reaction is spontaneous below a certain temperature.