To solve this problem it is necessary to apply the concepts related to Gibbs free energy and spontaneity
At constant temperature and pressure, the change in Gibbs free energy is defined as
[tex]\Delta G = \Delta H - T\Delta S[/tex]
Where,
H = Entalpy
T = Temperature
S = Entropy
When the temperature is less than that number it is negative meaning it is a spontaneous reaction. [tex]\Delta G[/tex] is also always 0 when using single element reactions. In numerical that implies [tex]\Delta G = 0[/tex]
At the equation then,
[tex]\Delta G = \Delta H - T\Delta S[/tex]
[tex]0 = \Delta H - T\Delta S[/tex]
[tex]\Delta H = T\Delta S[/tex]
[tex]T = \frac{\Delta H}{\Delta S}[/tex]
[tex]T = \frac{-93.8kJ}{-156.1J/K}[/tex]
[tex]T = \frac{-93.8*10^3J}{-156.1J/K}[/tex]
[tex]T = 600.89K[/tex]}
Therefore the temperature changes the reaction from non-spontaneous to spontaneous is 600.89K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
We want to know at what temperature a reaction changes from nonspontaneous to spontaneous, that is, at what temperature ΔG° = 0.
Given the standard enthalpy and entropy of the reaction, we can calculate that temperature using the following expression.
ΔG° = ΔH° - T . ΔS°
0 = -93.8 kJ - T . (-156.1 J/K)
T = 601 K
A particular reaction, with ΔH° = − 93.8 kJ and ΔS° = − 156.1 J/K, will change from nonspontaneous to spontaneous at 601 K.
Learn more about spontaneity here: https://brainly.com/question/9552459
