Answer: A:
ΔH = 10.5 kJ/mol, ΔS = 30.0 J/K·mol is non-spontaneous.
ΔH = 1.8 kJ/mol, ΔS = −113 J/K · mol is non-spontaneous.
Reaction A can become spontaneous
Reaction A is spontaneous at 76.85 °C
Explanation:
It's helpful to memorize that if:
-ΔS is greater than 0 and ΔH is less than 0; its spontaneous at all temperatures.
-ΔS is less than 0 and ΔH is greater than 0; its non-spontaneous at all temperature.
-ΔS is greater than 0 and ΔH is greater than 0; its spontaneous at high temperatures and non-spontaneous at low temperatures.
-ΔS is less than 0 and ΔH is less than 0; its spontaneous at low temperatures and non-spontaneous at high temperatures.
This comes from the equation ΔG=ΔH-TΔS
where ΔG is Gibbs free energy, ΔH is enthalpy, T is temperature (in Kelvin), and ΔS is entropy.
Without getting too in depth as to what each of those mean (you could take an entire class on entropy alone), the temperature at which the spontaneity changes is equal to ΔG (Gibbs free energy) at 0.
So take the above equation and set ΔG = 0, and rearrange the equation to solve for T.
ΔG=ΔH-TΔS
0=ΔH-TΔS
add TΔS to the other side
TΔS=ΔH
divide the right side by ΔS to find T (temperature)
T=ΔH/ΔS
Now we can find the temperature that the first reaction would occur at spontaneously.
We need to make sure that we have the same units for ΔH and ΔS, so divide 30 by 1000 to convert J/Kmol into kJ/kmol so that we have kJ for ΔH and ΔS.
30/1000 = 0.03 kJ
Plug in the values for the modified equation T=ΔH/ΔS
10.5 kJ/0.03 kJ = 350 K
The temperature is in Kelvin, so subtract 273.15 to convert it into Celsius
350-273.15 = 76.85 °C
