For the reaction H2(g) +S(s)->H2S(s), Triangle H= -20.2 kJ and triangle S= 43.1 J/K. WHen will the reaction be spontaneous?
I know that it has to be negative to be spontaneous but i don't exactly know how to answer this question.
To determine whether a reaction will be spontaneous or not, we can use the concept of Gibbs free energy (ΔG). The Gibbs free energy change (ΔG) is related to the enthalpy change (ΔH) and entropy change (ΔS) by 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, we have ΔH = -20.2 kJ and ΔS = 43.1 J/K. However, we need to express the enthalpy change in J to match the units of entropy (J/K). So, we convert -20.2 kJ to J by multiplying it by 1000:
ΔH = -20.2 kJ = -20,200 J
Now, we substitute the values into the equation:
ΔG = ΔH - TΔS
To determine when the reaction will be spontaneous, we need to find the temperature at which ΔG becomes negative. By setting ΔG < 0, we can calculate the temperature:
0 = -20,200 J - T * 43.1 J/K
Now, solve for T:
T = (-20,200 J) / (43.1 J/K)
T ≈ 469 K
Therefore, the reaction will be spontaneous (ΔG < 0) at temperatures higher than approximately 469 K.