for a particular reaction H=-28.4 KJ and S=-87.9 J/K. Assuming these values change very little with temperture, over what temperature range is the reaction spontaneous in the forward direction?
To determine the temperature range over which a reaction is spontaneous in the forward direction, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
Where:
ΔG is the Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
In this case, we have:
ΔH = -28.4 kJ
ΔS = -87.9 J/K
We need to convert the units of ΔH to J:
ΔH = -28.4 kJ × 1000 J/kJ = -28,400 J
Now, let's find the temperature range when the reaction is spontaneous in the forward direction. At equilibrium, ΔG = 0. This means that at any temperature where ΔG is negative, the reaction is spontaneous in the forward direction. Hence, we need to find at which temperature ΔG is negative.
0 = ΔH - TΔS
Rearranging the equation gives us:
TΔS = ΔH
Now, we can isolate T:
T = ΔH / ΔS
Substituting the values we have:
T = (-28,400 J) / (-87.9 J/K)
Calculating this gives us:
T ≈ 323 K
Therefore, the reaction is spontaneous in the forward direction at temperatures below approximately 323 Kelvin.