I don't understand what this question is asking .. i have already tried it and i got it wrong.
Assume that H^o and S^o do not change with temperature, determine the temperature above which you can theoretically prepare carbon disulfide (CS2)directly from it's elements. (for this reaction S=815 jK-1 and H=87.9kJ/mol)
To determine the temperature above which you can theoretically prepare carbon disulfide (CS2) directly from its elements, we need to use the concept of Gibbs free energy (ΔG). The equation relating Gibbs free energy to enthalpy (H) and entropy (S) is as follows:
ΔG = ΔH - TΔS
In this equation, T represents the temperature in Kelvin.
For a reaction to be spontaneous (i.e., proceed in the forward direction without external intervention), ΔG must be negative. At the temperature above which CS2 can be prepared directly from its elements, ΔG will be zero.
Since H° (standard enthalpy change) and S° (standard entropy change) are given, we can write the equation as:
0 = ΔH° - TΔS°
Now, we can solve for the temperature (T):
T = ΔH° / ΔS°
Substituting the values given:
T = (87.9 kJ/mol) / (815 J/K mol)
Note: We need to convert 87.9 kJ to J by multiplying it by 1000 since the units need to be consistent.
T = (87.9 kJ/mol) x (1000 J/kJ) / (815 J/K mol)
Simplifying:
T ≈ 107.8 K
Therefore, the temperature above which you can theoretically prepare carbon disulfide (CS2) directly from its elements is approximately 107.8 Kelvin.