I'm unbelievably stuck on this problem. Any help would be appreciated.
Red phosphorous is formed by heating white phosphorous. Calculate the temperature at which the two forms are at equilibrium, given
white P: (heat of formation)= 0.00 kj/mol; standard molar entropy= 41.09 J/mol K
red P: (heat of formation)= -17.6 kj/mol; standard molar entropy = 44.14 J/mol K
When they are in equilibrium, I believe the Gibbs free energy terms H - TS are equal. You could use that relationship to solve for T.
Review the subjects of Gibbs free energy, equilibrium and the Phase rule to verify this assertion.
To calculate the temperature at which white phosphorous (P) and red phosphorous (P) are at equilibrium, we can use the Gibbs free energy equation:
ΔG = ΔH - TΔS
where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy (heat of formation), T is the temperature in Kelvin, and ΔS is the change in entropy.
At equilibrium, ΔG = 0, so the equation becomes:
0 = ΔH - TΔS
Rearranging the equation gives:
T = ΔH/ΔS
Now we can substitute the values given:
ΔH = -17.6 kJ/mol (heat of formation)
ΔS = (44.14 J/mol K) - (41.09 J/mol K) = 3.05 J/mol K (change in entropy)
Before plugging these values into the equation, we need to convert ΔH to J/mol:
ΔH = -17.6 kJ/mol × 1000 J/1 kJ = -17,600 J/mol
Substituting the values into the equation:
T = (-17,600 J/mol) / (3.05 J/mol K) ≈ -5770 K
However, this answer doesn't make sense because the temperature should be positive. The negative value indicates that the reaction is spontaneous in the reverse direction (red phosphorus changing back into white phosphorus).
In reality, white phosphorus is heated to around 560-600 °C (833-873 K) to form red phosphorus. So, based on this information, it seems that the temperature at which the two forms are at equilibrium is approximately 833-873 K.
Note: It's always important to cross-check your results with other sources or known values in order to validate the answer.