At the normal boiling point of toluene,Δ Hvap=33.2 kJ/mol and ΔSvap= 86.5J/mol•K. What is the normal boiling point of toluene?
dG = dH-TdS
At equilibrium (boiling is an equilibrium process), dG = 0. Substitute dH and dS and solve for Tb.
A little lost but I'm getting 384...
To find the normal boiling point of toluene, we need to use the equation:
ΔG = ΔH - TΔS
where:
ΔG is the change in Gibbs free energy
ΔH is the enthalpy of vaporization
ΔS is the entropy of vaporization
T is the temperature
At the normal boiling point, ΔG is equal to zero,
So, the equation becomes:
0 = ΔH - TΔS
To find the normal boiling point, we need to rearrange the equation to solve for T:
TΔS = ΔH
T = ΔH / ΔS
Now, let's substitute the given values into the equation:
T = 33.2 kJ/mol / (86.5 J/mol•K)
First, we need to convert kJ to J by multiplying 33.2 kJ/mol by 1000:
T = (33.2 × 1000) J/mol / (86.5 J/mol•K)
T = 33200 J/mol / (86.5 J/mol•K)
Now, we can cancel out the units of J/mol:
T = 33200 / 86.5 K
After performing the calculation, we find:
T ≈ 383.24 K
Therefore, the normal boiling point of toluene is approximately 383.24 K.