Given that for the vaporization of benzene Hvap= 30.7 kj/mol and Svap= 87.0 J(K * mol) , calculate Delta G (in kJ/mol) for the vaporization of benzene at 13 Celcius.
To calculate ΔG (Gibbs free energy) for the vaporization of benzene at 13 degrees Celsius, we need to use the equation:
ΔG = ΔH - TΔS
where:
ΔG is the change in Gibbs free energy
ΔH is the change in enthalpy
T is the temperature in Kelvin
ΔS is the change in entropy
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 13°C + 273.15
T = 286.15 K
Now we can calculate ΔG:
ΔG = Hvap - TΔS
ΔG = 30.7 kJ/mol - (286.15 K * (87.0 J/(K·mol)))
To perform the calculation, we need to ensure that the units of the enthalpy and entropy match. Since 1 kJ = 1000 J, we convert the units of Hvap to J/mol:
ΔG = (30.7 kJ/mol * 1000 J/kJ) - (286.15 K * (87.0 J/(K·mol)))
Now we can calculate ΔG:
ΔG = (30700 J/mol) - (286.15 K * (87.0 J/(K·mol)))
ΔG = 30700 J/mol - 24960.05 J/mol
ΔG ≈ 5739.95 J/mol
To convert from J/mol to kJ/mol, divide the value by 1000:
ΔG ≈ 5.74 kJ/mol
Therefore, ΔG for the vaporization of benzene at 13 degrees Celsius is approximately 5.74 kJ/mol.
To calculate ΔG (Gibbs free energy) for the vaporization of benzene at 13 degrees Celsius, we can use the equation:
ΔG = ΔH - TΔS
Where:
ΔG = Gibbs free energy change
ΔH = Enthalpy change (Hvap)
T = Temperature in Kelvin
ΔS = Entropy change (Svap)
First, let's convert the temperature from Celsius to Kelvin:
T(Kelvin) = T(Celsius) + 273.15
T(Kelvin) = 13 + 273.15
T(Kelvin) = 286.15 K
Now, we can calculate ΔG:
ΔG = Hvap - TΔS
ΔG = 30.7 kJ/mol - (286.15 K) (87.0 J(K * mol)/1000)
ΔG = 30.7 kJ/mol - 24.89 kJ/mol
ΔG = 5.81 kJ/mol
Therefore, the ΔG for the vaporization of benzene at 13 degrees Celsius is 5.81 kJ/mol.
dG = dH - TdS
dG = 30.7 - (273+13)*0.087 = ? in kJ/mol.Note that dS is in J/mol*K so I changed that to kJ/mol*K