Benzene has a heat of vaporization of 30.72kJ/mol and a normal boiling point of 80.1 C. At what temperature does benzene boil when the external pressure is 490 ?
I think I have the equation setup, but I can't figure out how to solve for T2..please help
To solve for the boiling point of benzene (T2) when the external pressure is 490 mmHg, you can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 = Pressure at the normal boiling point (760 mmHg)
P2 = External pressure (490 mmHg)
ΔHvap = Heat of vaporization (30.72 kJ/mol)
R = Ideal Gas Constant (8.314 J/(mol·K))
T1 = Normal boiling point (80.1 °C + 273.15)
First, convert temperatures to Kelvin:
T1 = 80.1 °C + 273.15 = 353.25 K
Next, rearrange the equation to solve for T2:
ln(P2/P1) = -(ΔHvap/R) * (1/T2 - 1/T1)
Substitute the known values:
ln(490/760) = -(30.72 kJ/mol / (8.314 J/(mol·K))) * (1/T2 - 1/353.25)
Use the natural logarithm to solve for the logarithmic term on the left side:
-0.4089 = -(30.72 kJ/mol / (8.314 J/(mol·K))) * (1/T2 - 1/353.25)
Now, isolate the temperature term by rearranging the equation further:
(1/T2 - 1/353.25) = (0.4089 * 8.314 J/(mol·K)) / 30.72 kJ/mol
Evaluate the right side of the equation:
(1/T2 - 1/353.25) = 0.01113
Combine fractions on the left side:
(353.25 - T2) / (353.25 * T2) = 0.01113
Multiply through by 353.25 * T2 to eliminate the fractions:
353.25 - T2 = 0.01113 * (353.25 * T2)
353.25 - T2 = 3.92775 * T2
Combine like terms:
353.25 = 4.92775 * T2
Divide both sides by 4.92775 to isolate T2:
T2 = 353.25 / 4.92775
Calculate T2:
T2 = 71.62 °C
Therefore, benzene boils at approximately 71.62 °C when the external pressure is 490 mmHg.