The normal boiling point for acetone is 56.5 Celsius. The atmospheric pressure is 630torr. What would be the boiling point of acetone, delta Hvap = 32.0 KJ/mol. What would be the vapor pressure of acetone at 25.0 Celsius.

Please help me with this problem. Thank you

To find the boiling point of acetone at certain pressure, we can use the Clausius-Clapeyron equation. The equation is:

ln(P2/P1) = -(delta Hvap/R) * (1/T2 - 1/T1)

Where:
P2 = vapor pressure at the new temperature
P1 = vapor pressure at the known temperature
delta Hvap = molar enthalpy of vaporization
R = ideal gas constant (8.314 J/mol·K)
T2 = new temperature (in Kelvin)
T1 = known temperature (in Kelvin)

Given:
P1 = 630 torr
T1 = 56.5°C = 56.5 + 273.15 = 329.65 K
delta Hvap = 32.0 kJ/mol = 32.0 * 1000 J/mol

Convert the temperatures to Kelvin:
T2 = 25°C = 25 + 273.15 = 298.15 K

Now, let's plug the values into the Clausius-Clapeyron equation:

ln(P2/630) = -(32.0 * 1000 J/mol) / (8.314 J/mol·K) * (1/298.15 K - 1/329.65 K)

Simplify the equation:

ln(P2/630) = -38.56 + 41.25

ln(P2/630) = 2.69

Now, solve for P2 by taking the antilog of both sides:

P2/630 = e^2.69

P2 = 630 * e^2.69

Using a calculator or a scientific tool, you can calculate P2 which will give you the vapor pressure of acetone at 25.0°C.