2-propanol has a delta Hvp of 701 J/g and a vapor pressure of 31.6 mmHg at 20.0 degrees C. Estimate the normal boiling point of this alcohol in degrees C.
i used the clasius-clayperon equation and this is what i have so far.
ln (31.6 mmHg/760 mmHg)= (-701 J/g)/8.314 J/g C (1/ T2 - 1/20 C)
is that correct?
No, you have at least two errors. T must be converted to Kelvin and Hvap must be in J/mol. Therefore, T1 = 273.15 + 20.0 = ?? and Hvap for 2-propanol is 701 J/g x ??grams/mol = ??
Also, I think the equation should be ln(p2/p1). You have reversed the T1 and T2 but that is ok since you have a negativae sign for Hvap. But P2 should be on top and P1 on bottom.
Yes, you are on the right track in using the Clausius-Clapeyron equation to estimate the normal boiling point of 2-propanol. However, I see a small mistake in the rearrangement of the equation.
The correct equation is:
ln(P1/P2) = (-ΔHvap/R) * (1/T2 - 1/T1)
Where:
P1 is the vapor pressure at the known temperature T1
P2 is the vapor pressure at the unknown temperature T2
ΔHvap is the heat of vaporization (also known as the enthalpy of vaporization)
R is the ideal gas constant (8.314 J/(mol·K))
In this case, you are given the vapor pressure (P1) at 20.0 degrees C and the heat of vaporization (ΔHvap). You want to estimate the normal boiling point, which is equivalent to the unknown temperature (T2).
Therefore, the correct equation to use would be:
ln(31.6 mmHg/760 mmHg) = (-701 J/g)/(8.314 J/(g·C)) * (1/T2 - 1/20 C)
To solve for T2, rearrange the equation and solve for 1/T2:
1/T2 = (ln(31.6 mmHg/760 mmHg))/((-701 J/g)/(8.314 J/(g·C))) - 1/20 C
Then, solve for T2 by taking the reciprocal of both sides:
T2 = 1 / [(ln(31.6 mmHg/760 mmHg))/((-701 J/g)/(8.314 J/(g·C))) - 1/20 C]
Calculate the right side of the equation to find the estimated normal boiling point of 2-propanol in degrees Celsius.