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?
Yes, you are correct in using the Clausius-Clapeyron equation to estimate the normal boiling point of 2-propanol. However, there seems to be a minor error in your equation. Let's go step by step to clarify:
The Clausius-Clapeyron equation relates the natural logarithm of the ratio of vapor pressures to the enthalpy of vaporization and the temperatures:
ln(P2/P1) = -(ΔHvap/R)(1/T2 - 1/T1)
Where:
P1 and P2 are the initial and final vapor pressures, respectively (in this case, 760 mmHg is the normal boiling point vapor pressure);
ΔHvap is the enthalpy of vaporization per gram (701 J/g);
R is the ideal gas constant (8.314 J/g°C); and
T1 and T2 are the initial and final temperatures, respectively (20.0°C and the boiling point we want to estimate).
Now, let's plug in the values and solve for T2:
ln(31.6 mmHg/760 mmHg) = -(701 J/g)/ (8.314 J/g°C) * (1/T2 - 1/20.0°C)
Now, you need to simplify and solve for T2. Start by rearranging the equation:
ln(31.6/760) = -701/8.314 * (1/T2 - 1/20.0)
Next, calculate the natural logarithm of the ratio:
-3.361 = -84.31 * (1/T2 - 1/20.0)
Now, simplify further:
-3.361 = (-84.31/T2) + (84.31/20.0)
Now, isolate the 1/T2 term:
(-84.31/T2) = -3.361 - (84.31/20.0)
Next, combine the constants:
(-84.31/T2) = -3.361 - 4.216
Now, combine the constants on the right-hand side:
(-84.31/T2) = -7.577
Lastly, isolate T2 by taking the reciprocal of both sides:
T2 = -84.31/-7.577
Finally, calculate T2:
T2 = 11.12°C
So, according to the calculations, the estimated normal boiling point of 2-propanol is approximately 11.12°C.