0
$\begingroup$

After conducting a practical revolving around burning alcohols, I found that propan-1-ol heated the 50 mL of water to 60 °C about 1 min faster than propan-2-ol.

The experiment was controlled as much as possible, including the temperature escaping the setup.

I know that propan-2-ol is more volatile than propan-1-ol, so I have a reason to believe that this is tied to the slower heating. On the other hand, I have yet to find any sources to verify this.

If I do not find any more about this, I will have to call this out as an anomaly.

If anyone knows anything about the relationship between the position of the $\ce{OH}$ bond in an alcohol and the heat output/speed of heating, please explain this to me.

$\endgroup$
  • 3
    $\begingroup$ 1 min faster than what? 1 min vs 2 min is quite different from 100 min vs 101 min. $\endgroup$ – DHMO Sep 5 '16 at 10:03
  • $\begingroup$ You are aware of the difference in combustion enthalpy? -2021 kJ/mol compared to -2006 kJ/mol? $\endgroup$ – Karl Sep 5 '16 at 11:13
3
$\begingroup$

The enthalpy of combustion $\Delta_\mathrm cH$ of a substance can be calculated from the enthalpy of formation $\Delta_\mathrm fH$. For a compound containing only carbon, hydrogen, and oxygen, the general combustion reaction is

$$\ce{C_$a$H_$b$O_$c$ + ($a$ + 1/4 $b$ – 1/2 $c$) O2 -> $a$ CO2(g) + 1/2 $b$ H2O(l)}$$

The corresponding standard enthalpy of combustion is

$$\Delta_\mathrm cH^\circ=-a\Delta_\mathrm fH^\circ(\ce{CO2,g})-\frac12b\Delta_\mathrm fH^\circ(\ce{H2O,l})+\Delta_\mathrm fH^\circ(\ce{C_$a$H_$b$O_$c$})$$

This equation applies if the reactants start in their standard states and the products return to the same conditions. The standard state pressure is $p^\circ=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ (note that most data published before 1982 used a standard pressure of one ‘standard atmosphere’, i.e. $p=1\ \mathrm{atm}=101325\ \mathrm{Pa}$). The definition of standard state makes no reference to fixed temperature; thus, the standard enthalpy of formation and the standard enthalpy of combustion remain functions of temperature. The most widely used reference temperature is $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$.

The following values for the standard molar enthalpy of formation at $p=10^5\ \mathrm{Pa}=100\ \mathrm{kPa}=1\ \mathrm{bar}$ and $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$ are taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align} \Delta_\mathrm fH^\circ(\text{propan-1-ol},\ \mathrm l)&=-302.6\ \mathrm{kJ\ mol^{-1}}\\ \Delta_\mathrm fH^\circ(\text{propan-2-ol},\ \mathrm l)&=-318.1\ \mathrm{kJ\ mol^{-1}}\\ \Delta_\mathrm fH^\circ(\ce{H2O, l})&=-285.8\ \mathrm{kJ\ mol^{-1}}\\ \Delta_\mathrm fH^\circ(\ce{CO2, g})&=-393.5\ \mathrm{kJ\ mol^{-1}} \end{align}$$

During your experiment, the products certainly did not completely return to the initial temperature. Anyway, for the difference in enthalpy of combustion of propan-1-ol and propan-2-ol, the exact state of the products is not relevant (provided that the conditions are reproduced) since the produced amounts of $\ce{CO2}$ and $\ce{H2O}$ are identical. The remaining difference is caused by the difference in enthalpy of formation of propan-1-ol and propan-2-ol.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.