Answer to Question #150545 in Physical Chemistry for Tumelo

Question

Use Hess’s law to determine the heat of combustion of 1 mole of frozen propanol, C3H8O(s). Propanol is a liquid under standard conditions. You have to write down and use the appropriate chemical reactions. [Given: ∆Hfo[CO2(g)] = -210 kJ; ∆Hfo[(C3H8O(l)] = -495 kJ; ∆Hfo[H2O(l)] = -315 kJ; ∆Hocrystallisation[(C3H8O(s)] = -95 kJ].

Answer 1

The process of combustion of frozen propanol can be divided into two smaller processes:

1) melting of frozen propanol to liquid propanol, $\Delta H_{melting} = -\Delta H_{crystallization}$ = 95 kJ;

2) combustion of liquid propanol:

C₃H₈O(l) + 4.5O₂(g) $\rightarrow$ 3CO₂(g) + 4H₂O(l).

According to Hess's law, the enthalpy change of this reaction will be:

$\Delta H = 4\Delta H_f°_{H_2O(l)} + 3\Delta H_f°_{CO_2(g)} - \Delta H_f°_{C_3H_8O(l)}$

$\Delta H = 4\cdot (-315)+ 3\cdot (-210) +495 = -2385$ kJ.

Finally, we add the enthalpy changes of these two processes to get the enthalpy change for the combustion of frozen propanol:

$\Delta H = 95-2385 = -2290$ kJ.

Answer: the heat of combustion of 1 mole of frozen propanol, C3H8O(s) is -2290 kJ.