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\u00b0_{H_2O(l)} + 3\\Delta H_f\u00b0_{CO_2(g)} - \\Delta H_f\u00b0_{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.
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