Question

The enthalpy of combustion (ΔH°c) of cis-1-ethyl-4-methylcyclohexane (C9H18) is -5875.10 kJ/mol. Using the appropriate information given...

The enthalpy of combustion (ΔH°c) of cis-1-ethyl-4-methylcyclohexane (C9H18) is -5875.10 kJ/mol. Using the appropriate information given below, calculate the enthalpy of formation (ΔH°f), in kJ/mol, for cis-1-ethyl-4-methylcyclohexane.

Report your answer to two decimal places.



ΔH°f (CO2 (g)) = -393.51 kJ/mol

ΔH°f (H2O (l)) = -285.83 kJ/mol

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Answer #1

The required reaction is 9C(s) + 9H2 (g)-------> C9H18(g),   ^Hf =?? Now, from the given information the reactions can be written as:-   C9H18(g) + 27/2 O2(g) -----> 9 CO2( g) + 9H2O(g), ^Hc = -5875.10KJ/ mole. H2(g) + 1/2 O2(g)------> H2O(g), ^ Hf= -285.3 KJ/ mole. C(s) + O2(g) --------> CO2(g), ^Hf= -393.5KJ/ mole.      multiplying equation -3 by 9 and eqn-2 by 9 and then adding the two eqn,we get.    9C(s)+ 9H2 (g)+27/2 O2(g) -----> 9CO2(g) +9H2O(g), ^Hf= -6114.06KJ/mole ---eqn-4.    Now,by subtracting eqn-1 from eqn-4,we get. 9C(s)+ 9H2(g) -----> C9H18(g), ^ Hf= -238.96KJ/ mole

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