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Chemistry
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Write a balanced equation for the reaction of methyl octane, CH3C8H17, with oxygen, O2(g), forming the products carbon dioxide, CO2(g), and water, H2O(l). Now, if the standard heat of formation of methyl octane is -215.5 kJ/mol, what is the heat of reaction, the heat of reaction, in kJ, for the combustion reaction of methyl octane? The standard heats of formation of CO2,(g) and H2O,(l) are -394 kJ/mol and -286 kJ/mol respectively

Apr 21st, 2015

Balance the equation first by C and H atoms, then by O atoms. The balanced equation is

CH3C8H17 + 14O2 (g) → 9CO2 (g) + 10H2O (l)

ΔH0(CH3C8H17) = –215.5 kJ/mol

ΔH0(CO2 (g)) = –394 kJ/mol

ΔH0(H2O (l)) = –286 kJ/mol

ΔH0(O2 (g)) = 0 kJ/mol (since it is the most stable state of an element at normal conditions)

ΔH0rxn = 9ΔH0(CO2 (g)) + 10ΔH0(H2O (l)) – ΔH0(CH3C8H17) – 14 ΔH0(O2 (g)) =

9·(–394) + 4·(–286) – (–215.5) – 14·0 = –4474.5 kJ/mol = –4470 kJ/mol 


Apr 21st, 2015

Let's see. The equation is balanced (9 atoms of C on the left and on the right, 20 atoms of H on either side, and 28 atoms of O on either side). 

The heat of the reaction is found by the formula (zero means that the corresponding energy is for 1 mole):

ΔH0rxn = sum of stoichiometric coefficients times the standard heat of formation for the products - similar sum for the reactants

ΔH0rxn = 9ΔH0(CO2 (g)) + 10ΔH0(H2O (l)) – ΔH0(CH3C8H17) – 14 ΔH0(O2 (g)) =

9·(–394) + 4·(–286) – (–215.5) – 14·0 = –4474.5 kJ/mol = –4470 kJ/mol 

(or, in scientific notation - 4.47x10^3 kJ/mol).  

Apr 21st, 2015

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