Question

# What is the standard enthalpy of formation of     C H 3 C H 2 C H...

What is the standard enthalpy of formation of

C H 3 C H 2 C H 2 CHO( l )? 2C H 3 C H 2 C H 2 CHO( l )+5 O 2 ( g )→8 H 2 O( l )+8C O 2 ( g );

ΔH°=–4943.6 kJ

Substance ΔH ° f ( kJ/mol ) C O 2 ( g ) –393.5 H 2 O( l ) –285.8

Select one:

a.
–245.4 kJ/mol

b.
+245.4 kJ/mol

c.
–1792.5 kJ/mol

d.
–3151.1 kJ/mol

e.
+3151.1 kJ/mol

Correct answer is : (a) -245.4 kJ/mol

Explanation

The combustion reaction is : 2 CH3CH2CH2CHO (l) + 5 O2 8 H2O (l) + 8 CO2 (g)

Ho = Hof products - Hof reactants

Ho = [8 * Hof CO2 (g) + 8 * Hof H2O (l)] - [2 * Hof CH3CH2CH2CHO (l) + 5 * Hof O2]

Substituting the values

-4943.6 kJ = [8 * (-393.5 kJ/mol) + 8 * (-285.8 kJ/mol)] - [2 * Hof CH3CH2CH2CHO (l) + 5 * (0)]

-4943.6 kJ = -3148 kJ - 2286.4 kJ - 2 * Hof CH3CH2CH2CHO (l)

2 * Hof CH3CH2CH2CHO (l) = -3148 kJ - 2286.4 kJ + 4943.6 kJ

2 * Hof CH3CH2CH2CHO (l) = -490.8 kJ

Hof CH3CH2CH2CHO (l) = (-490.8 kJ) / 2

Hof CH3CH2CH2CHO (l) = -245.4 kJ/mol

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