Question

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

Answer 1

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

Explanation

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

H^o = H^o_f products - H^o_f reactants

H^o = [8 * H^o_f CO₂ (g) + 8 * H^o_f H₂O (l)] - [2 * H^o_f CH₃CH₂CH₂CHO (l) + 5 * H^o_f O₂]

Substituting the values

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

-4943.6 kJ = -3148 kJ - 2286.4 kJ - 2 * H^o_f CH₃CH₂CH₂CHO (l)

2 * H^o_f CH₃CH₂CH₂CHO (l) = -3148 kJ - 2286.4 kJ + 4943.6 kJ

2 * H^o_f CH₃CH₂CH₂CHO (l) = -490.8 kJ

H^o_f CH₃CH₂CH₂CHO (l) = (-490.8 kJ) / 2

H^o_f CH₃CH₂CH₂CHO (l) = -245.4 kJ/mol