1. The enthalpy of combustion (H° c ) of 1-tert-butoxybutane (C 8 H 18 O) is -53

Question

1. The enthalpy of combustion (H°c) of 1-tert-butoxybutane (C8H18O) is -5317.30 kJ/mol. Using the appropriate information given below, calculate the enthalpy of formation (H°f), in kJ/mol, for 1-tert-butoxybutane.

Report your answer to two decimal places.

H°f (CO2 (g)) = -393.51 kJ/mol
H°f (H2O (l)) = -285.83 kJ/mol

2. Determine the mass (in g) of 1-tert-butoxybutane produced, if H° was determined to be -270.18 kJ during an experiment in which 1-tert-butoxybutane was formed.

Report your answer to three significant figures.

Explanation / Answer

Q1.

Hcombustion = Hproducts - Hreactants

C8H18O +O2 = CO2 + H2O

C8H18O + 12 O2 = 8 CO2 + 9 H2O

so

Hcomb = 8*CO2 + 9*H2O - (C8H18O + 12*O2)

-5317.30 = 8*-393.51+ 9*-285.83- (C8H18O + 12*0)

C8H18O = 8*-393.51+ 9*-285.83+5317.30 = - 403.25 kJ/mol

Q2.

for

Q = -270.18

apply

Q = n*HRxn

n = Q/HRxn = -270.18/5317.30 = 0.05081 mol

mass = mol*MW = 0.05081*130.2279 = 6.62 g

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.