Question

The compound 1-propanol, C₃H₈O, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is

2 C₃H₈O(ℓ) + 9 O₂(g)6 CO₂(g) + 8 H₂O(g)  

The standard enthalpy of formation of liquid 1-propanol at 25 °C is -302.6 kJ mol^-1; other relevant enthalpy of formation values in kJ mol^-1 are:
  C₃H₈O(g) = -255.1 ; CO₂(g) = -393.5 ; H₂O(g) = -241.8

(a) Calculate the enthalpy change in the burning of 3.000 mol liquid 1-propanol to form gaseous products at 25°C. State explicitly whether the reaction is endothermic or exothermic.

ΔH° =  kJ

exo or endothermic

(b) Would more or less heat be evolved if gaseous 1-propanol were burned under the same conditions?

more or less

What is the standard enthalpy change for vaporizing 3.000 mol C₃H₈O() at 25°C?

ΔH° =  kJ

Calculate the enthalpy change in the burning of 3.000 mol gaseous 1-propanol to form gaseous products at 25°C.

ΔH° =  kJ

Answer 1

(a)

3.000 moles of liquid 1-propanol are burned to for gaseous products.

ΔΗ. = ΣΔΗ? ( products ) -ΣΔΗ? ( reactants )

ΔΗρη = 6ΔΗ (CO2(g))+8ΔΗ (H2O(g))-2ΔΗ (C3Hgo(1))-9ΔΗ (O2(g))

ΔΗ, = 67-393.5) + 8(-241.8) – 2 (-302.6) – 9 (0)

ΔΗ =-2361 – 1934.4 + 605.2

ΔΗ =-3690.2 kJ/mol

For three moles

AH = -3690.2 kJ/mol x 3.000

ᎪᎻ =-11070.6 kᎫ

Negative sign indicates exothermic reaction.

(b)

More heat will be evolved if gaseous 1-propanol were burned under same conditions.

This is because some heat is absorbed to vaporise liquid 1-propanol.

3.000 moles of liquid 1-propanol are vaporised

C3H2O(l) → C3H2O(9)

ΔΗ. = ΣΔΗ? ( products ) -ΣΔΗ? ( reactants )

ΔΗ» = ΔH (CHOg)) – ΔΗ? (CHNO(1))

ΔΗ, = -255.1 – (-302.6)

AH = 47.5 kJ/mol

For three moles

AH = 47.5 kJ/mol x 3.000

AH m = 142.5 kJ

3.000 moles of gaseous 1-propanol are burned to for gaseous products.

ΔΗ. = ΣΔΗ? ( products ) -ΣΔΗ? ( reactants )

ΔΗΜ = 6ΔΗ (CO2(g))+8ΔΗ (H2O(g))-2ΔΗ (C3H8O(g))-9ΔΗ (O2(g)

ΔΗ, = 67-393.5) + 8(-241.8) – 2 (-255.1) – 9 (0)

ΔΗ =-2361 – 1934.4 + 510.2

ΔΗ =-3785.2 kJ/mol

For three moles

AH = -3785.2 kJ/mol x 3.000

ᎪᎻ =-11355.6 kᎫ