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
(a)
3.000 moles of liquid 1-propanol are burned to for gaseous products.
For three moles
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
For three moles
3.000 moles of gaseous 1-propanol are burned to for gaseous products.
For three moles
The compound 1-propanol, C3H8O, is a good fuel. It is a liquid at ordinary temperatures. When...
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