Question

The compound 1-pentanol, C5H12O, is a good fuel. It is a liquid at ordinary temperatures. When the liquid is burned, the reaction involved is

2 C5H12O(ℓ) + 15 O2(g)10 CO2(g) + 12 H2O(g)


The standard enthalpy of formation of liquid 1-pentanol at 25 °C is -351.6 kJ mol-1; other relevant enthalpy of formation values in kJ mol-1 are:
C5H12O(g) = -294.6 ; CO2(g) = -393.5 ; H2O(g) = -241.8

(a) Calculate the enthalpy change in the burning of 4.000 mol liquid 1-pentanol to form gaseous products at 25°C. State explicitly whether the reaction is endothermic or exothermic.

ΔH° = ________ kJ

Answer 1

ΔH_rxn = 10ΔH( CO2) + 12ΔH( H2O ) - 2ΔH(C5H12O) = 10*(-393.5) + 12*(-241.8) – 2*(-351.6) = -6133.4  kJ/mol

The reaction is exothermic.

ΔH° = 4*(-6133.4) = -24533.6 kJ