Question

2. Calculate the standard enthalpy of formation of gaseous butane given that the heat of formation of water and gaseous CO2 are 285kJ and 393kJ respectively and heat of combustion of gaseous butane is 2874kJmol-1 all is exothermic.

Answer 1

balanced chemical equation for combustion is

2C₄H₁₀ + 13O₂ = 8CO₂ + 10 H₂O

Therefore  for formation of butane

8 CO₂ + 10 H₂O = 2C₄H₁₀ + 13O₂  is needed

enthalpy of formation of butane is = enthalpy of combustion - enthalpy of formation CO₂ + H₂O

it is     =-2874-( -10×285- 8×393 )IT is for two moles

for 2 mole=-2874-(- 5994)=3120

for 1 mole = 3120/2=1560