Question

A 200-L home heater tank contains butane gas (C4H10) at 26.0 °C. When some of the butane
is removed and burned in excess air, the pressure in the tank drops from 2.34 atm to 1.10 atm.
The energy generated in the combustion is then used to raise the temperature of 132.5 L of
water from 19.0°C to 71.3 °C.
Useful Information:
ΔH°f (CO2 (g)) = –393.5 kJ/mol ΔH°f (H2O (l)) = –285.8 kJ/mol Density of water = 1 g/mL

→Calculate the enthalpy of formation (ΔH°f) of butane gas.

Answer 1

enthalpy change of water = mass of water* specific heat of water* temperature difference =

Mass of water= density* Volume= 1000*132.5=132.5 gm*1000

enthalpy change = 132.5*1*4.18*(71.3-19) joules=28966.4*1000 Joules

since n= PV/RT

molea of butane in 200L at 26 deg.c temperature and 2.34 atm pressure = 2.34*200/(0.0821*(26+273.15)=19.05 moles

for constant volume, number of moles at 1.1 atm,

V= nRT/P and hence n1/P1= n2/P2

19.05/2.34= moles/1.1

moles at 1.11 atm = 19.05*1.1/2.34=8.95

Moles remaining at 1.1 atm = 8.95

Moles of butane combusted= moles initially- moles at 1.1 atm =19.05-8.95=10.1 moles

10.1 moles gives a change of enthalpy of water to 28966*1000/10.1 = 2868 J/mole

enthalpy of combustion of 1 mole of butane= 2868 Kj/mole

The combustion reaction is C4H10+6.5O2---> 4CO2 + 5H2O

Enthakpy change= enthalpy change of products- enthalpy change of reactants

enthalpy of formation of oxygen is =0

-2868 = 4*(-393.5)+5*(-285.8)- { enthalpy of fomration of C4H10)=

4 and 5 are coefficients of CO2 and water respectively.

Enthalpy of formation of C4H10 = -135 Kj/mole