Enthalpy cycles involving standard enthalpy changes of combustion.

Hess’s law

If a reaction can be carried out by two different pathways then the total enthalpy change for these two pathways will be the same provided that the starting and final conditions are the same for both pathways.

Using the data, calculate the standard enthalpy change of formation of butane.

We can carry out complete combustion of carbon and oxygen, forms carbon dioxide and water.

We can also combust butane, forming the same products.

Standard enthalpy change of combustion of carbon is -394, but there are 4 moles so it must be -1576 kJ

Standard enthalpy change of combustion of hydrogen is -286 kJ mol-1, we have 5 moles giving us value of -1430 kJ

Standard enthalpy change of combustion of butane is -2877 kJmol-1

Look at arrows , we are trying to work out Standard enthalpy change of combustion of butane, so we are going from carbon and hydrogen, to butane. Arrow on left is right, arrow on right hand side is pointing in wrong direction, si we must reverse the sign of butane.

Total Standard enthalpy change of combustion on left is -3006 kJ, on right hand side is +2877 kJ

Add numbers from both sides (-3006 + 2877) = -129 kJmol-1

a

a

a

a