# Calculating enthalpy change by manipulating equations

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Calculate the standard enthalpy of formation of ethanol given that its standard enthalpy of combustion is -1367 kJ mol-1 and that the standard enthalpies of formation of carbon dioxide and water are -394 and -286 kJ mol-1 respectively.

The standard enthalpy of formation of ethanol is given by the following equation:

2C(s) + 0.5O2(g) + 3H2(g) ---> C2H5OH(l)       ΔH = ? kJ mol-1

The equations for the known reactions are:

 Equation 1 C(s) + O2(g) ---> CO2(g) ΔH1 = -394 kJ mol-1 Equation 2 H2(g) + 0.5O2(g) ---> H2O(l) ΔH2 = -286 kJ mol-1 Equation 3 C2H5OH(l) + 3O2(g) ---> 2CO2(g) + 3H2O(l) ΔH3 = - 1367 kJ mol-1

The calculation can be performed diagramatically but we will use the 'manipulating equation' method here

The aim is to alter the three known equations to match the quantities and position of reactants and products in the 'unknown' equation.

Double equation 1 will give 2 mol of C, treble equation 2 will give 3 mol of H2 and reverse equation 3 will put 1 mol of C2H5OH on the right hand side.

 Equation 1 2C(s) + 2O2(g) ---> 2CO2(g) ΔH1 = (2 x -394) = -788 kJ mol-1 Equation 2 3H2(g) + 1.5O2(g) ---> 3H2O(l) ΔH2 = (3 x -286) = -858 kJ mol-1 Equation 3 2CO2(g) + 3H2O(l) ---> C2H5OH(l) + 3O2(g) ΔH3 = +1367 kJ mol-1

The equations can then be added:

2C(s) + 2O2(g) + 3H2(g) + 1.5O2(g) + 2CO2(g) + 3H2O(l) ---> 2CO2(g) + 3H2O(l) + C2H5OH(l) + 3O2(g)

ΔH = (- 788) + (- 858) + 1367 kJ mol-1

and quantities adjusted by cancelling identical amounts on different sides of the equation. DeltaH values are then added:

2C(s) + 0.5O2(g) + 3H2(g) ---> C2H5OH(l)   ΔH = - 279 kJ mol-1