Example one

The enthalpy changes for two reactions are shown below:

\(CO(g)+\frac{1}{2}O_{2}(g)\rightarrow CO_{2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H_{1}=-283\,\,kJ\,mol^{-1}\)

\(Cu(s)+\frac{1}{2}O_{2}(g)\rightarrow CuO(s)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H_{2}=-155\,\,kJ\,mol^{-1}\)

Using this data, calculate the enthalpy change for the following reaction:

\(CuO(s)+CO(g)\rightarrow Cu(s)+CO_{2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H=\, ?\)

The target equation (which we will denote (\(\Delta H_{T}\)) can be made by adding together equation one and the reverse of equation two.

Reversing equation two gives:

\(CuO(s)\rightarrow Cu(s)+\frac{1}{2}O_{2}(g)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta H=\,+155\,kJ\,mol^{-1}\)

\(\Delta H_{T}=\Delta H_{1}+(-\Delta H_{2})\)

\(\Delta H_{T}= (-283)-(-155)\)

\(\Delta H_{T}= -283 +155\)

\(\Delta H_{T}= -128\,\,kJ\,\,mol^{-1}\)

Revise: Chemical energyExample one