In chemical transformations energy can neither be created nor destroyed (the first law of thermodynamics).

  • Hess’s law states that the total enthalpy change for a reaction is independent of the route taken. It is a special case of the law of conservation of energy.

    eg:

    $\scriptstyle\rm \sum \Delta H^o_{form} (reactants)~ +~ \Delta H_{reaction} ~= ~\rm\sum \Delta H^o_{form} (products)$

Worked Example 

Determine the enthalpy change for reactions.

$\rm PbO_2(s) + C(s) \rightarrow CO_2(g) + Pb(s)$

From the data:

Solution

Method 1

From the energy cycle:

$\rm\Delta H_3 = -\Delta H_1 + \Delta H_2$
$= +248  + -394$
$\rm\Delta H_3 = -146~kJ~mol^{-1}$

Method 2

Write the equation with the enthalpy of formation data.

$\rm \underset{-248}{PbO_2(s)} + \underset{0}{C(s)} \rightarrow \underset{-394}{CO_2(g)} + \underset{0}{Pb(s)}$

$\scriptstyle \rm\Delta H_{reaction} = \sum \Delta H^o_{form} (products)~-~ \sum\Delta H^o_{form} \overset{\Delta H_{formation}}{(reactants)}$

$\displaystyle \rm\Delta H_{reaction} = -394 - -248$ $ = \bf-146~kJ~mol^{-1}$