State and prove Hess's law of constant heat summation.

(N/A) Hess's law states that if a reaction takes place in several steps,then its standard reaction enthalpy is the sum of the standard enthalpies of the intermediate reactions into which the overall reaction may be divided at the same temperature.
Example: Consider the formation of carbon monoxide:
$C_{(s)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{(g)} ; \Delta H^{\ominus} = ?$
Since this reaction cannot be measured directly due to the formation of $CO_2$,we use the following known reactions:
$(i) C_{(s)} + O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta_{r} H^{\ominus} = -393.5 \ kJ \ mol^{-1}$
$(ii) CO_{(g)} + \frac{1}{2} O_{2(g)} \longrightarrow CO_{2(g)} ; \Delta_{r} H^{\ominus} = -283.0 \ kJ \ mol^{-1}$
To obtain the target reaction,we reverse equation $(ii)$:
$(iii) CO_{2(g)} \longrightarrow CO_{(g)} + \frac{1}{2} O_{2(g)} ; \Delta_{r} H^{\ominus} = +283.0 \ kJ \ mol^{-1}$
Adding equations $(i)$ and $(iii)$:
$C_{(s)} + \frac{1}{2} O_{2(g)}$ $\longrightarrow CO_{(g)} ; \Delta_{r} H^{\ominus} = (-393.5) + (283.0) = -110.5 \ kJ \ mol^{-1}$
In general,for a reaction $A \longrightarrow B$,if the enthalpy change is $\Delta_{r} H$ along one route and $\Delta_{r} H_1, \Delta_{r} H_2, \dots$ along another route,then $\Delta_{r} H = \sum \Delta_{r} H_i$.