Which part of the reaction of potassium with water releases more energy?
$$\begin{array}{c}
\ce{2K(s) +2H2O (l)-&gt;2KOH(aq) +H2(g)}\\
\ce{2H2(g) + O2(g) -&gt; 2H2O(g)}\\
\end{array}$$

We have the two equations above, and we want to know which releases more energy. The answer is relatively simple to find: it's the standard enthalpy of formation of the products minus the standard enthalpy of formation of the reactants. This works because enthalpy is a state function and a systems enthalpy change independent of the path taken from reactants to products. So the way we do this is by imagining that we take our products, break them into their constituent elements (which have a $\Delta H_\mathrm{f}=0$ by definition), and convert those into the products. So:

$$(-482.37\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}-(-285.83\ \mathrm{kJ \ mol^{-1}})\cdot 2\ \mathrm{\ mol}=\frac{-393.08\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{K}}=-196.54\ \mathrm{kJ \ mol^{-1}}$$

$$(-241.82\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}=\frac{−483.64\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{H2}}=-241.82\ \mathrm{kJ \ mol^{-1}}$$

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)-&gt;2KOH(aq) +H2(g)} &amp; \Delta H=-196.54\ \mathrm{kJ \ mol^{-1}}\\
\ce{2H2(g) + O2(g) -&gt; 2H2O(g)} &amp; \Delta H=-241.82\ \mathrm{kJ \ mol^{-1}}^{[1]}\\
\end{array}$$

So the combustion of hydrogen gas produces more energy per mole than the reaction of potassium with water.




$^{[1]}$ Formation Enthalpies