# Which part of the reaction of potassium with water releases more energy?

When potassium is reacted with water, hydrogen is given off, which is normally ignited by the heat of the reaction and combusts. How much energy is given off by the burning hydrogen compared to the potassium reaction?

• I think this question contains every information necessary to calculate it yourself. If there are any open questions, you can edit your post and give the context here. – Martin - マーチン Oct 22 '15 at 10:00

## 1 Answer

$$\begin{array}{c} \ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)}\\ \ce{2H2(g) + O2(g) -> 2H2O(g)}\\ \end{array}$$

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:

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

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

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

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

$^{[1]}$ Formation Enthalpies

• I had badly misunderstood the question. My answer was truthful, it's just that the question had nothing to do with sodium!!! I deleted my answer accordingly and I consider this to be an excellent answer. – airhuff Feb 21 '17 at 8:35