Link to exam

Which metal reacts most vigorously with water?

(A) Ca (B) K (C) Mg (D) Na

The given answer is B and the reasoning is that is is the most electropositive since it the furthest left and down according to periodic trends.

However, isn't $\ce{K}$ the conjugate acid from a strong base -- $\ce{KOH}$? So isn't the conjugate base of a strong acid quite weak? So why would $\ce{K}$ react strongly?

  • $\begingroup$ $\ce{K}$, not $\ce{K^+}$. $\endgroup$
    – Shahar
    May 20, 2014 at 1:41

4 Answers 4


I interpret "reacts most vigorously" to mean which metal reacts the fastest. This is a question of kinetics, not thermodynamics (e.g. the answer is not based on the enthalpy of the reaction). Kinetic rates are determined by the height of the energy barrier that needs to be surmounted, we need to determine and compare the activation energies in this process for the various metals. The activation energy in this process is determined by the following two steps: $$\ce{M(s) -> M(g)}$$ $$\ce{M(g) -> M^+(g) + e^-}$$ $$\ce{or}$$ $$\ce{M(g) -> M^{+2}(g) + 2 e^-}$$ depending on whether we are analyzing a group I or group II metal. The following link provides a nice analysis and tabulation of the energies required for these two steps with the group I metals

Group I link

and this link does the same for the Group II metals

Group II link

I'll summarize their findings

Metal ... Activation Energy (kJ/mol)

K ... +508

Na ... +603

Mg ... +2200

Ca ... +1950

CONCLUSION: Potassium has the lowest activation energy and should react the fastest.

  • 1
    $\begingroup$ I believe that "most vigorously" would mean the explosive heat that is given off, e.g. when throwing an ingot of potassium into water. Seems thermodynamic to me. I prefer Martin's answer, which speaks also about the hydroxide, and points out that solvation energy is important. There is a significant effect of bulk transport, also -- the metal-hydroxide crust has to be dissolved into the water to allow for fresh metal to react. $\endgroup$
    – Eric Brown
    May 30, 2014 at 21:58
  • 1
    $\begingroup$ Another thing is that the full ionization process is never fully realized in solvent, so these values might be supportive of the observation, rather than explanatory. $\endgroup$
    – Eric Brown
    May 30, 2014 at 22:48

Please read ron's post before continuing here (too long for a comment). Its reasoning is very good and makes a lot of sense. The activation barrier is the defining component for how vigorously components react.

His presented reaction may cover the initial activation barrier, but subsequent reactions also have to be taken into account.

The overall reaction that will occur in the case of sodium can be formulated like this: \begin{aligned} \ce{2Na(s) + 2H2O(l) &-> 2Na+(aq) + H2(g) ^ + 2{}^{-}OH(aq)}& \Delta H &= -184~\mathrm{kJ/ mol} \end{aligned}

So the start of the reaction has to be considered as the atomisation energy of the metal, after that the ionisation energy, both of these reactions need energy. Energy is released to the system by the hydration of the ion. For sodium this boils down to \begin{aligned} \ce{Na(s) &-> Na(g)} & \Delta H &= +109~\mathrm{kJ/ mol}\\ \ce{Na(g) &-> Na+(g) + e-} & \Delta H &= +494~\mathrm{kJ/ mol}\\ \ce{Na+(g) &-> Na+(aq)} & \Delta H &= -406~\mathrm{kJ/ mol}\\[2ex]\hline \ce{Na(s) &->[\ce{H2O}] Na+(aq)} & \Delta H &= +197~\mathrm{kJ/ mol}.\\\hline\hline \end{aligned}

So why would this reaction happen, when there is no energy released? The cause of this are the secondary and tertiary reactions, that then again fuel the initial reactions. So the key question here is, what happens to the electron? \begin{aligned} \ce{2H2O + 2e- &<=> H2 ^ + 2{}^{-}OH}\\ \end{aligned} While I cannot find a value for this reaction, however I strongly suspect it to be negative. It more certainly gives us a hint on what to consider else, i.e. the hydration of hydroxide ions. (source) \begin{aligned} \ce{{}^{-}OH &-> {}^{-}OH(aq)} & \Delta H &= -460~\mathrm{kJ/mol} \end{aligned} So this releases a lot of energy to the system, helping to atomize and ionize the sodium.

A tertiary reaction also has to be considered. Because of the produced heat, hydrogen will immediately burn when coming in contact with air. (wikipedia) \begin{aligned} \ce{H2(g) + 1/2 O2(g) &-> H2O(g)} & \Delta H &= -242~\mathrm{kJ/mol}\\ \ce{2H2(g) + O2(g) &-> 2H2O(g)} & \Delta H &= -484~\mathrm{kJ/mol} \end{aligned}

So all these reactions successfully overcome the activation barrier once the reaction started. This also then has the effect of multiplying the initial reaction, causing a chain reaction and hence reacting even more furiously.

Watch Sodium burn in water and at one point you see the hydrogen ignite and how vigorous this reaction is.

The same applies to all other metals, but as ron stated the kinetics depend on the activation barrier.

  • $\begingroup$ The enthalpy of formation of $1\ \mathrm{mol}$ $\ce{H2O(g)}$ is $\Delta H^\circ=-242\ \mathrm{kJ}$; thus, the molar enthalpy of reaction for $\ce{H2(g) + 1/2O2(g) -> H2O(g)}$ is $\Delta H^\circ_\mathrm m = -242\ \mathrm{kJ/mol}$, whereas the molar enthalpy of reaction for $\ce{2H2(g) + O2(g) -> 2H2O(g)}$ is $\Delta H^\circ_\mathrm m = -484\ \mathrm{kJ/mol}$. $\endgroup$
    – user7951
    Oct 22, 2015 at 19:00
  • $\begingroup$ @Loong Thanks for pointing that out. I sometimes have issues with all that mol stuff.... :/ $\endgroup$ Oct 23, 2015 at 5:59

The other answers explain things in terms of the activation energy barrier, but they don't explain why the activation energy barrier is lowest for K. I will attempt to do that here.

When a metal reacts with water, several things must happen (not necessarily in this order):

  1. One or more electrons have to be removed from the metal
  2. One or more $O-H$ bonds have to break on water molecules (one per electron)
  3. The electron(s) must be "picked up" by the $OH$ to become $OH^-$

As @ron said, the "vigorousness" of this process depends on the kinetics of the reaction, which are in turn determined by the activation energy of the rate-limiting step.

In the links that he provided (1,2), they show that the activation energy is lowest for K, which means it would react most vigorously. (note - these are not real activation energies - the atoms are not actually atomized to the gas phase, and the electrons are not removed under vacuum to infinite distance, but they still give us a good relative estimate)

However, knowing that the kinetics of the reaction depends on the activation energy won't help you if you don't happen to have a chart of activation energies handy. On an exam, you certainly won't, and it is good to develop an intuition for these sorts of things anyhow. And so, you might ask:

Why is the activation energy lower?

Looking at the links, you can see that the activation energies are dominated by the ionization energy. In other words, when it takes more energy to remove electrons, the reaction slows down.

Why should K have the lowest ionization energy? First we need to look at the equation that governs the potential energy between charges:


Here E is electrostatic potential energy, $\kappa$ is Coulomb's constant, and $Q_1$ and $Q_2$ are charges.

You can see from this that if the charges have the opposite sign, then energy will be become more negative as they move closer together (inversely proportional to distance). As either of the charges increase, the energy will become more negative as well (directly proportional to charge). In other words, big charges are harder to pull apart, and charges that are close together are harder to pull apart than charges that are far apart. As a rough analogy, imagine pulling apart two magnets - it is more difficult when they are close together.

Now compare K to the other group I element (Na). K is further down the column. Since atomic radius increases with the row number (due to the fact that orbital radius increases with principle quantum number n), the valence electron on K is further from the nucleus (on average) than it is on Na. This means that it is easier to remove a valence electron from K. As an analogy, imagine trying to move a rock up a hill. You have to do less work (put in less energy) if the rock is already halfway up the hill. Since the electrons are further from the nucleus in K, they are already "further up the hill."

What about group II elements? As you progress from left to right, the atomic radius decreases - this is because the effective nuclear charge seen by the valence electrons increases. The combination of increased charge and decreased distance leads to a steeper "hill" that must be overcome. On top of that, now we have to remove two electrons for the reaction to work, which is roughly twice as difficult (technically we don't have to remove both, but the overall reaction thermodynamics won't work out if we don't). As a result, group I elements will always react more vigorously with water than group II will.

I think that it helps to imagine reactions as sort of a "fight" between elements over electrons. Who will win the fight and how it will go is determined by the number of protons in the nucleus and the electron orbital configuration. This approach is useful because it applies to more advanced chemistry as well.

For some demonstrations of alkali metal reactions with water, check out this video:

Alkali Metals

  • $\begingroup$ I'm inclined to believe that the liberation of heat is a thermodynamic effect. While there may be vestiges of the thermodynamics in the kinetic barrier, I would like to point out that the discrete ionization is probably not observed, rather being conflated with transfer to the hydroxide, as well as the enormous solvation energy. $\endgroup$
    – Eric Brown
    May 30, 2014 at 22:53
  • $\begingroup$ @Eric - I thought that too, until I read the references that ron posted. If you check those, you will see that potassium's net reaction enthalpy with water is less exothermic than either sodium or magnesium, both of which react much less "vigorously." $\endgroup$
    – thomij
    May 31, 2014 at 0:07

I might be wrong here, but I believe that the question asks about the metals in their atomic (non ionic) form.

In order for a $\ce{KOH}$ bond to form, the K must be ionized (since the $\ce{KOH}$ bond is ionic). This means that when $\ce{KOH}$ dissolves in water it'll form the $\ce{K}^+$ electrolyte, which is not the same thing as a pure $\ce{K}$ atom with its 19th electron.

A $\ce{K}$ atom, then, since it's the most electropositive of that list, is the one to react "more vigorously" (which I guess it refers to more speed/energy) with water (a polar molecule).

I hope this helped you. I'm sorry if I'm wrong!


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