Which of these compounds does not liberate heat when water is added to it: $\ce{KNO_3}$, $\ce{NaOH}$, $\ce{CaO}$, $\ce{H_2SO_4}$, $\ce{Na}$.

I know from experience that dropping sodium tablets into water has explosive results, but I am clueless about the method to solve this problem. I know the answer is $\ce{KNO_3}$, but I don't know why. If someone can provide the method, I think I can do the problem.


1 Answer 1


It is important to note that if a molecule or formula unit is capable of undergoing hydrogen bonding ($\ce{NaOH}$, $\ce{H2SO4}$), then the formation of this hydrogen bond will be extremely exothermic.

Also, if a substance can react to form a substance that will undergo hydrogen bonding ($\ce{CaO + H2O -> Ca(OH)2}$), this species will subsequently form these hydrogen bonds and liberate heat.

Lastly, as you have stated you know from personal experience, the reaction between a alkali metal ($\ce{Na}$) and water is extremely exothermic— so much so that many of these metals will melt from the heat liberated.

Otherwise, there is no fail-proof method for predicting the enthalpy of solution for any given substance. Generally, the enthalpies of solution are determined experimentally, so unless you are expected to perform all these reactions in a calorimeter yourself, I would recommend something like this: this.

  • $\begingroup$ Ok. A couple of things I don't understand. The first two compounds you pointed pointed out - you say they are capable of undergoing hydrogen bonding, but I don't understand what you mean because they are already held together by hydrogen bonding. The OH groups in NaOH and H2SO4 have hydrogen bonding, so when you put them in water the OH bonds break, and wouldn't this be highly endothermic? Second, when you put CaO in water, how do you know that Ca(OH)2 forms? Isn't Ca(OH)2 soluble? $\endgroup$
    – math_lover
    Apr 28, 2015 at 22:10
  • $\begingroup$ Hydrogen bonding is a special type of electrostatic attraction that occurs between $\ce{H}$ bonded to $\ce{O}$, $\ce{N}$, or $\ce{F}$ and the unbonded electron pairs of a highly electronegative atom. In the case of dissolving $\ce{NaOH}$, the $\ce{OH-}$ that dissociates and the $\ce{H2O}$ participate in hydrogen bonding. In the case of dissolving $\ce{H2SO4}$, the $\ce{HSO4-}$ and the $\ce{H2O}$ participate in hydrogen bonding. These bonds are the strongest intermolecular bond, and are very stable. The formation of these bonds is therefore extremely exothermic. $\endgroup$
    – ringo
    Apr 29, 2015 at 15:28
  • $\begingroup$ Don't you have to consider the energy absorbed in breaking of the hydrogen bonds which hold NaOH and H2SO4 together, compared to the energy released in forming H bonds with water molecules? $\endgroup$
    – math_lover
    Apr 29, 2015 at 20:01

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