2
$\begingroup$

Compare the heat liberated per mole for the following reactions:

$\ce{MgO + H2O -> Mg(OH)2}$

$\ce{CaO + H2O -> Ca(OH)2}$

$\ce{SrO + H2O -> Sr(OH)2}$

$\ce{BaO + H2O -> Ba(OH)2}$

Data for lattice enthalpy; $\ce{MO (s)->M^2+ (g) + O^2- (g)}$

$^1$Metal Oxide Lattice Enthalpy (kJ/mol)
$\ce{MgO}$ +3923
$\ce{CaO}$ +3517
$\ce{SrO}$ +3312
$\ce{BaO}$ +3120

Data for hydration enthalpy; $\ce{M^2+ (g)->M^2+ (aq)}$

$^1$Metal Ion Hydration Enthalpy (kJ/mol)
$\ce{Mg^2+}$ -1921
$\ce{Ca^2+}$ -1577
$\ce{Sr^2+}$ -1443
$\ce{Ba^2+}$ -1305

Data for enthalpy of formation; $\ce{M (s) + O2 (g) + H2 (g)->M(OH)2 (s)}$

$^2$Metal Hydroxide Formation Enthalpy (kJ/mol)
$\ce{Mg(OH)2}$ -925
$\ce{Ca(OH)2}$ -985
$\ce{Sr(OH)2}$ -959
$\ce{Ba(OH)2}$ -945

Energy required to break the lattice is positive, enthalpy of hydration is negative and enthalpy of formation is also negative. Hence, the net energy will sum up to a positive value. But these reactions are highly exothermic. Thus, there must be other energies to be considered.

Apart from lattice enthalpy, hydration enthalpy and enthalpy of formation what other factors to be considered?

P.S.: The highest value is for $\ce{Ca(OH)2}$.

References:

  1. Concise Inorganic Chemistry- 4E by J. D. Lee & Sudarshan Guha (ISBN 9788126566495)

  2. Table 6.3, Lange's Handbook of Chemistry- 15E

$\endgroup$
5
  • 2
    $\begingroup$ I miss the enthalpy of $\ce{O^2-}$ hydration and $\ce{M^2+(aq) + 2 OH-(aq)->M(OH)2(s)}$ precipitation, if the solid state is the final state. M(OH)2 formation enthalpy is for $\ce{M(s) + O2(g) + H2(g) -> M(OH)2(s)}$. $\endgroup$
    – Poutnik
    Jun 3 at 12:12
  • 1
    $\begingroup$ @Poutnik Now I think, it's very difficult to calculate by this method. Is there any short way? $\endgroup$
    – Apurvium
    Jun 3 at 12:28
  • 1
    $\begingroup$ Rather, it is difficult to obtain all needed input data. Shorter way is obtaining the results anywhere, or measuring it. BTW why the 1/2? 1/2 of O2 or H2 would not give 2 O or 2 H for M(OH)2. $\endgroup$
    – Poutnik
    Jun 3 at 12:53
  • 1
    $\begingroup$ I meant, by applying logic rather than calculating from data. My bad for 1/2. $\endgroup$
    – Apurvium
    Jun 4 at 4:11
  • 1
    $\begingroup$ As Poutnic said, you gotta account for O hydration energy too and its inevitable conversion to OH ion as aswell and its energy too. Frankly, if you need to measure any of these, don't get into theory. Just slake some of each in a calorimeter. $\endgroup$ Jun 7 at 16:10

1 Answer 1

4
+50
$\begingroup$

Some of the information pertains to the solid hydroxide and some pertains to the hydroxide in solution. As magnesium and calcium hydroxide actually have only limited solubility, I will assume the former applies.

You have enthalpies of formation for the solid hydroxides. From the Wikipedia articles for the respective compounds we can also get these enthalpies of formation, all in kilohoules per mole at 25°C:

$\ce{H2O}: -285$ (liquid)

$\ce{MgO}: -602$

$\ce{CaO}: -635$

$\ce{SrO}: -592$

$\ce{BaO}: -582$*

*There is surprising disagreeement over the heat of firmation of barium oxide, for instance NIST gives $-548$ kJ/mol.

From these values we may render all the hydroxide formations exothermic:

$\ce{MO + H2O -> M(OH)2(s)}$

$\ce{Mg}: \Delta H = -38$ kJ/mol

$\ce{Ca}: \Delta H = -65$ kJ/mol

$\ce{Sr}: \Delta H = -82$ kJ/mol

$\ce{Ba}: \Delta H = -78$ kJ/mol ($-112$ using NIST value for $\ce{BaO}$)

We see that on a molar basis, strontium or barium evolves the most heat, depending on which source is used for the barium data. But if we divide each of these results by the molecular weight of the product, we find that because of its lighter weight and still relatively high heat release calcium has the strongest heat effect per unit mass.

$\endgroup$
3
  • 1
    $\begingroup$ By using my second reference, the heat of formation of BaO = -548 kJ/mol from which the enthalpy of dissolution for BaO = -111 kJ/mol. So, the required enthalpy per mol is increasing from MgO<CaO<SrO<BaO. Can we explain this due to the increasing basic character from MgO to BaO? The order of liberated heat per gram is: BaO<SrO<MgO<CaO $\endgroup$
    – Apurvium
    Jun 12 at 7:16
  • 1
    $\begingroup$ That seems odd with barium oxide, perhaps another question is why there is such a difference. I tried to be consistent by using the current Wikipedia data. $\endgroup$ Jun 12 at 9:19
  • 1
    $\begingroup$ Except BaO, wiki data for other oxides matches with Lange's. $\endgroup$
    – Apurvium
    Jun 12 at 9:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.