2
$\begingroup$

I was doing a calorimetry experiment to calculate the change in enthalpy during neutralization reactions between metal carbonates and hydrochloric acid. I experimented with 5 metal carbonates: magnesium, zinc, copper(II), manganese(II), and calcium.

I noticed that when I reacted $\ce{MnCO3}$ with $\ce{HCl}$ and $\ce{CuCO3}$ with $\ce{HCl}$, the temperature of the solution increased unusually higher than magnesium, zinc, and calcium carbonates. After calculating the enthalpy change per mole of $\ce{MnO}$ and $\ce{CuCO3}$ when reacted with $\ce{HCl}$, it was unusually higher. For the enthalpy change in neutralization for magnesium, I obtained a value of $\pu{-45.0 kJ/mol}$ while I obtained a value of $\pu{-377 kJ/mol}$ for copper(II) carbonate and $\pu{-240 kJ/mol}$ for manganese(II) carbonate. I have double-checked my stoichiometry calculations and no mistakes were found. Could anyone please explain the large difference in change of enthalpy for copper(II) and manganese(II) carbonate?

$\endgroup$
6
  • $\begingroup$ Part of it will be differences of cation hydration ( following naked ion radius differences). Part of it will be formation of copper chlorocomplexes. Part of it may be differences in ionic lattice of the solid. But I am not sure if that explains such huge difference. $\endgroup$
    – Poutnik
    Aug 12, 2022 at 11:10
  • $\begingroup$ Just guessing if possibly stronger covalent character for Mn and Cu may play the role. BTW CuCO3 is very hard to obtain, more frequent are basic forms CuCO3 . 1(2)Cu(OH)2. $\endgroup$
    – Poutnik
    Aug 12, 2022 at 12:55
  • $\begingroup$ @Poutnik Is there any method to calculate the theoretical value for the enthalpy of neutralization for copper and manganese carbonates? $\endgroup$ Aug 14, 2022 at 12:30
  • $\begingroup$ Well there is a method to calculate theoretical value of mercury melting point. Chemists have waited like 50 years until there were available fast enough computers. Experimental determination is for both cases much easier. BTW, neutralisation enthalpy is just a part of the whole enthalpy change chain. $\endgroup$
    – Poutnik
    Aug 14, 2022 at 12:46
  • $\begingroup$ @Poutnik Thank you so much for your kind replies! Have a great day sir $\endgroup$ Aug 14, 2022 at 13:11

1 Answer 1

0
$\begingroup$

It is possible to calculate the enthalpy of reaction from enthalpy of formation of the reactants and products. Tables exist with the substances in the various states.

One would expect that as ionic radius gets larger (e.g. moving down a group of metals) the strength of the ionic bond would become weaker resulting in increasing enthalpy of reaction (less energy required to break the solid lattice). This trend is seen experimentally for metal oxides. Metal carbonates, however, display the opposite trend - as ionic radius of the metal increases, enthalpy of reaction decreases.

Part of the reaction of a metal carbonate with acid is the liberation of CO2 gas meaning one of the carbon-oxygen bonds from the carbonate ion has to be broken. Smaller metal ions (higher core charge) draw an oxygen and electrons from the carbonate ion towards it more strongly, allowing the remaining CO2 to be more easily liberated. Larger metal ions pull less on the carbonate (less polarized), meaning more energy is required to break the carbon - oxygen bond to liberate a molecule of carbon dioxide. This suggests that smaller metal ions will result in a greater (more exothermic) enthalpy of reaction.

Theoretically calculated enthalpies support this. As ionic radius increases, enthalpy of reaction of metal carbonate with hydrochloric acid decreases (becomes more positive). For reaction of the following metal carbonates with aqueous hydrochloric acid, calculated ΔHreaction values are: MgCO3 : -50.1 kJ/mol; CaCO3 : -15.3 kJ/mol; BaCO3 : 1.1 kJ/mol.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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