Question

Question #6bdcc

Answer 1

See explanation.

Explanation:

In every neutralization reaction in aqueous solution, the fundamental chemical reaction is:

`H^(+)(aq)+OH^(-)(aq)->H_2O(l)`

Since this reaction involves bond formation only, energy is released, (i.e. an exothermic reaction). This energy change between the product, `H_2O`, and the reactants, `H^+(aq)` and `OH^(-)(aq)`, is called the enthalpy of neutralization and is given the symbol, `ΔH_("neut.")`

Since the fundamental reaction is the same for all neutralisation reactions, the value of `ΔH_("neut.")` should be the same for all acid-base reactions in solution.

However, this is not the case and is only true for a neutralisation reaction between a strong acid and a strong base , in other words, where both the acid and the base are completely dissociated prior to any reaction.

For weak acids or bases , where the acid or base is only partly dissociated, the `ΔH_("neut.")` has a smaller value than for the strong acid-strong base reaction, but is still exothermic .

The main reason for this is that some of the enthalpy of neutralization is used to dissociate the molecular acid or molecular base , which is present in solution before any reaction has taken place.

Note that bond breakage is always an endothermic process.

The energy change taking place is,

Heat produced by the reaction = Heat gained by the neutralized solution.

Here is video that discusses this topic experimentally:
Lab Experiment #17: Heat of Neutralization.