Enthelpy of Neutralisation Strong acid + Strong base

[isaaclimdc]

Why is the enthalpy of neutralisation of a strong acid mixed with a strong base always the same, no matter what acid and base combination you use, provided they are 'strong' acids and bases?

Am I right to say that this value is accepted as -57.35 kJmol-1?

Thanks for any help,
Isaac.

 

[Kushal]

yeah...

neutralisation is more exactly H+ from acid reacting with OH- of alkali.

all strong acids and alkalis ionise completely in solution. so there is the same amount of H+ and OH- released by strong acids and alkalis.(i'm talking about monobasic acids and alkalis)

 

[isaaclimdc]

yeah...

neutralisation is more exactly H+ from acid reacting with OH- of alkali.

all strong acids and alkalis ionise completely in solution. so there is the same amount of H+ and OH- released by strong acids and alkalis.(i'm talking about monobasic acids and alkalis)

But why is it then, that H2SO4 and NaOH has the same enthalpy value of around -57 kJmol-1? Sulphuric acid has 2 H+ ions

 

[Kushal]

this is according to the stoichiometric equation.

0.5 H2SO4 + NaOH ----> 0.5 Na2SO4 + 0.5 H2O

when you react 0.5 mol H2SO4 with 1 mol NaOH, you get -57 kJmol-1. if you use 1 mol H2SO4 with 2 mol NaOH you will get about -114 kJmol-1.

you only have to consider H+ and OH- in stoichiometric proportions.

 

[GCT]

Why is the enthalpy of neutralisation of a strong acid mixed with a strong base always the same, no matter what acid and base combination you use, provided they are 'strong' acids and bases?

Am I right to say that this value is accepted as -57.35 kJmol-1?

Thanks for any help,
Isaac.

It's more or less the same because water is being formed and factors of variability include the changes in hydration of the resulting ion, for instance, sulfuric acid to its conjugate; with sulfuric acid it becomes deprotonated by the hydroxide, and the enthalpy of neutralisation in the context of which you mentioned it pertains to the latter reaction although the hydroxide would also deprotonate the conjugate.

 

[isaaclimdc]

But doesn't the formation of different metal halide salts produce different amounts of heat though?

 

[Kushal]

yes, this is different. there are diffrent types of enthalpies(energy change)

enthalpy of neutralisation
enthalpy of combustion
enthalpy of formation
enthalpy of solution
enthalpy of hydrogenation,...

enthalpy of formation is the energy change when 1 mol of a susbtance is formed from its elements in their standard states.

enthalpy of neutralisation is the energy released when 1 mol of H+ combines with 1 mol of OH-.

for example if you need enthalpy of formation of KCl

you will use K(s) and Cl2(g) as said by the definition (standard states).

now you will have to atomise K(s) to K(g). remove one electron from it(first ionisation energy). then you have to dissociate Cl2(g) into Cl(g). then add an electron(first electron affinity). then K+ will react with Cl- to give KCl. this is another energy change called lattice energy.

this would be different with other metal halides, because the energy involved (ionisation electron affinity, dissociation, atomisation, lattice energy,...) are different. you will get diffrent values for the enthalpy of formation.

 

[lightarrow]

Why is the enthalpy of neutralisation of a strong acid mixed with a strong base always the same, no matter what acid and base combination you use, provided they are 'strong' acids and bases?

This is only true for Brönsted acids and bases, if you are not talking about water (or similar solvents) dilute solutions.
In that case (of Brönsted acids and bases only) the reason is the bond between H+ and A- in the acid HA is very weak, as the bond between OH- and B+ in the base BOH. So the enthalpy of the reaction HA + BOH --> H2O + A- + B+, be it in water or in others similar solvents or between pures HA and BOH: HA + BOH --> H2O + BA, can be computed considering only the enthalpy of the reaction H+ + OH- --> H2O, without considering all the others enthalpies (which are low because of the weak bonds I said up).

If you are talking about acid-base reactions in diluted water solutions only, then the above result is not only approximately correct, but quite exact: every strong acid (and this time it can be Brönsted or Lewis acid) in water solution becomes, quantitatively, H3O+; any strong base (Brönsted or Lewis base) in water solution becomes, quantitatively, OH-, so, independently of which acid or base you initially had, the reaction will always be the same: H3O+ + OH- --> 2H2O and so the enthalpy of the reaction will always be the same.

 

[mysqlpress]

But why is it then, that H2SO4 and NaOH has the same enthalpy value of around -57 kJmol-1? Sulphuric acid has 2 H+ ions

By definition, it is defined as the enthalpy change when 1 mol of water is formed from complete neutralization of an acid and a base under standard conditions.
So even it has two ionizable H+
the reaction should be as follows

H2SO4+2NaOH-->2H2O+Na2SO4

 

[SMUDGY]

Does anyone know the actual value of the enthalpy change between sulphuric acid and sodium hydroxide. Also what volumes would be used when doing a calorimetric experiment when you put the chemicals into the polystyrene cup

 

[Kushal]

it depends on how much H2SO4 you add.

suppose both reagents are of the same conc. if you add equal volumes of both, the enthalpy will be about 57.3 kJ per mol of alkali. the H2SO4 has liberated only one mol of H+, it remains in the form of HSO4- now.

if you had added 2 mol of that NaOH, the HSO4- would further react with the alkali to give SO42-. This is another 57.3 kJ per mol of alkali. so, if had added 2 mol of alkali to 1 mol H2SO4, the enthalpy change would have been about 114.6 kJ per mol of alkali.

conc. has been kept constant. actually, you should be seeing the number of mol of reagents in solution. the first mole of NaOH will react with H+ from H2SO4 giving HSO4-. the second mole of NaOH will react with HSO4- to give SO42-.