the enthalpy of neutralisation of nh4oh with hcl is -51.46 kj /mol and the enthalpy of neautralisation of naoh with hcl is 55.90 kj/mol.the enthalpy of ionisation of nh4oh is:?????

the enthalpy of neutralisation of nh4oh with hcl is -51.46 kj /mol and the enthalpy of neautralisation of naoh with hcl is 55.90 kj/mol.the enthalpy of ionisation of nh4oh is:?????


4 Answers

Aman Bansal
592 Points
10 years ago

Dear Maheep,

enthalpy of neutralization of an acid by a base is defined as heat change when one gram equivalent of acid is neutralized by a base ,the reaction being carried out in dilute aqueous solution ... enthalpy of neutralization of base by an acid is defined in a similar manner... for example when 1 gram equivalent of HCl is neutralized with NaOH 57.1 kj of heat is produced .. HCl (aq) + NaOH (aq)----> NaCl (aq) + H2O (l)... H = -57.1 kj/mole hence enthalpy of neutralization of HCl with NaOH is 57.1 kj enthalpy of neutralization of any strong acid (like HCl,HNO3,H2SO4) with a strong base (like LiOH,NaOH,KOH) or vice versa is always the same i.e. 57.1 kj...this is because strong acids ,strong bases and salt that they form are all completely ionized in dilute aqueous solutions ...thus the reaction between any strong acid and strong base for example in the above case may be written as : NaOH (aq) + HCl(aq) -----> NaCl (aq) + H2O (l)... H = -57.1 kj/mole they will dissociate as : Na(+) (aq) + OH(-) (aq) + H(+) (aq) + Cl(-) (aq) ---> Na(+) (aq) + Cl(-) (aq) + H2O (l) common ions will cancel out.. H(+) (aq) + OH(-) (aq) ----> H2O (l) thus neutralization is simply a reaction between H(+) ions given by acids and OH(-) ions given by base to form one mole of H2O.....since strong acid and strong base completely ionize in aqueous solution number of H(+) and OH(-) produced by 1 gram equivalent of strong acid and strong base is always the same ...hence enthalpy of neutralization between a strong acid and strong base is always constant... in case either the acid or base or both are weak the enthalpy of neutralization is less than 57.1 kj ...the reason for this behavious can be explained by considering the neutralization between a strong base like NaOH and weak acid like acetic acid.. now acetic acid ionizes to a small extent whereas NaOH ionizes completely as : NaOH (aq) ----> Na(+) (aq) + OH(-) (aq) CH3COOH (aq)<--------> CH3COO(-)(aq) + H(+)(aq) (an equilibrium) when H(+) given by acid combine with OH(-) given by base the equilibrium shifts to right (in accordance with Le Chatellers principle) ,so more of acetic acid dissociates ...a part of heat produced during combination of H(+) and OH(-) ions is used up for complete dissociation of acetic acid ...the heat thus used up is called enthalpy of dissociation or enthalpy of ionization is 1.9 kj for acetic acid...hence net heat evolved in above reaction is 57.1 - 1. 9 = 55.2 kj ...which less than that evolved in strong base and strong acid reaction...

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Aman Bansal

Askiitian Expert

13 Points
6 years ago
If Enthalpy of neutralisation of hcl and naoh is -55.90 and that of hcl and nh4oh is -51.46 then enthalpy of ionisation of nh4oh is 4.44k.j/mole
13 Points
5 years ago
As it undergoes neutralization reaction it will products -reactant=-51.34-(-55.84)=4.5kj mol^-1hope u understand this cheers!!!
Yash Chourasiya
askIITians Faculty 256 Points
2 years ago
Dear Student

Given, Enthalpy of neutralisation of HCl and NaOH is -55.90 and that of HCl and NH4OH is -51.46
Therefore enthalpy of ionisation of NH4OH = -51.46 – (-55.90)

I hope this answer will help you.
Thanks & Regards
Yash Chourasiya

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