Question

For strong acid strong base neutralisation energy for $1$ mole $H_{2}O$ formation is $-57.1kJ$. If $0.25$ mole of strong monoprotic acid is reacted with $0.5$ mole of strong base then enthalpy of neutralisation is:

• A. $-(0.25\times 57.1)$
• B. $0.5\times 57.1$
• C. $57.1$
• D. $(-0.5\times 57.1)$

Answer 1

Answer: (A)

$-(0.25\times 57.1)$

The neutralization reaction of strong acid and strong base is :

$HA+BOH\to AB+H_{2}O$

If $\Delta H_{neut}=-57.1kJ/mol$ for 1 mol of the above reaction

Since the equimolar concentration of acid and base reacts, therefore whichever is in smaller amount will be the limiting factor for the reaction.

Since $\left[HA\right]=0.25moland\left[BOH\right]=0.5mol$, therefore $0.25$ mol of each (acid and base) will react to give $0.25$ mol water.

Hence enthalpy of neutralization will be= $-(0.25\times 57.1)kJ/mol$ or $-57.1kJ$

Therefore option A is correct.