Question

For strong acid strong base neutralisation energy for $1$ mole $H _{2} O$ formation is $-57.1 \,kJ$. 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)

$1$ mole of strong monoprotic acid reacts with $1$ mole of strong base to give $-57.1 \,kJ$
$\therefore 0.25 \,mol$ of strong acid will react with only $0.25\, mol$ of strong base (and not $0.5 \,mol$ )
$\therefore $ Energy involved $=-57.1 \,kJ \times 0.25$