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

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 × 57.1) (B) 0.5 × 57.1 (C) 57.1 (D) -(0.5 × 57.1)

Tardigrade Admin · Accepted Answer

1 mole of strong monoprotic acid reacts with 1 mole of strong base to give -57.1 kJ ∴ 0.25 mol of strong acid will react with only 0.25 mol of strong base (and not 0.5 mol ) ∴ Energy involved =-57.1 kJ × 0.25

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

$- (0.25 \times 57.1)$

$0.5 \times 57.1$

$57.1$

$- (0.5 \times 57.1)$

$1$ mole of strong monoprotic acid reacts with $1$ mole of strong base to give $- 57.1 k J$
$∴ 0.25 m o l$ of strong acid will react with only $0.25 m o l$ of strong base (and not $0.5 m o l$ )
$∴$ Energy involved $= - 57.1 k J \times 0.25$

Q. For strong acid strong base neutralisation energy for 1 mole H2​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

−(0.25×57.1)

0.5×57.1

57.1

−(0.5×57.1)