Question

for strong acid strong base neutralization energy for 1 MOLE H2O formatin is -57.1 kJ. if .25 mole of strong monoprotic acid is reacted with .5 mole of strong base than enthalpy of neutralization is...

Answer 1

We have a strong acid and a strong base so we can assume that the reactions goes to completion (as far as it can). Here's the general form of the reaction:

\[HA + BOH \rightarrow H_2O + BA\]

We know the enthalpy of formation per mole of H2O is -57.1 kJ. If we can figure out how much H2O is produced, we can figure out the enthalpy of our situation. Since the acid and base react 1:1 (we're assuming the reaction involves 1 equivalent of acid and base), already you can see that the acid will be the limiting reagent, and 0.25 mol of acid will react with 0.25 mol of base to produce 0.25 mol of H2O.

0.25 mol is a quarter of 1 mol, so you'll also be producing a quarter as much energy: -57.1 kJ/4 = -14.3 kJ.