Question

You mix 50.0 mL of a weak monoprotic acid with 50.0 mL of NaOH solution in a coffee cup calorimeter. Both solutions (and the calorimeter) were initially at 22.0C. The final temperature of the neutralization reaction was determined to be 22.5C.

a) What is the total amount of heat evolved in this reaction? Show all work.

b) If 0.135 moles of the monoprotic acid were neutralized in this reaction, what is the molar heat of neutralization (enthalpy) for this reaction?

Answer 1

Total volume of the solution = 100mL = 100g since density of water = 1g/mL

specific heat of water 4.18J/g

The toal heat evolved in this process is = sp. heat x mass x difference in temperature

                                                                       = 4.18 x 100x 0.5=209J

b) For one mole of monoprotic acid and monoacidic base the molar heat of neutralization is -57.2kJ/mol

If 0.135 moles were neutralized then the enthalpy of the reaction = -57.2x0.135=7.722 kJ

Alternatively in this reaction 100g of water is produced from acid and base . that is 100/18 moles of water produced. By definition, one equivalent of H⁺ and one equivalent of OH- ions combine to form one mole of water is the neutralization process and the heat liberated in this process is molar heat of neutralization.

Thus one mole of monobasic acid produces one mole of water.

0.135 moles of acid produces 0.135 moles of water.

100/18 moles of water production gives 209 J

0.135 moles of water gives 0.135x 209x18/100=5.0787 J

and molar heat of neutralization for the process = 5.0787/0.135= 37.62 J/mol