Question

In a constant-pressure calorimeter, 65.0 mL of 0.340 M Ba(OH)2 was added to 65.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 21.83 �C to 26.46 �C. If the solution has the same density and specific heat as water, what is ?H for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

Answer 1

In a constant-pressure calorimeter, 65.0 mL of 0.340 M Ba(OH)2 was added to 65.0 mL of 0.680 M HCl. The reaction caused the temperature of the solution to rise from 21.83 �C to 26.46 �C. If the solution has the same density and specific heat as water, what is ?H for this reaction (per mole of H2O produced)? Assume that the total volume is the sum of the individual volumes.

The reactio will be

Ba(OH)2 + 2HCl --> BaCl2 + 2H2O

So here one moleof Ba(OH)2 will react with two moles of HCl to give two moles of H2O

Moles of Ba(OH)2 present = Molarity X volume = 0.34 X 65 mL = 22.1 millimoles

Moles of HCl present = Moalrity X volume = 0.68 X 65 mL = 44.2 millimoles

Moles of H2O formed = 44.2 millimoles

Heat generated by reaction will be absorbed by water

The heat absorbed by water = Mass X change in temperature X Specific heat of water

Mass of water = Volume of solution X Density = (65+65) X 1g / mL = 130 grams

Change in temperature = 26.46 - 21.83 = 4.63 ⁰C

Specific heat = 4.184 J / g ⁰C

Putting values

Heat absrobed by water = 130 x 4.184 X 4.63 = 2518.35 Joules

This heat is evolved when 44.2 millimoles of water is produced

The heat released when 1 mole of water is produced = 2518.35 X 1000 / 44.2 Joules = 56976.244 Joules / mole of water

Or 56.976 KJ / mole of water

The heat will be released hence it will be -56.976 KJ / mole of water