Question

A calorimeter temperature increases by 0.45 °C when 30 J of energy is added to it by electrical heating. When 0.10 grams of HCl is neutralized in the same calorimeter, the temperature increased by 7.3 °C. What is the ΔH of neutralization of HCl in unit?

Answer 1

The heat of neutralization is -180 kJ/mol.

In Part 1, you determine the specific heat of the calorimeter.

`q = C_"Cal"ΔT`

`C_"cal" = q/(ΔT) = "30 J"/"0.45 °C" = "66.7 J/°C"`

Now we can use this number in Part 2 of the problem.

There are two heats to consider:

heat of neutralization + heat to warm calorimeter = 0

`q_1 + q_2 = 0`

`mΔH + C_"Cal"ΔT = 0`

`"0.10 g" ×ΔH + 66.7 cancel("°C")·"g⁻¹" × 7.3 cancel("°C") = 0`

`ΔH = "-487 J"/"0.10 g" = "-4870 J/g"`

Your instructor probably wants the molar heat of neutralization.

`ΔH = "-4870 J"/(1 cancel("g HCl")) × (36.46 cancel("g HCl"))/"1 mol HCl" = "-180 000 J/mol" = "-180 kJ/mol"`

Note: The answer can have only 2 significant figures, because that is all you gave for all your measurements.

This does not agree with the accepted value for the heat of neutralization of HCl, which is -57.1 kJ/mol.