the enthalpy of dissolution (dissolving) of sodium hydroxide,

Question

A student wants to determine the enthalpy change associated with dissolving solid sodium hydroxide,

NaOH, in water. The student dissolves 1.96 g of NaOH in 100 mL of water in a coffee-cup calorimeter.

The initial temperature of the water was 23.4 o C. After the NaOH dissolves, the temperature of the water

rises to 28.7 o C. Use this data to determine the enthalpy of dissolution (dissolving) of sodium hydroxide,

in kJ/mol of NaOH. Assume that the heat capacity of the calorimeter is negligible.

Answer 1

The formula that connects heat (Q), mass (m), specific heat capacity (c), and change in temperature (ΔT) is as follows:

Q = mcΔT

The specific heat capacity of water is approximately 4.186 J/g°C, the mass of water is 100 grams, and the change in temperature (ΔT) from 23.4°C to 28.7°C is 5.3°C.

So, the energy released when NaOH dissolved is:

Q = 100g * 4.186 J/g°C * 5.3°C = 2,218.58 J

Since the enthalpy of dissolution is defined as the energy released per mole, we will divide the released energy by the mass of NaOH in g, and then convert the result from Joules / g to Joules / mol using the formula weight of NaOH (39.997 g/mol) as a conversion factor

H = (2,218.58 J / 1.96 g NaOH) x (39.997 g NaOH / mol NaOH) = 45,273.75 J/mol

Rounding the result to 3 significant figures, we get

H = 45.3 kJ/mol