When a solid dissolves in water, the solution may become hotter or colder. The d

Question

When a solid dissolves in water, the solution may become hotter or colder. The dissolution enthalpy (dissolving) can be determined using a coffee cup calorimeter. In the laboratory a general chemistry student finds that when 2.81 g KClO4(s) is dissolved in 111.30 g water, the temperature of the solution drops from 24.18 to 21.80 °C. The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.69 J/°C. Based on the student's observation, calculate the dissolution enthalpy of KClO4(s) in kJ/mol. Assume the specific heat capacity of the solution is equal to the specific heat capacity of water.

disH = _______kJ/mol

Explanation / Answer

Ans. # Mass of solution = 2.81 g (mass of KClO4) + 111.30 g (mass of water) = 114.11 g

It’s assumed that the specific heat of solution remains the same as that of pure water.

# Step 1: Heat lost by Calorimeter is given by-

            Qc = C x dT

                        Where, C = Calorimeter constant   , dT = (Final – Initial) temperature

            Or, Qc = (1.69 J 0C-1) x (21.80 – 24.18)0C

            Hence, Qc = -4.0222 J

            The –ve sign of Qc indicates heat loss.

# Step 2: Heat lost by Solution is given by-

q = m s dT                           

Where,

q = heat

m = mass

s = specific heat

dT = Final temperature – Initial temperature

It’s assumed that the specific heat of solution is same as that of water.

            Or, Qs = 114.11 g x (4.184 J g-10C-1) x (21.80 – 24.18)0C

            Hence, Qs = -1136.2982512 J

# Step 3: Total heat lost by (solution and calorimeter) must be equal to the total amount of heat gained during dissolution of KClO4 –

Or,

            Heat gain during dissolution of KClO4, Q = - (Qc + Qs)

            Or, Q = - (-4.0222 J - 1136.2982512 J)

            Hence, Q = 1140.3204512 J

That is, total 1140.3204512 J heat is absorbed during dissolution of KClO4.

# Step 4: Moles of KClO4 = 2.81 g / (138.5486 g mol-1) = 0.0202817 mol

Now,

            dHdissolution = Total heat absorbed during dissolution / Moles of KClO4

                                    = 1140.3204512 J / 0.0202817 mol

                                    = 56224.106 J/ mol

                                    = 56.224 kJ/ mol

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