In: Chemistry
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.
ΔdisH = _______kJ/mol
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