Question 5.4.3: Use constant-pressure calorimetry to determine enthalpy chan......
Use constant-pressure calorimetry to determine enthalpy change.
Ammonium chloride is very soluble in water. When 4.50 g NH_{4}Cl is dissolved in 53.00 g of water, the temperature of the solution decreases from 20.40 ºC to 15.20 ºC. Calculate the enthalpy of dissolution of NH_{4}Cl (in kJ/mol).
Assume that the specific heat of the solution is 4.18 J/g · ºC and that the heat absorbed by the calorimeter is negligible.
You are asked to calculate the enthalpy change for the dissolution of ammonium chloride.
You are given the masses of the solid and water, and the temperature change that occurs when the two are combined.
First calculate the energy change for the surroundings (q_{solution}) in the coffee cup calorimeter.
c_{solution} = 4.18 J/g · ºC
m_{solution} = 4.50 g + 53.00 g = 57.50 g
∆T = T_{final} − T_{initial} = 15.20 ºC − 20.40 ºC = −5.20 ºC
q_{solution} = m × c_{solution} × ∆T = (57.50 g)(4.18 J/g · ºC)(−5.20 ºC) = −1250 J
Next calculate q for the dissolution of NH_{4}Cl, q_{system}.
q_{dissolution} + q_{solution} = 0
q_{dissolution} = − q_{solution} = 1250 J
Finally, calculate the amount of NH_{4}Cl dissolved (in mol) and ∆H for the dissolution of NH_{4}Cl (kJ/mol)
4.50 g NH_{4}Cl × \frac{1\text{ mol NH}_{4}\text{Cl}}{53.49\text{ g}} = 0.0841 mol NH_{4}Cl
ΔH_{dissolution} = \frac{q_{dissolution}}{\text{mol NH}_{4}Cl} = \frac{1250\text{ J}}{0.0841\text{ mol}} \times \frac{1\text{ KJ}}{10^{3}\text{ J}} = 14.9 KJ/mol
Is your answer reasonable? The temperature of the solution decreased, indicating an endothermic process and a positive value for ∆H_{dissolution}.