Answer:
ΔH = 20468 J/mol = 20.5 kJ/mol
Explanation:
Step 1 : Data given
Mass of sodium nitrate NaNO3 = 15.3 grams
The temperature fell from 25.00 °Celsius to 21.56 °Celsius
The heat capacity of the solution and the calorimeter is 1071 J/°C
Step 2: Calculate Q
Q = Cp * ΔT
⇒with Q = the heat transfer = TO BE DETERMINED
⇒with Cp = The heat capacity of the solution and the calorimeter is 1071 J/°C
⇒with ΔT = the change of temperature = 25.00 - 21.56 = 3.44 °C
Q= 1071 J/°C * 3.44 °C
Q = 3684.24 J
Step 3: Calculate moles NaNO3
Moles NaNO3 = mass / molar mass NaNO3
Moles NaNO3 = 15.3 grams / 84.99 g/mol
Moles NaNO3 = 0.180 moles
Step 4: Calculate the enthalpy change when 1 mol of sodium nitrate dissolves in water.
ΔH = Q / moles
ΔH = 3684.24 J/ 0.180 moles
ΔH = 20468 J/mol = 20.5 kJ/mol
Since the temperature decreases, this is an endothermic process.
For an endothermic process, the enthalpy change is positive.