Answer:
20468J / mol
Explanation:
The dissolution in water of NaNO₃(s) is:
NaNO₃(s) → Na⁺(aq) + NO₃⁻(aq)
Now, the equation of a calorimeter is:
Q = -C × ΔT
Where Q is heat, C is heat capacity (1071 J/°C) and ΔT is change in temperature (21.56°C - 25.00°C = -3.44°C)
Replacing:
Q = -1071 J/°C × -3.44°C
Q = 3684.24 J is change in enthalpy per 15.3g of sodium nitrate.
Moles of sodium nitrate are:
15.3g × (1mol / 85g) = 0.18 moles
Thus, enthalpy change per mole of sodium nitrate is:
3684.24J / 0.18mol = 20468J / mol
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.
