Question

When 15.3 g of sodium nitrate, NaNO₃, was dissolved in water in a constant-pressure calorimeter, the temperature fell from 25.00 °Celsius to 21.56 °Celsius. If the heat capacity of the solution and the calorimeter is 1071 J/oC, what is the enthalpy change when 1 mol of sodium nitrate dissolves in water?
The solution process is NaNO₃(s) _______.

Answer 1

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.

Answer 2

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