50.0 mL (50.0g) of water was transferred to a polystyrene calorimeter. The initial temperature was recorded as 21.6°C. Then 1.0g of ammonium nitrate weighed and transferred to the calorimeter. The mixture was stirred and the final temperature was recorded as 20.2°C.
Calculate the heat of dissolution of ammonium nitrate in kJ/mol.
q = [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = ? in J if Cp H2O is used as 4.184 J/g*C
mols NH4NO3 = grams/molar mass = ? mols
Then heat solution is q/mol NH4NO3.
Convert to kJ/mol
Note: Since the temperature of the H2O cooled you know the dHsoln is + (endothermic)
I got 0.0037017 kj/mol. Is that correct?
No, Not close. Post your work if you want me to find the error.
To calculate the heat of dissolution of ammonium nitrate, we need to use the equation:
q = m × c × ΔT
- q represents the heat exchange in Joules
- m represents the mass of the substance in grams
- c represents the specific heat capacity of the substance in J/g°C
- ΔT represents the change in temperature in °C
First, let's find the heat exchange (q) for the water:
q_water = m_water × c_water × ΔT_water
Since the water is the solvent, we assume that the specific heat capacity is 4.18 J/g°C, which is the specific heat capacity of water.
m_water = 50.0 g (given)
c_water = 4.18 J/g°C (specific heat capacity of water)
ΔT_water = final temperature - initial temperature = 20.2°C - 21.6°C = -1.4°C
q_water = 50.0 g × 4.18 J/g°C × (-1.4°C) = -292.6 J
The negative sign indicates that the water released heat.
Next, let's find the heat exchange (q) for ammonium nitrate:
q_ammonium_nitrate = m_ammonium_nitrate × c_ammonium_nitrate × ΔT_ammonium_nitrate
To find the specific heat capacity of ammonium nitrate (c_ammonium_nitrate), we can use the molar heat of dissolution equation:
ΔH_dissolution = q / n
- ΔH_dissolution represents the heat of dissolution in J/mol
- q represents the heat exchange in Joules
- n represents the number of moles of the substance
Weighing 1.0 g of ammonium nitrate, we can calculate the number of moles:
n = m_ammonium_nitrate / M_ammonium_nitrate
- n represents the number of moles
- m_ammonium_nitrate represents the mass of ammonium nitrate in grams (1.0 g, given)
- M_ammonium_nitrate represents the molar mass of ammonium nitrate (NH4NO3 = 80.043 g/mol)
n = 1.0 g / 80.043 g/mol ≈ 0.0125 mol
Now we can calculate the molar heat of dissolution:
ΔH_dissolution = q_ammonium_nitrate / n
However, we still need to find q_ammonium_nitrate.
q_ammonium_nitrate = q_calorimeter + q_water
Since the polystyrene calorimeter has a negligible heat capacity, we assume q_calorimeter = 0. Therefore, q_ammonium_nitrate = q_water.
Finally, we can calculate the heat of dissolution:
ΔH_dissolution = q_ammonium_nitrate / n = q_water / n
ΔH_dissolution = -292.6 J / 0.0125 mol = -23408 J/mol
To convert the answer from Joules (J) to kilojoules (kJ), divide the value by 1000:
ΔH_dissolution = -23408 J/mol / 1000 = -23.408 kJ/mol
Therefore, the heat of dissolution of ammonium nitrate is approximately -23.408 kJ/mol.