Create the Born-Haber cycle and calculate the lattice energy of lithium oxide (Li₂O) from the following data:
Quantity
Magnitude (kJ/mol)
519
603
147
499
-20
Ionization energy of Li(g)
Electron affinity of O(g) for 2e
Vaporization energy of Li(s)
Bond energy of O2(g)
Reaction enthalpy: 2 Li (s) +1/2O2(g) → Li₂O(s)
a. -3081 kJ/mol
d. -2972 kJ/mol
b. -2586 kJ/mol
e. -2831 kJ/mol
c. -2205 kJ/mol
The freezing point of ammonia (NH3, 17.04 g/mol) is -77.73°C. In an experiment, 367.2 grams of solid NH3 at-77.73°C is used to cool 250.0 grams of liquid NH3 from 25.00°C to the freezing point. Assume that the specific heat capacity of liquid NH3 is 4.742 J/(g °C). What is the enthalpy of fusion of ammonia?
a. 5.652 kJ/mol
b. -5.652 kJ/mol
c. -2.894 kJ/mol
d. 0.3315 kJ/mol
e. 2.894 kJ/mol
The Correct Answer and Explanation is :
Born-Haber Cycle Calculation (Lattice Energy for Li₂O):
The lattice energy of lithium oxide (Li₂O) can be calculated using the Born-Haber cycle, which accounts for the steps involved in forming the solid from its ions in the gas phase. From the data provided:
- Ionization energy of Li: 519 kJ/mol
- Electron affinity of O for 2e: 603 kJ/mol
- Vaporization energy of Li: 147 kJ/mol
- Bond energy of O₂: 499 kJ/mol
- Reaction enthalpy for the formation of Li₂O: -3081 kJ/mol
By using the Born-Haber cycle equation:
[
Lattice Energy = \Delta H_{\text{reaction}} – (2 \times IE_{\text{Li}} + 2 \times VE_{\text{Li}} + \frac{1}{2} \times BE_{\text{O}2} + EA{\text{O}})
]
The lattice energy comes out to be approximately -2586 kJ/mol (answer choice b).
Enthalpy of Fusion of Ammonia:
For the enthalpy of fusion of ammonia, we can use the heat absorbed by the liquid ammonia to reach its freezing point from the initial temperature. The heat absorbed by the liquid ammonia is calculated using the formula:
[
Q_{\text{liquid NH}3} = m{\text{liquid NH}3} \times C{\text{liquid NH}_3} \times \Delta T
]
Where:
- ( m_{\text{liquid NH}_3} = 250.0 \, \text{g} )
- ( C_{\text{liquid NH}_3} = 4.742 \, \text{J/(g °C)} )
- ( \Delta T = (25.00 – (-77.73)) = 102.73 \, \text{°C} )
This gives the total heat absorbed by the liquid ammonia. From there, we calculate the enthalpy of fusion by dividing the heat absorbed by the moles of solid ammonia.
The calculated enthalpy of fusion for ammonia is approximately 5.65 kJ/mol, which corresponds to answer choice a.
Final Answers:
- Lattice energy of Li₂O: -2586 kJ/mol (Answer choice b).
- Enthalpy of fusion of ammonia: 5.65 kJ/mol (Answer choice a).
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