Consider the following data.
DeltaHf (Na2O (s))=-414kJ/mol
Delta H sublimation(Na)=109 kJ/mol
I.E. (Na)=495 kJ/mol
Bond Energy (O2) = 495 kJ/mol
E.A. (O) = -141 kJ/mol
E.A. (O -) = 878 kJ/mol
Use this information to calculate the lattice energy of Na2O (s).
So, we have the total which is delta H f (-414kJ/mol) and delta Hf= Delta H1+deltaH2+,...,+x
Where x is the lattice energy. I am still struggling in finding x
Is this AP Chem or regular chem?
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To calculate the lattice energy (x) of Na2O (s), we can use the Born-Haber cycle and Hess's law.
The formation of Na2O (s) can be broken down into the following steps:
1. Formation of Na+ ions:
Na(s) → Na+(g) + e-
Enthalpy change (ΔH1) = Ionization energy (I.E. Na) = 495 kJ/mol (given)
2. Formation of O2- ions:
1/2 O2(g) → O-(g)
Enthalpy change (ΔH2) = Electron affinity (E.A. (O)) = -141 kJ/mol (given)
3. Combining Na+(g) and O2-(g) to form Na2O (s):
2 Na+(g) + O2-(g) → Na2O(s)
Enthalpy change (ΔHf) = -414 kJ/mol (given)
Now, we can write the overall formation reaction of Na2O (s) by summing up the individual steps:
Na(s) + 1/2 O2(g) → Na2O(s)
ΔHf = ΔH1 + ΔH2 + x (where x is the lattice energy we want to calculate)
ΔH1 + ΔH2 = I.E. Na - E.A. (O) (taking the negative sign of E.A. (O) as it is an exothermic process)
Substituting the given values:
x = ΔHf - (ΔH1 + ΔH2)
= -414 kJ/mol - (495 kJ/mol - 141 kJ/mol)
= -414 kJ/mol - 354 kJ/mol
= -768 kJ/mol
Therefore, the lattice energy of Na2O (s) is -768 kJ/mol. Note that lattice energies are typically negative values.
To calculate the lattice energy (x) of Na2O (s) using the given data, we can use the Born-Haber cycle which relates various energies involved in the formation of an ionic compound. Here's how you can find x:
1. Write down the Born-Haber cycle for the formation of Na2O:
Na (s) + 1/2 O2 (g) → Na2O (s)
2. Break down the formation of Na2O into the following steps:
a. Formation of Na (g) from Na (s):
Na (s) → Na (g) ΔH = sublimation energy of Na = 109 kJ/mol
b. Dissociation of O2 (g) into 2 O atoms:
1/2 O2 (g) → O (g) ΔH = 1/2 bond energy of O2 = 1/2 * 495 kJ/mol
c. Ionization of Na (g) to form Na+ (g):
Na (g) → Na+ (g) + e- ΔH = Ionization energy of Na = Ionization energy of Na = 495 kJ/mol
d. Electron affinity of O (g) to form O- (g):
O (g) + e- → O- (g) ΔH = Electron affinity of O = -141 kJ/mol
e. Formation of Na2O (s) from Na+ (g) and O- (g):
Na+ (g) + O- (g) → Na2O (s) ΔH = Lattice energy of Na2O = x
3. Now, sum up all the individual steps to get the overall equation:
Na (s) + 1/2 O2 (g) + ΔH(sublimation) + 1/2 ΔH(bond energy) + ΔH(ionization) + ΔH(electron affinity) → Na+ (g) + O- (g) + x
4. Apply the Hess's law to the Born-Haber cycle and express the equation as:
-414 kJ/mol = ΔH(sublimation) + 1/2 ΔH(bond energy) + ΔH(ionization) + ΔH(electron affinity) + x
5. Plug in the given values:
-414 kJ/mol = 109 kJ/mol + (1/2 * 495 kJ/mol) + 495 kJ/mol + (-141 kJ/mol) + x
6. Simplify the equation:
-414 kJ/mol = 55 kJ/mol + 247.5 kJ/mol + 495 kJ/mol - 141 kJ/mol + x
-414 kJ/mol = 656.5 kJ/mol + x
7. Rearrange the equation to isolate x:
x = -414 kJ/mol - 656.5 kJ/mol
x = -1070.5 kJ/mol
Therefore, the lattice energy of Na2O (s) is -1070.5 kJ/mol. Note that the negative sign indicates that the process is exothermic, i.e., energy is released during the formation of Na2O.