I keep finding all sorts of different enthalpy of vaporizations for bromine. I used: 254.66
So factoring 254.66/2 into my equation below i ended up with a lattice energy of -863.33.
Can you confirm this?
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What is the enthalpy change (in kJ/mol) for the following process?
Na+(g) + Br-(g) --> NaBr (s)
Hint: use the Born-Haber cycle
Note: the reference state for Br2 is Br2(l) not Br2(g)
Data (in kJ/mol):
enthalpy of sublimation of Na(s): 107
standard enthalpy of formation of Br2(g): 32
bond dissociation energy of Br2(g): 194
ionization energies (1, 2, 3) of Na(g): 496
ionization energies Br(g): 140, 103, 473
first electron affinity of Br(g): -326
standard enthalpy of formation of NaBr(s): -351
Sooo..
deltaH1=+107
deltaH2=194/2=+97
deltaH3=+32
deltaH4=+496
deltaH5=-326
deltaH6=lattice energy=?
deltaHf= -351
Sooo..
-351=107+97+32+496-326+lattice energy
lattice energy= -757
This is not the correct answer.
Can someone please tell me where I've gone wrong please!
Chemistry - DrBob222, Wednesday, February 2, 2011 at 12:24am
The problem gave you a hint that the standard state for Br2 was liquid and not gas. Therefore, you must include enthalpy for Br2(l) to Br2(gas) (and 1/2 that)
Based on the information provided, it appears that you have made an error in including the enthalpy change for the transition from liquid bromine (Br2(l)) to gaseous bromine (Br2(g)) in the Born-Haber cycle.
To calculate the enthalpy change for the formation of NaBr(s) from its constituent elements, you need to consider the following steps:
1. First, we need to calculate the enthalpy change for the sublimation of sodium (Na).
ΔH1 = enthalpy of sublimation of Na(s) = +107 kJ/mol
2. Then, we need to calculate the enthalpy change for the bond dissociation of bromine (Br2).
ΔH2 = 1/2 × bond dissociation energy of Br2(g) = 1/2 × +194 kJ/mol = +97 kJ/mol
3. Next, we need to calculate the enthalpy change for the ionization of sodium (Na).
ΔH3 = Ionization energy of Na(g) = +496 kJ/mol
4. Then, we need to calculate the enthalpy change for the first electron affinity of bromine (Br).
ΔH4 = First electron affinity of Br(g) = -326 kJ/mol
5. Finally, we need to calculate the lattice energy of NaBr(s) using the given enthalpy of formation.
ΔH6 = -351 kJ/mol
Now, the missing step is the enthalpy change for the transition of bromine from liquid to gas.
ΔH5 = Enthalpy change for the transition Br2(l) to Br2(g)
This value is not provided in the given data, so we cannot proceed with the calculation without it.
To calculate the enthalpy change for the given process using the Born-Haber cycle, you need to consider the individual enthalpy changes involved in the formation of NaBr(s) from its constituent elements.
First, let's correct the given data by including the enthalpy change for the conversion of Br2(l) to Br2(g). Based on the hint provided, we can assume that the enthalpy change for this process is half of the enthalpy of vaporization of bromine. Let's refer to this as deltaH0:
deltaH0 = 254.66 kJ/mol / 2 = 127.33 kJ/mol
Now, let's revise the enthalpy changes for the formation of NaBr(s) using the corrected data:
deltaH1 = enthalpy of sublimation of Na(s) = +107 kJ/mol
deltaH2 = 1/2 enthalpy of vaporization of Br2(l) to Br2(g) = 1/2 x 127.33 kJ/mol = +63.67 kJ/mol
deltaH3 = standard enthalpy of formation of Br2(g) = +32 kJ/mol
deltaH4 = ionization energy of Na(g) = +496 kJ/mol
deltaH5 = first electron affinity of Br(g) = -326 kJ/mol
deltaH6 = lattice energy (to be determined)
deltaHf = standard enthalpy of formation of NaBr(s) = -351 kJ/mol
Now, let's apply the Born-Haber cycle equation:
deltaHf = deltaH1 + deltaH2 + deltaH3 + deltaH4 + deltaH5 + deltaH6
Plugging in the values:
-351 kJ/mol = +107 kJ/mol + 63.67 kJ/mol + 32 kJ/mol + 496 kJ/mol - 326 kJ/mol + deltaH6
Simplifying the equation:
deltaH6 = -351 kJ/mol - (107 kJ/mol + 63.67 kJ/mol + 32 kJ/mol + 496 kJ/mol - 326 kJ/mol)
= -757.67 kJ/mol
So, the correct lattice energy value is -757.67 kJ/mol, not -863.33 kJ/mol as you calculated.
To summarize, the enthalpy change for the process Na+(g) + Br-(g) --> NaBr(s) is -757.67 kJ/mol.