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!
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).
Your calculations are mostly correct, but there is an error in the calculation of the lattice energy. The lattice energy should be calculated as the sum of the ionization energies of Na(g), the electron affinity of Br(g), and the enthalpy of formation of NaBr(s), all multiplied by their respective coefficients in the balanced equation.
Here's the correct calculation:
ΔH1 = enthalpy of sublimation of Na(s) = +107 kJ/mol
ΔH2 = bond dissociation energy of Br2(g) / 2 = +97 kJ/mol
ΔH3 = standard enthalpy of formation of Br2(g) = +32 kJ/mol
ΔH4 = ionization energy of Na(g) = +496 kJ/mol
ΔH5 = first electron affinity of Br(g) = -326 kJ/mol
ΔH6 = lattice energy (to be calculated)
ΔHf = standard enthalpy of formation of NaBr(s) = -351 kJ/mol
Using the Born-Haber cycle:
ΔHf = ΔH1 + ΔH2 + ΔH3 + ΔH4 + ΔH5 + ΔH6
-351 = 107 + 97 + 32 + 496 - 326 + ΔH6
ΔH6 = -757 kJ/mol
Therefore, the correct value for the lattice energy is -757 kJ/mol.
It appears that you made an error in the sign of the lattice energy in your previous calculation.
Based on the information you provided, there seems to be a mistake in your calculations. Let's go through it again step by step.
First, let's calculate the enthalpy change for the formation of the gas phase ions Na+(g) and Br-(g) from their elements in their standard states.
For sodium:
deltaH1 = ionization energy of Na(g) = 496 kJ/mol
For bromine:
deltaH2 = first ionization energy of Br(g) = 140 kJ/mol
deltaH3 = second ionization energy of Br(g) = 103 kJ/mol
deltaH4 = third ionization energy of Br(g) = 473 kJ/mol
deltaH5 = first electron affinity of Br(g) = -326 kJ/mol
Next, let's consider the formation of solid NaBr:
deltaH6 = standard enthalpy of formation of NaBr(s) = -351 kJ/mol
Now, using the Born-Haber cycle, we can write the following equation:
deltaH6 = deltaH1 + deltaH2 + deltaH3 + deltaH4 + deltaH5 + lattice energy
Given the values:
deltaH1: 496 kJ/mol
deltaH2: 140 kJ/mol
deltaH3: 103 kJ/mol
deltaH4: 473 kJ/mol
deltaH5: -326 kJ/mol
deltaH6: -351 kJ/mol
Let's substitute these values into the equation:
-351 = 496 + 140 + 103 + 473 - 326 + lattice energy
Now, rearrange the equation to solve for the lattice energy:
lattice energy = -351 - 496 - 140 - 103 - 473 + 326
Calculating this, we get:
lattice energy = -237 kJ/mol
So, the correct value for the lattice energy is -237 kJ/mol.