How do you calculate lattice energy?
I have to place the following in order of decreasing magnitude of lattice energy, but I don't know how to do it.
KF MgS RbI
Is there a formula or some simple way of knowing which has the greatest energy?
2. I also have the same question, but with regard to vapor pressure. Given NF3, NH3, and BCl3, how do I figure out their vapor pressures? Thanks.
For 1, the smaller the ions the greater the lattice energy so RbI would be lowest and KF the highest (of those two). Mg and S are divalent and the attractive forces are higher than for monovalent cations and anions. So MgS I would expect to be the highest of the three.
For 2, the boiling point are
NH3 = -33
NF3 = -129
BCl3 = 12.5
You would expect vapor pressure to track (in reverse) the boiling point; i.e., higher b.p. gives lower v.p.
BCl3 has the highest molar mass and the IM forces are greatest; therefore b.p. is highest and vapor pressure the least.
NH3 has lowest molar mass BUT has hydrogen bonding so expect higher b.p. and lower v.p.
NF3 has no H bonding and a b.p. of -129 so vapor pressure should be the highest.
To calculate lattice energy, you can use the Born-Haber cycle or use the Born-Lande equation. Here's a step-by-step process using the Born-Haber cycle:
1. Write the balanced equation for the formation of the ionic compound from its constituent elements, including their respective enthalpy changes (such as formation energies or ionization energies).
2. Estimate the ionization energy and electron affinity of the metal and non-metal ions involved in the compound.
3. Calculate the lattice energy using the equation:
Lattice energy = (|-Lattice enthalpy|) + (sum of ionization energy) + (sum of electron affinity) - (sum of enthalpies of formation)
For the given compounds KF, MgS, and RbI, you need to compare their lattice energies to determine the order of decreasing magnitude.
To compare lattice energies, look at the charges of the ions involved. Higher charges lead to stronger attractions and higher lattice energies. In other words, greater differences in electronegativity or greater charges result in stronger interactions.
In this case, KF has a +1 charge for K and a -1 charge for F. MgS has a +2 charge for Mg and a -2 charge for S. RbI has a +1 charge for Rb and a -1 charge for I.
Comparing the charges, MgS has the highest magnitude of lattice energy because of the double charge on both Mg and S ions. KF would have the lowest lattice energy because its ions have the smallest charges. RbI would have an intermediate lattice energy.
For the question regarding vapor pressure, to determine the relative vapor pressures of NF3, NH3, and BCl3, you can use the boiling points of these compounds as an indicator. Higher boiling points correspond to lower vapor pressures.
NF3 has a boiling point of -129°C, NH3 has a boiling point of -33°C, and BCl3 has a boiling point of 12.5°C. Therefore, the order of decreasing vapor pressure is BCl3 > NH3 > NF3.
Remember, the lower the boiling point, the higher the vapor pressure.
To calculate lattice energy, you can use the Born-Haber cycle or the Coulomb's Law, depending on the information given. Here's a general approach to calculating lattice energy:
1. Determine the ionic compounds involved and their respective ions.
2. Determine the charges of the ions.
3. Use Coulomb's Law or the Born-Haber cycle to calculate the lattice energy.
1. For the given ionic compounds, KF, MgS, and RbI, we can calculate the lattice energy by comparing the charges of the ions and their sizes.
- KF: The potassium ion (K+) and fluoride ion (F-) have charges of +1 and -1, respectively. The ions are relatively small in size. The lattice energy will be relatively high.
- MgS: The magnesium ion (Mg2+) and sulfide ion (S2-) have charges of +2 and -2, respectively. The ions are larger in size compared to K+ and F-. The lattice energy will be lower than KF.
- RbI: The rubidium ion (Rb+) and iodide ion (I-) have charges of +1 and -1, respectively. The ions are larger in size compared to K+ and F-, but smaller than Mg2+ and S2-. The lattice energy of RbI will be lower than KF but higher than MgS.
Therefore, placing them in order of decreasing magnitude of lattice energy would be: KF > RbI > MgS.
2. To compare the vapor pressures of NF3, NH3, and BCl3, you need to consider the intermolecular forces between the molecules.
- NF3 is a polar molecule with relatively strong dipole-dipole interactions. It also has higher molecular weight compared to NH3 and BCl3. Therefore, it would have the highest vapor pressure among the given compounds.
- NH3 is also a polar molecule, but it is smaller and lighter than NF3. It would have a lower vapor pressure compared to NF3.
- BCl3 is a non-polar molecule and is lighter in weight. It would have the lowest vapor pressure among the given compounds.
Therefore, placing them in order of decreasing vapor pressure would be: NF3 > NH3 > BCl3.