What is ΔrH° for the following phase change? LiCl(s) → LiCl(l)
1 answer:
Answer:
Explanation:
Hello!
In this case, since the enthalpy change for any process is computed by subtracting the enthalpy of the final state and the enthalpy of the initial state, for the given phase change, we subtract the enthalpy of the liquid (final state) and the enthalpy of the solid (initial state) considering this a melting process:
Which makes sense because this process absorbs energy.
Best regards!
You might be interested in
Answer:
Carbon dioxide
Explanation:
Neither helium nor carbon dioxide has a molecular dipole, so their strongest van der Waals attractive forces are London forces.
Helium is a small spherical atom with only a two electrons, so its atoms have quite weak attractions to each other.
CO₂ is a large linear molecule. It has more electrons than helium, so the attractive forces are greater. Furthermore, the molecules can align themselves compactly side-by-side and maximize the attractions (see below).
For example. CO₂ becomes a solid at -78 °C, but helium must be cooled to -272 °C to make it freeze (that's just 1 °C above absolute zero).
