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).
Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :
[butane]=1.0 M , [isobutane]=2.5 M
If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?
Answer:
The equilibrium concentration of each gas:
[Butane] = 1.14 M
[isobutane] = 2.86 M
Explanation:
Butane ⇄ Isobutane
At equilibrium
1.0 M 2.5 M
After addition of 0.50 M of butane:
(1.0 + 0.50) M -
After equilibrium reestablishes:
(1.50-x)M (2.5+x)
The equilibrium expression will wriiten as:
![K_c=\frac{[Isobutane]}{[Butane]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BIsobutane%5D%7D%7B%5BButane%5D%7D)
x = 0.36 M
The equilibrium concentration of each gas:
[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M
[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M
The alcohol on board is cognac 2
