Of the following compounds, which would you expect to have the lowest vapor pressure at room temperature: CHCl3 , CH2Cl2, CH3Cl or CCl4 ? Hint: base your decision mainly on dispersion forces.
I believe the answer is CH3Cl but I am not quit sure. Please help
ch3cl does have the lowest vapour pressure because the molecule has space for 3 hydrogen bonds which require a lot of energy to break up the molecule so that it can evaporate and thus hydrogen bonding causes a low vapour pressure
To determine which compound is expected to have the lowest vapor pressure at room temperature, we need to consider the intermolecular forces present in each compound. In this case, the main intermolecular force to consider is dispersion force.
Dispersion forces are the weakest intermolecular forces and result from temporary fluctuations in electron distribution around atoms or molecules. They are present in all molecules, but their strength depends on the size of the molecule or atom, as well as the shape of the molecule.
In general, larger and more symmetrical molecules tend to have stronger dispersion forces. This is because larger molecules have more electrons, which can create temporary dipoles, leading to stronger dispersion forces.
Now, let's compare the compounds:
1. CHCl3 (chloroform) - It is a relatively small molecule, and although it has chlorine atoms that can increase dispersion forces, it is outweighed by the presence of three hydrogen atoms. The hydrogen atoms are smaller and have fewer electrons. Hence, CHCl3 has weaker dispersion forces.
2. CH2Cl2 (dichloromethane) - This molecule is larger than CHCl3 because it has an additional chlorine atom. So, the dispersion forces are stronger due to the increased number of electrons.
3. CH3Cl (chloromethane) - This molecule is smaller and has fewer electrons compared to CH2Cl2. Though it has one chlorine atom, the overall weaker dispersion forces make it the most likely candidate to have the lowest vapor pressure at room temperature.
4. CCl4 (carbon tetrachloride) - This molecule is the largest among the given compounds. It has four chlorine atoms, resulting in stronger dispersion forces compared to the rest. Therefore, CCl4 is expected to have the highest vapor pressure at room temperature.
Based on this information, your initial guess is correct. CH3Cl would be expected to have the lowest vapor pressure at room temperature among the given compounds due to its weaker dispersion forces.
To determine which compound would have the lowest vapor pressure at room temperature based on dispersion forces, we need to compare the molecular weights and molecular shapes of the compounds.
Dispersion forces, also known as London forces, are the primary intermolecular forces between nonpolar molecules. They arise due to temporary changes in electron distribution, creating temporary dipoles that attract neighboring molecules. Larger, more polarizable molecules tend to have stronger dispersion forces.
In this case, we can compare the molecules CHCl3, CH2Cl2, CH3Cl, and CCl4.
1. CHCl3 (chloroform)
Molecular weight: 119.38 g/mol
Molecular shape: Trigonal pyramid
Number of carbon atoms: 1
2. CH2Cl2 (dichloromethane)
Molecular weight: 84.93 g/mol
Molecular shape: Tetrahedral
Number of carbon atoms: 1
3. CH3Cl (chloromethane)
Molecular weight: 50.49 g/mol
Molecular shape: Tetrahedral
Number of carbon atoms: 1
4. CCl4 (carbon tetrachloride)
Molecular weight: 153.82 g/mol
Molecular shape: Tetrahedral
Number of carbon atoms: 1
Looking at these compounds, CH3Cl has the lowest molecular weight, indicating weaker dispersion forces compared to the other compounds. Additionally, as all the compounds have the same number of carbon atoms, the size of the molecule also affects the strength of dispersion forces. Since CCl4 is the largest molecule of the four, it is expected to have the strongest dispersion forces.
Therefore, based on the molecular weight and molecular shape, we can conclude that CH3Cl would have the lowest vapor pressure at room temperature compared to the other compounds (CHCl3, CH2Cl2, and CCl4).