Select the pair of substances in which the one with the higher vapor pressure at a given temperature is listed first.
1. C7H16, C5H12
2. CCl4, CBr4
3. H2O, H2S
4. CH3CH2OH, CH3-O-CH3
5. Xe, Kr
Use the molar mass; the lower molar mass should have the higher vp.
That doesn't work for c because of H bonding in H2O. And 4 has the same molar mass the the OH on the alcohol has H bonding also. H bonding makes the boiling point higher and the vp lower.
4r
The pair of substances in which the one with the higher vapor pressure at a given temperature is listed first is:
1. C7H16, C5H12
To determine which substance has the higher vapor pressure at a given temperature, we need to consider the intermolecular forces between the molecules.
1. C7H16, C5H12: These are both hydrocarbon compounds. Generally, as the molecules become larger, the intermolecular forces increase, leading to lower vapor pressure. Therefore, C5H12 (pentane) would have higher vapor pressure compared to C7H16 (heptane). The answer is C5H12, C7H16.
2. CCl4, CBr4: Both substances are halocarbons. Chlorine (Cl) is more electronegative than bromine (Br), so CCl4 would have stronger intermolecular forces compared to CBr4. Hence, CCl4 would have a lower vapor pressure at a given temperature than CBr4. The answer is CBr4, CCl4.
3. H2O, H2S: Both substances are hydrides. Water (H2O) has stronger hydrogen bonds compared to hydrogen sulfide (H2S), which leads to stronger intermolecular forces and lower vapor pressure at a given temperature. Hence, H2S would have a higher vapor pressure than H2O. The answer is H2S, H2O.
4. CH3CH2OH, CH3-O-CH3: Both substances are alcohols. CH3CH2OH (ethanol) has stronger intermolecular forces due to the presence of a hydrogen bond compared to CH3-O-CH3 (dimethyl ether). Therefore, CH3-O-CH3 would have a higher vapor pressure at a given temperature. The answer is CH3-O-CH3, CH3CH2OH.
5. Xe, Kr: Both substances are noble gases. As we move down the noble gas group, the size of the atoms increases, resulting in weaker intermolecular forces and higher vapor pressure at a given temperature. Therefore, Kr (krypton) would have a higher vapor pressure compared to Xe (xenon). The answer is Kr, Xe.