Question

In which of the following pairs, the critical temperature of latter gascous species is higher than the first? (a) \(\mathrm{CO}_{2}, \mathrm{H}_{2}\) (b) \(\mathrm{H}_{2}, \mathrm{NH}_{3}\) (c) \(\mathrm{NH}_{3}, \mathrm{He}\) (d) \(\mathrm{CO}_{2}, \mathrm{He}\)

Short answer

The correct pair is (b) \(\mathrm{H}_2, \mathrm{NH}_3\).

Step-by-step solution

Understanding Critical Temperature

The critical temperature of a substance is the temperature above which a substance cannot exist in the liquid phase, regardless of pressure. The stronger the intermolecular forces, the higher the critical temperature.

Identify Intermolecular Forces

Examine each gas: 1. \( ext{CO}_2\) has dispersion forces. 2. \( ext{H}_2\) has dispersion forces, but weaker than \( ext{CO}_2\). 3. \( ext{NH}_3\) has hydrogen bonding, giving it a much higher critical temperature. 4. \( ext{He}\) has very weak dispersion forces since it is a noble gas.

Analyze Each Pair

For each pair:(a) \( ext{CO}_2\) vs \( ext{H}_2\): \( ext{CO}_2\) has higher forces.(b) \( ext{H}_2\) vs \( ext{NH}_3\): \( ext{NH}_3\) has significantly stronger hydrogen bonds.(c) \( ext{NH}_3\) vs \( ext{He}\): \( ext{NH}_3\) has higher due to hydrogen bonding.(d) \( ext{CO}_2\) vs \( ext{He}\): \( ext{CO}_2\) has stronger forces.

Conclusion

Compare each pair and check if the critical temperature of the second is higher than the first: only in pair (b) \( ext{H}_2, ext{NH}_3\) does \( ext{NH}_3\) have a higher critical temperature.

Key concepts

Intermolecular Forces

Intermolecular forces are the forces that bind molecules together and play a crucial role in determining the physical properties of substances, such as boiling and melting points, and critical temperatures. These forces dictate how molecules interact and stick together. The stronger the intermolecular forces, the more energy is required to separate the molecules, leading to higher critical temperatures. Intermolecular forces can be divided into several types:

• Hydrogen bonding
• Dipole-dipole interactions
• Dispersion (London) forces

Each substance has different intermolecular forces impacting its critical temperature.
In general, stronger intermolecular forces correspond to higher critical temperatures, which explains why different substances behave uniquely when they reach their critical temperature.

Hydrogen Bonding

Hydrogen bonding is one of the strongest types of intermolecular forces, occurring between molecules where a hydrogen atom is directly bonded to a highly electronegative atom like nitrogen, oxygen, or fluorine. Consider a molecule such as ammonia ( NH_3 ) that exhibits hydrogen bonding. This bond is much stronger compared to dispersion forces.
In the context of critical temperature, substances with hydrogen bonding, such as ammonia, tend to have higher critical temperatures due to these strong attractions. This is why, when comparing nH_2 and NH_3 , ammonia's critical temperature surpasses that of hydrogen because of these significant hydrogen bonds. Hydrogen bonds increase the energy required to transition from liquid to gas phases, elevating the critical temperature.

Dispersion Forces

Dispersion forces, also known as London forces, are the weakest type of intermolecular forces. They are present in all molecules but are the only type of intermolecular force in nonpolar substances like noble gases. These forces arise from temporary fluctuations in the electron distribution within molecules, creating transient dipoles. Heavier and larger atoms or molecules usually have stronger dispersion forces. For gases like carbon dioxide ( CO_2 ) and helium ( He ), dispersion forces play a pivotal role in determining their critical temperatures. CO_2 , being a larger molecule than He , has stronger dispersion forces and hence a higher critical temperature. Dispersion forces require less energy to overcome compared to hydrogen bonds, explaining why substances reliant on dispersion forces generally have lower critical temperatures compared to those that exhibit hydrogen bonding.