Q12.51P - Chemistry: Molecular Nature Of Matter And Change

Question

Which substance has the lower boiling point? Explain.

a.${\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2{\mathbf{CH}}_3$ OR 

b.$\mathbf{NaBr}\mathbf{ }\mathbf{ }\mathbf{OR}\mathbf{ }{\mathbf{PBr}}_3$

c.${\mathbf{H}}_2\mathbf{O}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{OR}\mathbf{ }\mathbf{ }\mathbf{HBr}$

Step-by-Step Solution

Answer

a.${\mathbf{CH}}_3{\mathbf{CH}}_2{\mathbf{CH}}_2{\mathbf{CH}}_3$

b.${\mathbf{PBr}}_3$

c.$\mathbf{HBr}$

Step 1: Boiling point

Boiling point is the temperature at which the phase transition takes place from liquid to gaseous phase. At this temperature, the vapor pressure becomes equal to the atmospheric pressure. Also, boiling point of any liquid depends on the interactions between molecules.

 

Step 2: Relative strengths

1. Both butane and cyclobutane has weak London dispersive forces for their intermolecular interactions which depend on the surface area.
2. Br atom being larger in size have a dipolar behavior. Now, Na⁺ being an ion with a comparatively smaller size has ion-dipolar interactions in the compound $\mathbf{NaBr}$ , whereas  ${\mathbf{PBr}}_3$ has dipole-dipole interactions.
3. Considering both   ${\mathbf{H}}_2\mathbf{O}$ and  $\mathbf{HBr}$ , both O and Br are electronegative in nature, but considering their electronegative strength, O is more electronegative than Br. Also, O being smaller in size than Br, ionic interactions are more in the case of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ than  $\mathbf{HBr}$

Step 3: Effect of these interactions on boiling point

1. As the ring shape allows for a larger contact area, the interactions will be more for cyclobutane than n-butane. Thus, the former has a high boiling point than the latter.
2. As ion-dipole interactions are stronger than dipole-dipole interactions, $\mathbf{NaBr}$ has higher boiling points than ${\mathbf{PBr}}_3$ .
3. As H-bonding in   ${\mathbf{H}}_2\mathbf{O}$ are stronger than $\mathbf{HBr}$ , the boiling point of  ${\mathbf{H}}_2\mathbf{O}$ is stronger than $\mathbf{HBr}$.