Chapter 12, Problem 12.55UTC

Interpretation Introduction

Interpretation:

The given properties must relate to the two given compounds.

Concept introduction:

• $\text{KCl}$ is an ionic compound and butane is an organic covalent compound.
• $\text{KCl}$ is solid whereas butane at room temperature is a gas.

Answer to Problem 12.55UTC

Solution: a) Butane melts at $-{138}^0\text{C}$

b) Butane burns vigorously in air.

c) $\text{KCl}$ melts at ${770}^0\text{C}$

d) $\text{KCl}$ contains ionic bonds.

e) Butane is a gas at room temperature.

Explanation of Solution

a)

Butane being organic covalent compound, its melting and boiling point is low. There is very weak van der Waal force. So, it melts at very low temperature which is $-{138}^0\text{C}$

b)

As butane is a fuel so it undergoes combustion in presence of oxygen. It produces carbon dioxide and water along with huge amount of heat energy as per the following equation.

  $$

  ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}+6.5{\text{O}}_{\text{2}}\to 4{\text{CO}}_{\text{2}}+5{\text{H}}_{\text{2}}\text{O}$

c)
$\text{KCl}$ being ionic compound, it has very strong force of electrostatic attraction and so it melts at very high temperature which is ${770}^0\text{C}$

d)
$\text{KCl}$ is an ionic compound having ionic bond between ${\text{K}}^{+}$ and ${\text{Cl}}^{-}$

$\text{KCl}$ is formed by complete electron transfer from $\text{K}$ to $\text{Cl}$ .

e)

Butane has very weak intermolecular force which is van der Waal force. Thus, it is a gas at room temperature.

Conclusion

Thus, all the properties are matched with respective compounds.