Chapter 17, Problem 26E

(a)

Interpretation Introduction

Interpretation:

The sign and magnitude of $\Delta H_1,\Delta H_2,\Delta H_3\text{ and }\Delta H_{so\ln }$ for the solute-solvent combinations needs to be determined.

  

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

Answer to Problem 26E

The values for $\Delta H_1=Small;\Delta H_2=\text{Large; }\Delta H_3=Small$

Therefore $\Delta H_{so\ln }=\text{Large positive}$ and hence no solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

1. The solutes separating into individual components require energy making it an endothermic reaction.
2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as

  $\Delta H_{so\ln }=\Delta H_1+\Delta H_2+\Delta H_3$

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Result
Non polar solute, Non polar solvent	Small	Small	Small	Small	Solution will form
Non-polar solute, polar solvent	Small	Large	Small	Large, positive	No-solution will form
Polar solute, Non polar solvent	Large	Small	Small	Large negative	No solution will form
Polar solute-polar solvent	Large	Large	Large, negative	small	Solution will form

In this reaction, acetone reacts with water and as water is polar in nature, the values of each enthalpy changes are depicted as

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Outcome
Polar solute, Non polar solvent	Large	Small	Small	Large negative	No solution will form

(b)

Interpretation Introduction

Interpretation:

The sign and magnitude of $\Delta H_1,\Delta H_2,\Delta H_3\text{ and }\Delta H_{so\ln }$ for the solute-solvent combinations needs to be determined.

  

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

Answer to Problem 26E

The values for $\Delta H_1=\text{Large};\Delta H_2=\text{Large; }\Delta H_3=\text{Large (negative)}$

Therefore $\Delta H_{so\ln }=\text{Small}$ and hence solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

1. The solutes separating into individual components require energy making it an endothermic reaction.
2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as

  $\Delta H_{so\ln }=\Delta H_1+\Delta H_2+\Delta H_3$

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Result
Non polar solute, Non polar solvent	Small	Small	Small	Small	Solution will form
Non-polar solute, polar solvent	Small	Large	Small	Large, positive	No-solution will form
Polar solute, Non polar solvent	Large	Small	Small	Large negative	No solution will form
Polar solute-polar solvent	Large	Large	Large, negative	small	Solution will form

In this reaction, ethanol reacts with water and as water is polar in nature, the values of each enthalpy changes are depicted as

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Outcome
Polar solute-polar solvent	Large	Large	Large, negative	small	Solution will form

(c)

Interpretation Introduction

Interpretation:

The sign and magnitude of $\Delta H_1,\Delta H_2,\Delta H_3\text{ and }\Delta H_{so\ln }$ for the solute-solvent combinations needs to be determined.

  

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This is possible when there is disruption between the solute -solute and solvent-solvent interaction.

Answer to Problem 26E

The values for $\Delta H_1=\text{Small};\Delta H_2=\text{Small; }\Delta H_3=\text{Small}$

Therefore $\Delta H_{so\ln }=\text{Small}$ and hence solution will form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

1. The solutes separating into individual components require energy making it an endothermic reaction.
2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which is depicted as:

  $\Delta H_{so\ln }=\Delta H_1+\Delta H_2+\Delta H_3$

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Result
Non polar solute, Non polar solvent	Small	Small	Small	Small	Solution will form
Non-polar solute, polar solvent	Small	Large	Small	Large, positive	No-solution will form
Polar solute, Non polar solvent	Large	Small	Small	Large negative	No solution will form
Polar solute-polar solvent	Large	Large	Large, negative	small	Solution will form

In this reaction, heptane reacts with hexane and as both is non-polar in nature; the values of each enthalpy changes are depicted as:

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Outcome
Non polar solute, Non polar solvent	Small	Small	Small	Small	Solution will form

(d)

Interpretation Introduction

Interpretation:

The sign and magnitude of $\Delta H_1,\Delta H_2,\Delta H_3\text{ and }\Delta H_{so\ln }$ for the solute-solvent combinations needs to be determined.

  

Concept Introduction :

The intermolecular forces in a solute can be broken down by new interactions from the solution as each solute particle will be surrounded by solvent particles in a solution. This can happen when there is disruption between the solute-solute and solvent-solvent interaction.

Answer to Problem 26E

The values for $\Delta H_1=\text{Small};\Delta H_2=\text{Large; }\Delta H_3=\text{Large positive}$

Therefore $\Delta H_{so\ln }=\text{Large positive }$ and hence solution will not form

Explanation of Solution

The process of formation of solution takes place in 3 main steps

1. The solutes separating into individual components require energy making it an endothermic reaction.
2. The intermolecular forces in solvent must be such that it can make space for solute which requires energy making it an endothermic reaction.
3. To allow the solvent and solute molecules to interact which absorbs energy making it an exothermic reaction.

The formation of solution involves enthalpy changes which are depicted as:

  $\Delta H_{so\ln }=\Delta H_1+\Delta H_2+\Delta H_3$

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Result
Non polar solute, Non polar solvent	Small	Small	Small	Small	Solution will form
Non-polar solute, polar solvent	Small	Large	Small	Large, positive	No-solution will form
Polar solute, Non polar solvent	Large	Small	Small	Large negative	No solution will form
Polar solute-polar solvent	Large	Large	Large, negative	small	Solution will form

In this reaction, heptane reacts with water wherein heptane is non-polar and water is polar in nature. The values of each enthalpy changes are depicted as:

	$\Delta H_1$	$\Delta H_2$	$\Delta H_3$	$\Delta H_{so\ln }$	Result
Non-polar solute, polar solvent	Small	Large	Small	Large, positive	Solution will not form.