Chapter 9, Problem 103AE

(a)

Interpretation Introduction

Interpretation:The work done by or on the system or surrounding or no work done for the given reaction needs to be determined:

  ${\text{N}}_{\text{2}}{}_{\text{(g) }}\stackrel{}{\to }{\text{ N}}_{\text{2}}{}_{\text{(l)}}$

Concept Introduction: Thermodynamic is a branch of chemistry that deals with the energy change with the system and surroundings. It indicates the energy conversion and transfer between system and surroundings.

At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V.

At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p. The relation between $\text{ΔH}$ and $\text{ΔE}$ can be written as:

  $\text{ΔH = ΔE + }\Delta \text{(PV) = ΔE + }\Delta \text{nRT}$

Answer to Problem 103AE

Nitrogen gas changes to liquid nitrogen hence volume decrease that causes surrounding work on the system.

Explanation of Solution

Given reaction: ${\text{N}}_{\text{2}}{}_{\text{(g) }}\stackrel{}{\to }{\text{ N}}_{\text{2}}{}_{\text{(l)}}$

To predict the sign of w and its direction, the coefficients of all the gaseous products and reactants must be checked. If reaction forms more moles of gaseous products compared to gaseous reactants the reaction will expand the volume and the system does work on the surroundings. When the reaction forms more moles of gaseous reactant compare to gaseous products, the reaction will contract the volume and the surrounding will work on the system. If the gaseous reactant and product numbers are same, it means no change in volume and no work will be done.

Here, nitrogen gas changes to liquid nitrogen hence volume decrease that causes surrounding work on the system.

(b)

Interpretation Introduction

Interpretation:The work done by or on the system or surrounding or no work done for the given reaction needs to be determined.

  ${\text{CO}}_{\text{(g) }}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(g)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{}_{\text{(g)}}{\text{ + CO}}_{\text{2(g)}}$

Concept Introduction: Thermodynamic is a branch of chemistry that deals with the energy change with the system and surroundings. It indicates the energy conversion and transfer between system and surroundings.

At constant volume, the change in heat for a system to change the internal energy is represented as ΔE or q_V.

At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p. The relation between $\text{ΔH}$ and $\text{ΔE}$ can be written as:

  $\text{ΔH = ΔE + }\Delta \text{(PV) = ΔE + }\Delta \text{nRT}$

Answer to Problem 103AE

In the given reaction,the number of gaseous reactants and products is same hence no change in volume. It means no work is done on the system and surroundings.

Explanation of Solution

Given reaction: ${\text{CO}}_{\text{(g) }}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(g)}}\stackrel{}{\to }{\text{ H}}_{\text{2}}{}_{\text{(g)}}{\text{ + CO}}_{\text{2(g)}}$

To predict the sign of w and its direction, the coefficients of all the gaseous products and reactants must be checked. If reaction forms more moles of gaseous products compare to gaseous reactants the reaction will expand the volume and the system does work on the surroundings. When the reaction forms more moles of gaseous reactant compared to gaseous products the reaction will contract the volume and the surrounding will work on the system. If the gaseous reactant and product numbers are same, it means no change in volume and no work will be done.

Here,number of gaseous reactants and products is same hence no change in volume. It means no work is done on the system and surroundings.

(c)

Interpretation Introduction

Interpretation:The work done by or on the system or surrounding or no work done for the given reaction needs to be determined.

  ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{}_{\text{(s) }}{\text{+ 6 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ 3 Ca(OH)}}_{\text{2(s)}}{\text{ + 2 PH}}_{\text{3(g)}}$

Concept Introduction: Thermodynamic is a branch of chemistry that deals with the energy change with the system and surroundings. It indicates the energy conversion and transfer between system and surroundings.

At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V.

At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p. The relation between $\text{ΔH}$ and $\text{ΔE}$ can be written as:

  $\text{ΔH = ΔE + }\Delta \text{(PV) = ΔE + }\Delta \text{nRT}$

Answer to Problem 103AE

The number of gaseous products is formed from solid and liquid reactant hence volume increases during the reaction that causes expansion of system and work done on the surroundings.

Explanation of Solution

Given reaction: ${\text{Ca}}_{\text{3}}{\text{P}}_{\text{2}}{}_{\text{(s) }}{\text{+ 6 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{ 3 Ca(OH)}}_{\text{2(s)}}{\text{ + 2 PH}}_{\text{3(g)}}$

To predict the sign of w and its direction, the coefficients of all the gaseous products and reactants must be checked. If reaction forms more moles of gaseous products compare to gaseous reactants the reaction will expand the volume and the system does work on the surroundings. When the reaction forms more moles of gaseous reactant compare to gaseous products the reaction will contract the volume and the surrounding will work on the system. If the gaseous reactant and product numbers are same it means no change in volume and no work will be done.

Here number of gaseous products is formed from solid and liquid reactant hence volume increases during the reaction that causes expansion of system and work is done on the surroundings.

(d)

Interpretation Introduction

Interpretation:The work done by or on the system or surrounding or no work done for the given reaction needs to be determined.

  ${\text{2 CH}}_{\text{3}}{\text{OH}}_{\text{(l) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Concept Introduction: Thermodynamic is a branch of chemistry that deals with the energy change with the system and surroundings. It indicates the energy conversion and transfer between system and surroundings.

At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V.

At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p. The relation between $\text{ΔH}$ and $\text{ΔE}$ can be written as:

  $\text{ΔH = ΔE + }\Delta \text{(PV) = ΔE + }\Delta \text{nRT}$

Answer to Problem 103AE

  ${\text{2 CH}}_{\text{3}}{\text{OH}}_{\text{(l) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

Here the number of gaseous products is less than reactant hence volume decreases during the reaction that causes contraction of system and work done by the surroundings.

Explanation of Solution

Given reaction: ${\text{2 CH}}_{\text{3}}{\text{OH}}_{\text{(l) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

To predict the sign of w and its direction, the coefficients of all the gaseous products and reactants must be checked. If reaction forms more moles of gaseous products compare to gaseous reactants the reaction will expand the volume and the system does work on the surroundings. When the reaction forms more moles of gaseous reactant compared to gaseous products the reaction will contract the volume and the surrounding will work on the system. If the gaseous reactant and product numbers are same, it means no change in volume and no work will be done.

Here number of gaseous products are less than reactant hence volume decreases during the reaction that causes contraction of system and work done by the surroundings.

(e)

Interpretation Introduction

Interpretation:The work done by or on the system or surrounding or no work done for the given reaction needs to be determined.

  ${\text{I}}_{\text{2}}{}_{\text{(s) }}\stackrel{}{\to }{\text{ I}}_{\text{2(g)}}$

Concept Introduction: Thermodynamic is a branch of chemistry that deals with the energy change with the system and surroundings. It indicates the energy conversion and transfer between system and surroundings.

At constant volume the change in heat for a system to change the internal energy is represented as ΔE or q_V.

At constant pressure the change in heat for a system to change the enthalpy is represented as ΔH or q_p. The relation between $\text{ΔH}$ and $\text{ΔE}$ can be written as:

  $\text{ΔH = ΔE + }\Delta \text{(PV) = ΔE + }\Delta \text{nRT}$

Answer to Problem 103AE

  ${\text{I}}_{\text{2}}{}_{\text{(s) }}\stackrel{}{\to }{\text{ I}}_{\text{2(g)}}$

Here the number of gaseous products increases than reactant hence volume increases during the reaction that causes expansion of system and work done by the system.

Explanation of Solution

Given reaction: ${\text{2 CH}}_{\text{3}}{\text{OH}}_{\text{(l) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{ 2 CO}}_{\text{2(g)}}{\text{ + 4 H}}_{\text{2}}{\text{O}}_{\text{(l)}}$

To predict the sign of w and its direction, the coefficients of all the gaseous products and reactants must be checked. If reaction forms more moles of gaseous products compared to gaseous reactants the reaction will expand the volume and the system will work on the surroundings. When the reaction forms more moles of gaseous reactant compared to gaseous products, the reaction will contract the volume and the surrounding will work on the system. If the gaseous reactant and product numbers are same it means there is no change in volume and no work will be done.

  ${\text{I}}_{\text{2}}{}_{\text{(s) }}\stackrel{}{\to }{\text{ I}}_{\text{2(g)}}$

Here, the number of gaseous products increases than reactant hence volume increases during the reaction that causes expansion of system and work done by the system.