Chapter 9, Problem 35E

(a)

Interpretation Introduction

Interpretation: The heat released during the reaction of 4.00 moles of iron to react with excess of O₂ for the given reaction needs to be determined.

  ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

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 35E

Energy released = 1652 kJ

Explanation of Solution

Given reaction: ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

According to given reaction; 4 moles of Fe react to release 1652 kJ energy.

(b)

Interpretation Introduction

Interpretation: The heat released during the reaction of 1.00 moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is produced needs to be determined.

  ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

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 35E

Energy released = $\text{ -826 kJ}$

Explanation of Solution

Given reaction: ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

Moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ = 1.0 moles

According to given reaction; 2 moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ produced with release of 1652 kJ energy. Hence for 1.00 moles the energy must be:

  $\text{1}{\text{.0 moles Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{×}\frac{\text{-1652kJ}}{\text{2moles}}\text{= -826 kJ}$

(c)

Interpretation Introduction

Interpretation: The heat released during the reaction of 1.00 g of $\text{Fe}$ reacts with excess of ${\text{O}}_{\text{2}}$ needs to be determined.

  ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

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 35E

Energy released = $\text{-7}\text{.39 kJ}$

Explanation of Solution

Given reaction: ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

Mass of Fe = 1.00 g

Molar mass of Fe = 55.8 g/mol

Calculate moles of Fe = $\frac{\text{1}\text{.00g}}{\text{55}\text{.8g/mol}}\text{= 0}\text{.0179moles}$

According to given reaction; 4 moles of Fe reacts to release of 1652 kJ energy. Hence for 0.0179 moles the energy must be:

  $\text{0}{\text{.0179 moles Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{× }\frac{\text{-1652kJ}}{\text{4 moles}}\text{= -7}\text{.39 kJ}$

(d)

Interpretation Introduction

Interpretation: The heat released during the reaction of 10.0 g of $\text{Fe}$ and 2.00 g of ${\text{O}}_{\text{2}}$ are reacted needs to be determined.

  ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

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 35E

Since less energy is released from ${\text{O}}_{\text{2}}$ and it is in less amount hence it must be limiting reagent and will release 34.4 kJ energy.

Explanation of Solution

Given reaction: ${\text{4Fe}}_{\text{(s) }}{\text{+ 3 O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 Fe}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{ ΔH = -1652 kJ}$

Mass of Fe = 10.0 g

Molar mass of Fe = 55.8 g/mol

Calculate moles of Fe = $\frac{\text{10}\text{.0g}}{\text{55}\text{.8g/mol}}\text{= 0}\text{.179moles}$

Molar mass of ${\text{O}}_{\text{2}}$ = 32 g/mol

Calculate moles of ${\text{O}}_{\text{2}}$ = $\frac{\text{2}\text{.0g}}{\text{32 g/mol}}\text{= 0}\text{.0625 moles}$

According to given reaction; 4 moles of Fe reacts to release of 1652 kJ energy. Hence energy from both given reactants:

  $\text{0}{\text{.179 moles Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{× }\frac{\text{-1652kJ}}{\text{4 moles}}\text{= -73}\text{.9 kJ}$

  $\text{0}{\text{.0625 moles O}}_{\text{2}}\text{× }\frac{\text{-1652kJ}}{\text{3 moles}}\text{= -34}\text{.4 kJ}$

Since less energy is released from ${\text{O}}_{\text{2}}$ and it is in less amount hence it must be limiting reagent and will release 34.4 kJ energy.