Chapter 9, Problem 34E

(a)

Interpretation Introduction

Interpretation: The heat evolved for the production of 1.00 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$ for the given reaction needs to be determined.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

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 34E

Energy released = $\text{-286 kJ}$

Explanation of Solution

Given reaction: ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

According to given reaction; 2 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$ are formed by the release of 572 kJ energy.

Hence the energy required for 1.00 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$ :

  $\text{1}{\text{.0 moles H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ×}\frac{\text{-572 kJ}}{\text{2 moles}}\text{= -286 kJ}$

(b)

Interpretation Introduction

Interpretation: The heat evolved during the reaction of 4.03 g hydrogen gas with excess of oxygen for the given reaction needs to be determined.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

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 34E

Energy released = $\text{- 572 kJ}$

Explanation of Solution

Given reaction: ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

Mass of ${\text{H}}_{\text{2}}$ = 4.03 g

Molar mass of ${\text{H}}_{\text{2}}$ = 2.01 g/mol

Calculate moles of ${\text{H}}_{\text{2}}$ = $\frac{\text{4}\text{.03 g}}{\text{2}\text{.01 g/mol}}\text{= 2}\text{.00 moles}$

According to given reaction; 2 moles of ${\text{H}}_{\text{2}}$ reacted to release of 572 kJ energy.

Hence the energy required for 2.00 moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{(l)}}$ = 572 kJ

(c)

Interpretation Introduction

Interpretation: The heat evolved during the reaction of 186 g ${\text{ O}}_{\text{2}}$ gas with excess of ${\text{H}}_{\text{2}}$ for the given reaction needs to be determined.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

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 34E

Energy released = $\text{ -3}\text{.32 }\times {\text{10}}^{\text{3}}\text{ kJ}$

Explanation of Solution

Given reaction: ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

Mass of ${\text{ O}}_{\text{2}}$ = 186 g

Molar mass of ${\text{ O}}_{\text{2}}$ = 32.0 g/mol

Calculate moles of ${\text{ O}}_{\text{2}}$ = $\frac{\text{186 g}}{\text{32}\text{.0 g/mol}}\text{= 5}\text{.81 moles}$

According to given reaction; 1 moles of ${\text{ O}}_{\text{2}}$ reacted to release of 572 kJ energy. Hence energy released from 5.81 moles of ${\text{ O}}_{\text{2}}$ :

  $\text{5}{\text{.81 moles O}}_{\text{2}}{}_{\text{(g)}}\text{ ×}\frac{\text{-572 kJ}}{\text{1 moles}}\text{= -3}\text{.32 }\times {\text{10}}^{\text{3}}\text{ kJ}$

(d)

Interpretation Introduction

Interpretation: The heat evolved during the explosion of Hindenburg explosion which contains $\text{2}\text{.0 }\times {\text{ 10}}^{\text{8}}\text{ L}$ of hydrogen gasat 1.0 atm and $\text{25°C}$ needs to be determined.

  ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

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 34E

Energy released = $\text{-2}\text{.34 }\times {\text{10}}^9\text{ kJ}$

Explanation of Solution

Given reaction: ${\text{2H}}_{\text{2}}{}_{\text{(g)}}{\text{+ O}}_{\text{2}}{}_{\text{(g)}}\stackrel{}{\to }{\text{2 H}}_{\text{2}}{\text{O}}_{\text{(l)}}\text{ ΔH = -572 kJ}$

Volume of ${\text{H}}_{\text{2}}$ = $\text{2}\text{.0 }\times {\text{ 10}}^{\text{8}}\text{ L}$

Pressure = 1.0 atm

Temperature = $\text{25°C}\text{.}$

Molar mass of ${\text{H}}_{\text{2}}$ = 2.01 g/mol

Calculate moles of ${\text{H}}_{\text{2}}$ = $\text{n=}\frac{\text{P×V}}{\text{R×T}}\text{=}\frac{\text{1}\text{.0atm×2}{\text{.0 × 10}}^{\text{8}}\text{ L}}{\text{0}\text{.0821L}\text{.atm/K}\text{.mol×298K}}=\text{8}{\text{.17× 10}}^{\text{6}}\text{moles}$

According to given reaction; 2 moles of ${\text{H}}_{\text{2}}$ reacted to release of 572 kJ energy. Hence energy released from $\text{8}{\text{.17× 10}}^{\text{6}}\text{moles}$ of ${\text{H}}_{\text{2}}$ :

  $\text{8}{\text{.17× 10}}^{\text{6}}{\text{moles H}}_{\text{2}}{}_{\text{(g)}}\text{ ×}\frac{\text{-572 kJ}}{\text{2 moles}}\text{= -2}\text{.34 }\times {\text{10}}^9\text{ kJ}$