Chapter 5, Problem 58PS

(a)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ for the overall process is the sum of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ values of all those reactions-Hess’s law.

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\Delta \text{H}}{\text{number of moles }}$

Answer to Problem 58PS

The enthalpy change for the formation is $\text{-4}\text{.6kJ}$

Explanation of Solution

Given mass is $\text{1}\text{.0}\text{g}$

The enthalpy of formation is $\text{ -296}\text{.8 KJ/Mol}$ 

$\text{S}\text{+}{\text{O}}_{\text{2}}\to {\text{SO}}_{\text{2}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{ -296}\text{.8 KJ/Mol}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\Delta \text{H}}{\text{number of moles }}$

${\text{Δ}}_{\text{r}}\text{H}$=$\text{1}\text{.0g}\text{(1mol÷}64.066 \text{g)(-296}\text{.8kJ/mol}÷\text{1mol)}$ =$\text{-4}\text{.6kJ}$

So, the enthalpy change for the formation is $\text{-4}\text{.6kJ}$

(b)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ for the overall process is the sum of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ values of all those reactions-Hess’s law.

Answer to Problem 58PS

The enthalpy change for the formation is $\text{18}\text{.14kJ}$

Explanation of Solution

Given mass is $\text{0}\text{.20}\text{mol}$

The enthalpy of formation is $\text{90}\text{.7kJ}$ 

  $\text{HgO}\stackrel{}{\to }\text{2Hg}\text{+}{\text{O}}_{\text{2}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$= $\text{90}\text{.7kJ}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\Delta \text{H}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$= $\text{90}\text{.7kJ/mol}$ (reverse rxn)

  $\left(\frac{\text{90}\text{.7kJ}}{\text{1mol}}\right)\text{(0}\text{.20mol)}$ $\text{=}$$\text{18}\text{.14kJ}$

So, the change in enthalpy is $\text{18}\text{.14kJ}$

(c)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ for the overall process is the sum of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ values of all those reactions-Hess’s law

Answer to Problem 58PS

The change in enthalpy is $\text{-7}\text{.3kJ}$

Explanation of Solution

Given mass is $\text{2}\text{.4}\text{g}$

The enthalpy of formation is ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-411}\text{.12kJ/mol}$

  $\text{2Na +}{\text{Cl}}_{\text{2}}\stackrel{}{\to }\text{NaCl}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-411}\text{.12kJ/mol}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\Delta \text{H}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-46}\text{.0 kJ/mol}$

  $\text{2}\text{.40g×}\frac{\text{1mol}}{\text{15}\text{.01464g}}\text{=}0.159\text{mol}$

Energy change = $\text{-46}\text{.0 kJ/mol}\times \text{0}\text{.159mol}$=$\text{-7}\text{.3kJ}$

So, the change in enthalpy is $\text{-7}\text{.3kJ}$

(d)

Interpretation Introduction

Interpretation:

The enthalpy change for the formation the following reaction has to be determined.

Concept Introduction:

If a reaction proceeds in two or more other reaction,${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ for the overall process is the sum of ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$ values of all those reactions-Hess’s law.

Answer to Problem 58PS

The change in enthalpy is $\text{-4}\text{.13 kJ}$

Explanation of Solution

Given mass is $\text{1}\text{.05}\text{×}{\text{10}}^{\text{-2}}\text{mol}$

The enthalpy of formation is ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-393}\text{.5 kJ/mol }$

  $\text{C +}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{CO}}_{\text{2}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-393}\text{.5 kJ/mol }$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\Delta \text{H}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-393}\text{.5 kJ/mol }$

Energy change =$\left(\frac{\text{-393}\text{.5 kJ/mol }}{\text{1mol}}\right)\text{(1}{\text{.05×10}}^{\text{-2}}\text{mol)}$=$\text{-4}\text{.13 kJ}$

So, the change in enthalpy is $\text{-4}\text{.13 kJ}$