Chapter 5, Problem 57PS

Use standard enthalpies of formation in Appendix L to calculate enthalpy changes for the following:

1. (a) 1.0 g of white phosphorus burns, forming P₄O_l0(s)
2. (b) 0.20 mol of NO(g) decomposes to N₂(g) and O₂(g)
3. (c) 2.40 g of NaCl(s) is formed from Na(s) and excess Cl₂(g)
4. (d) 250 g of iron is oxidized with oxygen to Fe₂O₃(s)

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 57PS

The enthalpy change for the formation is $\text{-24}\text{.26}\text{kJ}$

Explanation of Solution

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

The enthalpy of formation is $\text{-2984}\text{kJ/mol}$ 

${\text{P}}_{\text{4}}\text{+}{\text{5O}}_{\text{2}}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-2984kJ/mol}$

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

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

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

${\text{Δ}}_{\text{r}}\text{H}$=$\text{1}\text{.0g}\text{(1mol÷1}\text{.23g)(-2984kJ/mol}÷\text{1mol)}$ =$\text{-24}\text{.26}\text{kJ}$

So, the enthalpy change for the formation is $\text{-24}\text{.26}\text{kJ}$

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 57PS

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

Explanation of Solution

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

The enthalpy of formation is -$\text{90}\text{.29kJ/mol}$ 

  $\text{NO}\stackrel{}{\to }{\text{N}}_{\text{2}}\text{+}{\text{O}}_{\text{2}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=-$\text{90}\text{.29kJ/mol}$

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{-}$$\text{90}\text{.29kJ/mol}$ (reverse rxn)

${\text{Δ}}_{\text{r}}\text{H}$= $\left(\frac{\text{-90}\text{.29kJ}}{\text{1mol}}\right)\text{(0}\text{.20mol)}$ $\text{=}$$\text{-18kJ}$

So, the change in enthalpy is $\text{-18}\text{J}$

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 57PS

The change in enthalpy is $\text{16}\text{.9kJ}$

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:

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

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

  $\text{2}\text{.40g×}\frac{\text{1mol}}{\text{58}\text{.44g}}\text{=}\text{0}\text{.0411mol}$

Energy change =$\text{-411}\text{.12kJ/mol×0}\text{.0411mol}$ =$\text{-16}\text{.9}\text{kJ}$

So, the change in enthalpy is $\text{16}\text{.9kJ}$

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 57PS

The change in enthalpy is $\text{-1}{\text{.8×10}}^{\text{3}}\text{kJ}$

Explanation of Solution

Given mass is $\text{250}\text{g}$

The enthalpy of formation is ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-824}\text{.2kJ}$

  $\text{Fe +}{\text{O}}_{\text{2}}\stackrel{}{\to }{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-824}\text{.2kJ}$

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

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

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

  ${\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}$=$\text{-824}\text{.2kJ/mol}$

Energy change =$\text{250g×}\frac{\text{1mol}\text{Fe}}{\text{55}\text{.8847}}\text{×}\frac{\text{1mol}{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{\text{2mol}}\text{×}\frac{\text{-824}\text{.2kJ}}{\text{1mol}}$=$\text{-1}{\text{.8×10}}^{\text{3}}\text{kJ}$

So, the change in enthalpy is $\text{-1}{\text{.8×10}}^{\text{3}}\text{kJ}$