Chapter 8, Problem 43QAP

Use Table 8.3 to obtain AHO for the following thermochemical equations:

(a) $\text{Mg}{\left(\text{OH}\right)}_2\left(s\right)+2{\text{NH}}_4{}^{+}\left(aq\right)\to {\text{Mg}}^{2+}\left(aq\right)+2{\text{NH}}_3\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

(b) $\text{PbO}\left(s\right)+\text{C}\left(s\right)\to \text{CO}\left(g\right)+\text{Pb}\left(s\right)$

(c) $\text{Mn}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+{\text{SO}}_4{}^{2-}\to {\text{Mn}}^{2+}\left(aq\right)+{\text{SO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(l\right)$

Interpretation Introduction

(a)

Interpretation:

The value of ΔH° for the given chemical equation needs to be calculated.

${\text{Mg(OH)}}_{\text{2}}{\text{(s) + 2NH}}_4^{+}\text{(aq)}\to {\text{Mg}}^{\text{2+}}{\text{(aq) + 2NH}}_{\text{3}}{\text{(g) + 2H}}_{\text{2}}\text{O(l)}$

Concept introduction:

The change in standard enthalpy for a reaction, ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

$\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(1)}$

Where, n_p and n_r are the number of moles of the products and reactants

${\text{The values of ΔH}}_{\text{f}}^{\text{0}}$ for different reactant and products can be calculated under standard temperature and pressure conditions.

Answer to Problem 43QAP

ΔH° = +58.9 kJ

Explanation of Solution

The given chemical equation is as follows:

${\text{Mg(OH)}}_{\text{2}}{\text{(s) + 2NH}}_4^{+}\text{(aq)}\to {\text{Mg}}^{\text{2+}}{\text{(aq) + 2NH}}_{\text{3}}{\text{(g) + 2H}}_{\text{2}}\text{O(l)}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1):

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mg}}^{\text{2+}}{\text{(aq)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(NH}}_{\text{3}}{\text{(g)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_2{\text{O(l))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mg(OH)}}_{\text{2}}{\text{(s)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(NH}}_4^{+}\text{(aq)) ]}\end{array}$

Putting the values from table 8.3: $\begin{array}{l}\Delta {\text{H}}^0\text{ = [1 mole }\times (-466.8)\text{ kJ/mol + 2 mole }\times \text{(-46}\text{.1 kJ/mol) + 2 mole }\times \text{(-285}\text{.8 kJ/mol)]}\\ \text{ - [1 mole }\times \text{(}-924.5\text{ kJ/mol) + 2 mole }\times (-132.5\text{ kJ/mol)]}\\ \text{ = + 58}\text{.9 kJ }\end{array}$

Interpretation Introduction

(b)

Interpretation:

The value of ΔH° for the given chemical equation needs to be calculated.

$\text{PbO(s) + C(s)}\to \text{CO(g) + Pb(s)}$

Concept introduction:

The change in standard enthalpy for a reaction, ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

$\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(1)}$

Where, n_p and n_r are the number of moles of the products and reactants.

${\text{The values of ΔH}}_{\text{f}}^{\text{0}}$ for different reactant and products can be calculated under standard temperature and pressure conditions.

Answer to Problem 43QAP

ΔH° = +108.5 kJ

Explanation of Solution

The given chemical equation is:

$\text{PbO(s) + C(s)}\to \text{CO(g) + Pb(s)}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1), we have:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [1ΔH}}_{\text{f}}^{\text{0}}{\text{(CO(g)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(Pb(s))] - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(PbO(s)) + 1ΔH}}_{\text{f}}^{\text{0}}\text{(C(s)) ]}\\ \end{array}$

Putting the values from table 8.3 thus,

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\left[\begin{array}{l}\left(\text{1 mole }\times (-110.5)\text{ kJ/mol + 1 mole }\times \text{(0 kJ/mol)}\right)\\ -\left(\text{1 mole }\times \text{(}-219.0\text{ kJ/mol) + 1 mole }\times (0\text{ kJ/mol}\right)\end{array}\right]\\ \text{ = + 108}\text{.5 kJ }\end{array}$

Interpretation Introduction

(c)

Interpretation:

The value of ΔH° for the given chemical equation needs to be calculated.

${\text{Mn(s) + 4H}}^{+}{\text{(aq) + SO}}_4^{2-}(\text{aq})\to {\text{Mn}}^{\text{2+}}{\text{(aq) + SO}}_2{\text{(g) + 2H}}_{\text{2}}\text{O(l)}$

Concept introduction:

The change in standard enthalpy for a reaction, ${\text{ΔH}}^{\text{0}}$ is given in terms of the difference in the standard enthalpy of formation of the products and that of reactants.

$\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) ------(1)}$

Where, n_p and n_r are the number of moles of the products and reactants.

${\text{The values of ΔH}}_{\text{f}}^{\text{0}}$ for different reactant and products can be calculated under standard temperature and pressure conditions.

Answer to Problem 43QAP

ΔH° = -179.9 kJ

Explanation of Solution

The given chemical equation is:

${\text{Mn(s) + 4H}}^{+}{\text{(aq) + SO}}_4^{2-}(\text{aq})\to {\text{Mn}}^{\text{2+}}{\text{(aq) + SO}}_2{\text{(g) + 2H}}_{\text{2}}\text{O(l)}$

The amount of heat absorbed or evolved can be given in terms of the enthalpy change for the above reaction. Based on equation (1), we have:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mn}}^{\text{2+}}{\text{(aq)) + ΔH}}_{\text{f}}^{\text{0}}{\text{(SO}}_2{\text{(g)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_2\text{O(l))] }\\ {\text{ - [1ΔH}}_{\text{f}}^{\text{0}}{\text{(Mn(s)) + 4ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 1ΔH}}_{\text{f}}^{\text{0}}{\text{(SO}}_4^{2-}\text{(aq))]}\end{array}$

Putting the values from table 8.3:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = [1 mole }\times (-220.8)\text{ kJ/mol + 1 mole }\times \text{(-296}\text{.8 kJ/mol) + 2 mole }\times \text{(-285}\text{.8 kJ/mol)]}\\ \text{ - [1 mole }\times \text{(}0\text{ kJ/mol) + 2 mole }\times (0\text{ kJ/mol) + 1 mole }\times \text{(}-909.3\text{ kJ/mol)]}\\ \text{ = - 179}\text{.9 kJ }\end{array}$