Chapter 8, Problem 44QAP

Use Table 8.3 to obtain ΔH° for the following chemical equations:

(a) $\text{Zn}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{H}}_2\left(g\right)$

(b) ${\text{2H}}_2\text{S}\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{SO}}_2\left(g\right)+2{\text{H}}_2\text{O}\left(g\right)$

(c) $\text{3Ni}\left(s\right)+2{\text{NO}}_3{}^{-}\left(aq\right)+{\text{8H}}^{+}\left(aq\right)\to 3{\text{Ni}}^{2+}\left(aq\right)+2\text{NO}\left(g\right)+4{\text{H}}_2\text{O}\left(l\right)$

Interpretation Introduction

(a)

Interpretation:

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

${\text{Zn (s) + 2H}}^{\text{+}}\text{(aq)}\to {\text{Zn}}^{\text{2+}}{\text{(aq) + H}}_2\text{(g) }$

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 44QAP

ΔH° = -153.9 kJ

Explanation of Solution

The given chemical equation is:

${\text{Zn (s) + 2H}}^{\text{+}}\text{(aq)}\to {\text{Zn}}^{\text{2+}}{\text{(aq) + H}}_2\text{(g) }$

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

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

Putting the values from table 8.3:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = [1 mole }\times (-153.9)\text{ kJ/mol + 1 mole }\times \text{(0 kJ/mol)]}\\ \text{ - [1 mole }\times \text{(}0\text{ kJ/mol) + 1 mole }\times (0\text{ kJ/mol)] = -153}\text{.9 kJ }\end{array}$

Interpretation Introduction

(b)

Interpretation:

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

${\text{2H}}_{\text{2}}{\text{S(g) + 3O}}_{\text{2}}\text{(g)}\to 2{\text{SO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g)}$

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 44QAP

ΔH° = -1036 kJ

Explanation of Solution

The given chemical equation is:

${\text{2H}}_{\text{2}}{\text{S(g) + 3O}}_{\text{2}}\text{(g)}\to 2{\text{SO}}_{\text{2}}{\text{(g) + 2H}}_{\text{2}}\text{O(g)}$

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:

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

Putting the values from table 8.3:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = [2 mole }\times (-296.8)\text{ kJ/mol + 2 mole }\times \text{(-241}\text{.8 kJ/mol)]}\\ \text{ - [2 mole }\times \text{(}-20.6\text{ kJ/mol) + 1 mole }\times (0\text{ kJ/mol)] = -1036 kJ }\end{array}$

Interpretation Introduction

(c)

Interpretation:

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

${\text{3Ni(s) + 8H}}^{+}{\text{(aq) + 2NO}}_3^{-}(\text{aq})\to 3{\text{Ni}}^{\text{2+}}\text{(aq) + }2{\text{NO(g) + 4H}}_{\text{2}}\text{O(l)}$

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 44QAP

ΔH° = -714.8 kJ

Explanation of Solution

The given chemical equation is:

${\text{3Ni(s) + 8H}}^{+}{\text{(aq) + 2NO}}_3^{-}(\text{aq})\to 3{\text{Ni}}^{\text{2+}}\text{(aq) + }2{\text{NO(g) + 4H}}_{\text{2}}\text{O(l)}$

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

$\begin{array}{l}\Delta {\text{H}}^0\text{ = }\sum {\text{n}}_{\text{p}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(products) - }\sum {\text{n}}_{\text{r}}{\text{ΔH}}_{\text{f}}^{\text{0}}\text{(reactants) }\\ {\text{= [3ΔH}}_{\text{f}}^{\text{0}}{\text{(Ni}}^{\text{2+}}{\text{(aq)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(NO(g)) + 4ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}_2\text{O(l))] }\\ {\text{ - [3ΔH}}_{\text{f}}^{\text{0}}{\text{(Ni(s)) + 8ΔH}}_{\text{f}}^{\text{0}}{\text{(H}}^{\text{+}}{\text{(aq)) + 2ΔH}}_{\text{f}}^{\text{0}}{\text{(NO}}_3^{-}\text{(aq))]}\end{array}$

Putting the value from table 8.3:

$\begin{array}{l}\Delta {\text{H}}^0\text{ = [3 mole }\times (-54.0)\text{ kJ/mol + 2 mole }\times \text{(90}\text{.2 kJ/mol) + 4 mole }\times \text{(-285}\text{.8 kJ/mol)]}\\ \text{ - [3 mole }\times \text{(}0\text{ kJ/mol) + 8 mole }\times (0\text{ kJ/mol) + 2 mole }\times \text{(}-205.0\text{ kJ/mol)]}\\ \text{ = - 714}\text{.8 kJ }\end{array}$