Chapter 4, Problem 4.23P

(a)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=91.88\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{N}}_{\text{2}}(g)+{\text{3H}}_2\left(g\right)\to 2{\text{NH}}_{\text{3}}\text{(}g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{N{H}_3}=-42.94\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_2}=0\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}={\nu }_{N{H}_3}\Delta H_{N{H}_3|298f}-\left({\nu }_{H_2}\Delta H_{H_2|298f}+{\nu }_{N_2}\Delta H_{N_2|298f}\right)\\ \\ \Delta H_{298|r}=2\times (-45.94\frac{\text{kJ}}{\text{mol}})-\left(3\times 0\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=91.88\frac{\text{kJ}}{\text{mol}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-906.396\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  $4{\text{NH}}_{\text{3}}(g)+{\text{5O}}_2(g)\to 4\text{NO}(g)+6{\text{H}}_{\text{2}}\text{O}\left(g\right)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{N{H}_3}=-42.94\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{NO}=90.2\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O}=-241.826\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{NO}\Delta H_{NO|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O|298f}\right)-\left({\nu }_{N{H}_3}\Delta H_{N{H}_3|298f}+{\nu }_{O_2}\Delta H_{O_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(4\times 90.2\frac{\text{kJ}}{\text{mol}}+6\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(4\times -45.94\frac{\text{kJ}}{\text{mol}}+5\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-906.396\frac{\text{kJ}}{\text{mol}}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-71.77\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{3NO}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(l)\to 2{\text{HNO}}_{\text{3}}(l)+\text{NO}(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{N{O}_2}=33.2\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{HN{O}_3}=-174.1\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{NO}=90.2\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.830\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{HN{O}_3}\Delta H_{HN{O}_3(l)|298f}+{\nu }_{NO}\Delta H_{NO|298f}\right)-\left({\nu }_{H_{2}O}\Delta H_{H_{2}O(l)|298f}+{\nu }_{N{O}_2}\Delta H_{N{O}_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(2\times -174.1\frac{\text{kJ}}{\text{mol}}+1\times 90.2\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -285.83\frac{\text{kJ}}{\text{mol}}+3\times 33.2\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-71.77\frac{\text{kJ}}{\text{mol}}\end{array}$

(d)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-62.59\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CaC}}_{\text{2}}(s)+{\mathrm{H}}_{2}O(l)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{2}}(g)+\text{CaO}(s)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{Ca{C}_2(s)}=-59.8\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{C_2{H}_2}=226.7\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{CaO(s)}=-634.92\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.830\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C_2{H}_2}\Delta H^{\circ }{}_{C_2{H}_2(g)|298f}+{\nu }_{CaO(s)}\Delta H_{CaO(s)|298f}\right)-\left({\nu }_{H_{2}O}\Delta H_{H_{2}O(l)|298f}+{\nu }_{Ca{C}_2}\Delta H_{Ca{C}_2(s)|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times 226.7\frac{\text{kJ}}{\text{mol}}+1\times -634.92\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -285.83\frac{\text{kJ}}{\text{mol}}+1\times -59.8\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-62.59\frac{\text{kJ}}{\text{mol}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=1334.852\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  $2\text{Na}(s)+2{\text{H}}_{\text{2}}\text{O}(g)\to 2\text{NaOH}(s)+{\text{H}}_2(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{NaOH(s)}=425.6\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{Na(s)}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{NaOH}\Delta H^{\circ }{}_{NaOH(s)|298f}+{\nu }_{H_2}\Delta H_{H_2(g)|298f}\right)-\left({\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}+{\nu }_{Na}\Delta H_{Na(s)|298f}\right)\\ \\ \Delta H_{298|r}=\left(2\times 425.6\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(2\times -241.826\frac{\text{kJ}}{\text{mol}}+2\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=1334.852\frac{\text{kJ}}{\text{mol}}\end{array}$

(f)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-2733.592\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  $6{\text{NO}}_{\text{2}}(g)+8N{H}_3(g)\to 7{\text{N}}_{\text{2}}(g)+12{\text{H}}_{\text{2}}\text{O}(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{N{H}_3(g)}=-45.94\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N_2}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N{O}_2(g)}=33.2\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{H_{2}O}\Delta H^{\circ }{}_{H_{2}O(g)|298f}+{\nu }_{N_2}\Delta H_{N_2(g)|298f}\right)-\left({\nu }_{N{O}_2}\Delta H_{N{O}_2(g)|298f}+{\nu }_{N{H}_3}\Delta H_{N{H}_3|298f}\right)\\ \\ \Delta H_{298|r}=\left(12\times -241.826\frac{\text{kJ}}{\text{mol}}+7\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(6\times 33.2\frac{\text{kJ}}{\text{mol}}+8\times -45.94\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-2733.592\frac{\text{kJ}}{\text{mol}}\end{array}$

(g)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-104.9\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}(g)+\frac{1}{2}{\text{O}}_2(g)\to {\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{O}(g)$

The catalytic oxidation of ethylene gives ethylene oxide as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4(g)}=52.3\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{\langle {\left(C{H}_2\right)}_2\rangle O(g)}=-52.6\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{\langle {\left(C{H}_2\right)}_2\rangle O(g)}\Delta H^{\circ }{}_{\langle {\left(C{H}_2\right)}_2\rangle O(g)|298f}\right)-\left({\nu }_{C_2{H}_4}\Delta H_{C_2{H}_4(g)|298f}+{\nu }_{O_2}\Delta H_{O_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -52.6\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times 52.3\frac{\text{kJ}}{\text{mol}}+\frac{1}{2}\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-104.9\frac{\text{kJ}}{\text{mol}}\end{array}$

(h)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-151.074\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{2}}(g)+{\text{H}}_{\text{2}}\text{O}(g)\to {\left({\text{CH}}_{\text{2}}\right)}_{\text{2}}\text{O}(g)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_2(g)}=226.7\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O}=-241.826\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{\langle {\left(C{H}_2\right)}_2\rangle O(g)}=-166.2\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{\langle {\left(C{H}_2\right)}_2\rangle O(g)}\Delta H^{\circ }{}_{\langle {\left(C{H}_2\right)}_2\rangle O(g)|298f}\right)-\left({\nu }_{C_2{H}_2}\Delta H_{C_2{H}_2(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -166.2\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times 226.7\frac{\text{kJ}}{\text{mol}}+1\times -241.826\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-151.074\frac{\text{kJ}}{\text{mol}}\end{array}$

(i)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=164.942\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CH}}_4(g)+{\text{2H}}_{\text{2}}\text{O}(g)\to {\text{CO}}_{\text{2}}(g)+{\text{4H}}_2(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_4}=-74.8\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C{O}_2(g)}=-393.51\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{O}_2(g)}\Delta H^{\circ }{}_{C{O}_2(g)}+{\nu }_{H_2}\Delta H_{H_2|298f}\right)-\left({\nu }_{C{H}_4}\Delta H_{C{H}_4(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -393.51\frac{\text{kJ}}{\text{mol}}+4\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -74.8\frac{\text{kJ}}{\text{mol}}+2\times -241.826\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=164.942\frac{\text{kJ}}{\text{mol}}\end{array}$

(j)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-49.316\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CO}}_{\text{2}}(g)+3{\mathrm{H}}_2(g)\to {\text{CH}}_{\text{3}}\text{OH(}g)+{\text{H}}_{\text{2}}\text{O}(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_{3}OH(g)}=-201\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C{O}_2(g)}=-393.51\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{H}_{3}OH(g)}\Delta H_{C{H}_{3}OH(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}\right)-\left({\nu }_{C{O}_2(g)}\Delta H^{\circ }{}_{C{O}_2(g)}+{\nu }_{H_2}\Delta H_{H_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -201\frac{\text{kJ}}{\text{mol}}+1\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -393.51\frac{\text{kJ}}{\text{mol}}+3\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-49.316\frac{\text{kJ}}{\text{mol}}\end{array}$

(k)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-149.396\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CH}}_{\text{3}}\text{OH}(g)+\frac{1}{2}{\text{O}}_2(g)\to \text{HCHO}(g)+{\mathrm{H}}_{2}O(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_{3}OH(g)}=-201\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{HCHO(g)}=-108.570\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{HCHO}\Delta H_{HCHO(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}\right)-\left({\nu }_{C{H}_{3}OH(g)}\Delta H_{C{H}_{3}OH(g)|298f}+{\nu }_{O_2}\Delta H_{O_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -108.570\frac{\text{kJ}}{\text{mol}}+1\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -201\frac{\text{kJ}}{\text{mol}}+\frac{1}{2}\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-149.396\frac{\text{kJ}}{\text{mol}}\end{array}$

(l)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-1036.052\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

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

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{H_{2}S(g)}=-20.63\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{S{O}_2(g)}=-296.83\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{S{O}_2(g)}\Delta H_{S{O}_2(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}\right)-\left({\nu }_{H_{2}S(g)}\Delta H_{H_{2}S(g)|298f}+{\nu }_{O_2}\Delta H_{O_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(2\times -296.83\frac{\text{kJ}}{\text{mol}}+2\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(2\times -20.63\frac{\text{kJ}}{\text{mol}}+3\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-1036.052\frac{\text{kJ}}{\text{mol}}\end{array}$

(m)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=207.452\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

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

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{H_{2}S(g)}=-20.63\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{S{O}_2(g)}=-296.83\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{S{O}_2(g)}\Delta H_{S{O}_2(g)|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O(g)|298f}\right)-\left({\nu }_{H_{2}S(g)}\Delta H_{H_{2}S(g)|298f}+{\nu }_{O_2}\Delta H_{O_2|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -296.83\frac{\text{kJ}}{\text{mol}}+3\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -20.63\frac{\text{kJ}}{\text{mol}}+2\times -241.826\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=207.452\frac{\text{kJ}}{\text{mol}}\end{array}$

(n)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-180.5\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{N}}_{\text{2}}(g)+{\text{O}}_2\to 2\text{NO(}g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{NO}=-90.25\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{O_2}=0\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}={\nu }_{NO}\Delta H_{NO|298f}-\left({\nu }_{O_2}\Delta H_{O_2|298f}+{\nu }_{N_2}\Delta H_{N_2|298f}\right)\\ \\ \Delta H_{298|r}=2\times (-90.25\frac{\text{kJ}}{\text{mol}})-\left(1\times 0\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-180.5\frac{\text{kJ}}{\text{mol}}\end{array}$

(o)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=178.321\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CaCO}}_3(s)\to \text{CaO}(s)+{\text{CO}}_2(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{CaC{O}_3}=-1206.92\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{CaO}=-635.09\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C{O}_2(g)}=-393.509\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{O}_2(g)}\Delta H^{\circ }{}_{C{O}_2(g)}+{\nu }_{CaO(s)}\Delta H^{\circ }{}_{CaO(s)}\right)-\left({\nu }_{CaC{O}_3(s)}\Delta H^{\circ }{}_{CaC{O}_3(s)}\right)\\ \\ \Delta H_{298|r}=\left(1\times -393.509\frac{\text{kJ}}{\text{mol}}+1\times -635.09\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -1206.92\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=178.321\frac{\text{kJ}}{\text{mol}}\end{array}$

(p)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-132.439\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{SO}}_{\text{3}}(g)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(l)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{S{O}_3(g)}=-395.72\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.83\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}S{O}_4(l)}=-813.989\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{H_{2}S{O}_4(l)}\Delta H^{\circ }{}_{H_{2}S{O}_4(l)}\right)-\left({\nu }_{S{O}_3(g)}\Delta H_{S{O}_3(g)|298f}+{\nu }_{H_{2}O(l)}\Delta H_{H_{2}O(l)|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -813.989\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -395.72\frac{\text{kJ}}{\text{mol}}+1\times -285.83\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-132.439\frac{\text{kJ}}{\text{mol}}\end{array}$

(q)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=60.65\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}\text{(}g)+{\text{H}}_{\text{2}}\text{O}(l)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}(l)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_4(g)}=-52.51\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.83\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C_2{H}_{5}OH(l)}=-277.69\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C_2{H}_{5}OH(l)}\Delta H^{\circ }{}_{C_2{H}_{5}OH(l)}\right)-\left({\nu }_{C_2{H}_4(g)}\Delta H_{C_2{H}_4(g)|298f}+{\nu }_{H_{2}O(l)}\Delta H_{H_{2}O(l)|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -277.69\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -52.51\frac{\text{kJ}}{\text{mol}}+1\times -285.83\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=60.65\frac{\text{kJ}}{\text{mol}}\end{array}$

(r)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-68.91\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{CH}}_{\text{3}}\text{CHO}(g)+{\text{H}}_{\text{2}}(g)\to {\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}(g)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_{3}CHO(g)}=-166.19\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2(g)}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C_2{H}_{5}OH(g)}=-235.1\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C_2{H}_{5}OH(g)}\Delta H^{\circ }{}_{C_2{H}_{5}OH(g)}\right)-\left({\nu }_{C{H}_{3}CHO(g)}\Delta H_{C{H}_{3}CHO(g)|298f}+{\nu }_{H_2(g)}\Delta H_{H_2(g)|298f}\right)\\ \\ \Delta H_{298|r}=\left(1\times -235.1\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -166.19\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-68.91\frac{\text{kJ}}{\text{mol}}\end{array}$

(s)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-492.64\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}(l)+{\text{O}}_2(g)\to {\text{CH}}_{\text{3}}\text{COOH}(l)+{\text{H}}_{\text{2}}\text{O}(l)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_{3}COOH(l)}=-484.5\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.83\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C_2{H}_{5}OH(l)}=-277.69\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{H}_{3}COOH(l)}\Delta H_{C{H}_{3}COOH(l)|298f}+{\nu }_{H_{2}O(l)}\Delta H_{H_{2}O(l)|298f}\right)-\left({\nu }_{C_2{H}_{5}OH(l)}\Delta H^{\circ }{}_{C_2{H}_{5}OH(l)}+{\nu }_{O_2(g)}\Delta H_{O_2(g)|298f}\right)\\ \Delta H_{298|r}=\left(1\times -484.5\frac{\text{kJ}}{\text{mol}}+1\times -285.83\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -277.69\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-492.64\frac{\text{kJ}}{\text{mol}}\end{array}$

(t)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=109.78\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH:CH}}_{\text{2}}(g)\to {\text{CH}}_{\text{2}}{\text{:CHCH:CH}}_{\text{2}}(g)+{\text{H}}_{\text{2}}(g)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_2:CHCH:C{H}_2(g)}=109.24\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, and $\Delta H^{\circ }{}_{C_2{H}_{5}CH:C{H}_2(g)}=-0.540\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{H}_2:CHCH:C{H}_2(g)}\Delta H_{C{H}_2:CHCH:C{H}_2(g)|298f}+{\nu }_{H_2(g)}\Delta H_{H_2(g)|298f}\right)-\left({\nu }_{C_2{H}_{5}CH:C{H}_2(g)}\Delta H^{\circ }{}_{C_2{H}_{5}CH:C{H}_2(g)}\right)\\ \Delta H_{298|r}=\left(1\times 109.24\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -0.540\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=109.78\frac{\text{kJ}}{\text{mol}}\end{array}$

(u)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=235.03\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}(g)\to {\text{CH}}_{\text{2}}{\text{:CHCH:CH}}_{\text{2}}(g)+{\text{2H}}_2(g)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_2:CHCH:C{H}_2(g)}=109.24\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, and $\Delta H^{\circ }{}_{C_4{H}_{10}(g)}=-125.79\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{H}_2:CHCH:C{H}_2(g)}\Delta H_{C{H}_2:CHCH:C{H}_2(g)|298f}+{\nu }_{H_2(g)}\Delta H_{H_2(g)|298f}\right)-\left({\nu }_{C_4{H}_{10}(g)}\Delta H^{\circ }{}_{C_4{H}_{10}(g)}\right)\\ \Delta H_{298|r}=\left(1\times 109.24\frac{\text{kJ}}{\text{mol}}+2\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -125.79\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=235.03\frac{\text{kJ}}{\text{mol}}\end{array}$

(v)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-132.046\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{CH:CH}}_{\text{2}}(g)+\frac{1}{2}{\text{O}}_{\text{2}}(g)\to {\text{CH}}_{\text{2}}{\text{:CHCH:CH}}_{\text{2}}\text{(}g)+{\text{H}}_{\text{2}}\text{O}(g)$

The acetylene reacts with water to give acetaldehyde as a product.

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C{H}_2:CHCH:C{H}_2(g)}=109.24\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{O_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(g)}=-241.826\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{C_2{H}_{5}CH:C{H}_2(g)}=-0.540\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C{H}_2:CHCH:C{H}_2(g)}\Delta H_{C{H}_2:CHCH:C{H}_2(g)|298f}+{\nu }_{H_{2}O(g)}\Delta H_{H_{2}O(g)|298f}\right)-\left({\nu }_{C_2{H}_{5}CH:C{H}_2(g)}\Delta H^{\circ }{}_{C_2{H}_{5}CH:C{H}_2(g)}+{\nu }_{O_2(g)}\Delta H_{O_2(g)|298f}\right)\\ \Delta H_{298|r}=\left(1\times 109.24\frac{\text{kJ}}{\text{mol}}+1\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times -0.540\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-132.046\frac{\text{kJ}}{\text{mol}}\end{array}$

(w)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=-1808.396\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  $4{\text{NH}}_{\text{3}}(g)+6\text{NO}(g)\to {\text{5N}}_{\text{2}}(g)+{\text{6H}}_{\text{2}}\text{O(}g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{N{H}_3}=-42.94\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N_2}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{NO}=90.2\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{H_{2}O}=-241.826\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{N_2}\Delta H_{N_2|298f}+{\nu }_{H_{2}O}\Delta H_{H_{2}O|298f}\right)-\left({\nu }_{N{H}_3}\Delta H_{N{H}_3|298f}+{\nu }_{NO(g)}\Delta H_{NO(g)|298f}\right)\\ \\ \Delta H_{298|r}=\left(5\times 0\frac{\text{kJ}}{\text{mol}}+6\times -241.826\frac{\text{kJ}}{\text{mol}}\right)-\left(4\times -45.94\frac{\text{kJ}}{\text{mol}}+6\times 90.2\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=-1808.396\frac{\text{kJ}}{\text{mol}}\end{array}$

(x)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=43.5\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{N}}_{\text{2}}(g)+{\text{C}}_{\text{2}}{\text{H}}_{\text{2}}\to 2\text{HCN}(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_2{H}_2(g)}=226.7\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{N_2}=0\frac{\text{kJ}}{\text{mol}}$ and $\Delta H^{\circ }{}_{HCN(g)}=135.1\frac{\text{kJ}}{\text{mol}}$

Since The standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}={\nu }_{HCN(g)}\Delta H_{HCN(g)|298f}-\left({\nu }_{C_2{H}_2}\Delta H_{C_2{H}_2|298f}+{\nu }_{N_2}\Delta H_{N_2|298f}\right)\\ \\ \Delta H_{298|r}=2\times (135.1\frac{\text{kJ}}{\text{mol}})-\left(1\times 226.7\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=43.5\frac{\text{kJ}}{\text{mol}}\end{array}$

(y)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=109.78\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{.C}}_{\text{2}}{\text{H}}_{\text{5}}(g)\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{.CH:CH}}_2(g)+{\text{H}}_2(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C_6{H}_5.CH:C{H}_2(g)}=147.36\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, and $\Delta H^{\circ }{}_{C_6{H}_5.C_2{H}_5(g)}=29.92\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{C_6{H}_5.CH:C{H}_2(g)}\Delta H_{C_6{H}_5.CH:C{H}_2(g)|298f}+{\nu }_{H_2(g)}\Delta H_{H_2(g)|298f}\right)-\left({\nu }_{C_6{H}_5.C_2{H}_5(g)}\Delta H^{\circ }{}_{C_6{H}_5.C_2{H}_5(g)}\right)\\ \Delta H_{298|r}=\left(1\times 147.36\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times 29.92\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=117.44\frac{\text{kJ}}{\text{mol}}\end{array}$

(z)

Interpretation Introduction

Interpretation:

Calculate the standard heat of given equation at ${\text{25}}^{\text{o}}\text{C}$ .

Concept Introduction:

The standard heat of reaction is calculated as:

  $\Delta H_{298}=H_P-H_R$

....(1)

Where $\Delta H_{298}$ is standard heat of reaction at temperature ${\text{25}}^{\text{o}}\text{C}$ .

Answer to Problem 4.23P

  $\Delta H_{298|r}=175.305\frac{\text{kJ}}{\text{mol}}$

Explanation of Solution

Given information:

The reaction is given as:

  $\text{C}(s)+{\text{H}}_{\text{2}}\text{O}(l)\to \text{CO}(g)+{\text{H}}_{\text{2}}(g)$

The standard enthalpies of formation at standard temperature $\text{298}\text{K}$ are written from appendix C table C.4.

Values are:

$\Delta H^{\circ }{}_{C(s)}=0\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_{2}O(l)}=-285.83\frac{\text{kJ}}{\text{mol}}$, $\Delta H^{\circ }{}_{H_2(g)}=0\frac{\text{kJ}}{\text{mol}}$, and $\Delta H^{\circ }{}_{CO(g)}=-110.525\frac{\text{kJ}}{\text{mol}}$

Since the standard heat of reaction is calculated as:

  $\begin{array}{l}\Delta H_{298}{}_{|r}=H_P-H_R\\ \\ \Delta H_{298|r}=\left({\nu }_{CO(g)}\Delta H_{CO(g)|298f}+{\nu }_{H_2(g)}\Delta H_{H_2(g)|298f}\right)-\left({\nu }_{C(s)}\Delta H^{\circ }{}_{C(s)}+{\nu }_{H_{2}O(l)}\Delta H_{H_{2}O(l)|298f}\right)\\ \Delta H_{298|r}=\left(1\times -110.525\frac{\text{kJ}}{\text{mol}}+1\times 0\frac{\text{kJ}}{\text{mol}}\right)-\left(1\times 0\frac{\text{kJ}}{\text{mol}}+1\times -285.83\frac{\text{kJ}}{\text{mol}}\right)\\ \\ \Delta H_{298|r}=175.305\frac{\text{kJ}}{\text{mol}}\end{array}$