Chapter 5, Problem 5.139AP

(a)

Interpretation Introduction

Interpretation: Hydrogen is given to be an attractive fuel as it has high value and produces no ${\text{CO}}_2$. However, the production and storage of hydrogen fuel remain problematic. On the basis of this information the given questions are to be answered.

Concept introduction: The fuel density of hydrogen is calculated by the formula,

$\text{Fuel}\text{density}\text{of}\text{hydrogen}=\frac{\text{Fuel}\text{value}\text{of}\text{hydrogen}}{\text{density}\text{of}\text{hydrogen}\text{liquid}/\text{gas}}$

The enthalpy change for a balanced reaction $\left(\Delta H_{\text{balanced rxn}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$

The amount of ammonia borane $\left(m_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}\right)$ needed is calculated by the formula,

$m_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}=\frac{m_{{\text{H}}_{\text{2}}}\times M_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}}{M_{{\text{H}}_{\text{2}}}}$

To determine: The fuel density of hydrogen in gaseous and liquid phase.

Answer to Problem 5.139AP

Solution

The fuel densities of hydrogen in gaseous and liquid phases are $\underset{\_}{1,668.52kJ/L}$ and $\underset{\_}{2.118kJ/L}$.

Explanation of Solution

Explanation

Given

The density of hydrogen gas is $0.0899\text{g}/\text{L}$.

The density of hydrogen liquid is $70.8\text{g}/\text{L}$.

The fuel value of hydrogen is $150\text{kJ}/\text{g}$.

The fuel density of hydrogen is calculated by the formula,

$\text{Fuel}\text{density}\text{of}\text{hydrogen}=\frac{\text{Fuel}\text{value}\text{of}\text{hydrogen}}{\text{density}\text{of}\text{hydrogen}\text{liquid}/\text{gas}}$

So, the fuel density of hydrogen in gaseous phase is calculated by the formula,

$\text{Fuel}\text{density}\text{of}\text{hydrogen}\left(g\right)=\frac{\text{Fuel}\text{value}\text{of}\text{hydrogen}}{\text{density}\text{of}\text{hydrogen}\text{gas}}$ (1)

Where,

• Fuel density of hydrogen $\left(g\right)$ represents fuel density of hydrogen in gaseous phase.

Substitute the fuel value of hydrogen and density of hydrogen gas in equation (1).

$\begin{array}{c}\text{Fuel}\text{density}\text{of}\text{hydrogen}\left(g\right)=\frac{150\text{kJ}/\text{g}}{0.0899\text{g}/\text{L}}\\ =1,668.52\text{kJ}/\text{L}\end{array}$

Similarly, the fuel density of hydrogen in liquid phase is calculated by the formula,

$\text{Fuel}\text{density}\text{of}\text{hydrogen}\left(l\right)=\frac{\text{Fuel}\text{value}\text{of}\text{hydrogen}}{\text{density}\text{of}\text{hydrogen}\text{liquid}}$ (2)

Where,

• Fuel density of hydrogen $\left(l\right)$ represents fuel density of hydrogen in liquid phase.

Substitute the fuel value of hydrogen and density of hydrogen liquid in equation (1).

$\begin{array}{c}\text{Fuel}\text{density}\text{of}\text{hydrogen}\left(l\right)=\frac{150\text{kJ}/\text{g}}{70.8\text{g}/\text{L}}\\ =2.118\text{kJ}/\text{L}\end{array}$

Hence, the fuel densities of hydrogen in gaseous and liquid phases are $\underset{\_}{1,668.52kJ/L}$ and $\underset{\_}{2.118kJ/L}$.

(b)

Interpretation Introduction

To determine: The enthalpy change $\left(\Delta H\right)$ for the given reactions.

Answer to Problem 5.139AP

Solution

The enthalpy change $\left(\Delta H\right)$ for the given reactions is $\underset{\_}{197.2kJ}$.

Explanation of Solution

Explanation

Given

The value of $\Delta H_{\text{f,}{\text{H}}_3{\text{NBH}}_{\text{3}}}^{\text{ο}}$ is $-38.1\text{kJ}/\text{mol}$.

The value of $\Delta H_{\text{f,}{\text{BH}}_{\text{3}}}^{\text{ο}}$ is $+110.2\text{kJ}/\text{mol}$.

The value of $\Delta H_{\text{f,}{\text{NH}}_3}^{\text{ο}}$ is $-46.1\text{kJ}/\text{mol}$.

The value of $\Delta H_{\text{f,}\text{HNBH}}^{\text{ο}}$ is $+56.9\text{kJ}/\text{mol}$.

The stated reactions are,

$\begin{array}{c}{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)\to {\text{NH}}_{\text{3}}\left(g\right)+{\text{BH}}_{\text{3}}\left(g\right)\\ {\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{H}}_2{\text{NBH}}_2\left(g\right)\\ {\text{H}}_2{\text{NBH}}_2\left(g\right)\to {\text{H}}_2\left(g\right)+\text{HNBH}\left(g\right)\end{array}$

The overall balanced reaction is the summation of above stated reaction.

$\begin{array}{c}{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)\to {\text{NH}}_{\text{3}}\left(g\right)+{\text{BH}}_{\text{3}}\left(g\right)\\ {\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)\to {\text{H}}_2\left(g\right)+{\text{H}}_2{\text{NBH}}_2\left(g\right)\\ {\text{H}}_2{\text{NBH}}_2\left(g\right)\to {\text{H}}_2\left(g\right)+\text{HNBH}\left(g\right)\\ \underset{\_}{\overline{{\text{2H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)\to 2{\text{H}}_2\left(g\right){\text{+NH}}_{\text{3}}\left(g\right)+{\text{BH}}_{\text{3}}\left(g\right)+\text{HNBH}\left(g\right)}}\end{array}$ (3)

The enthalpy change for a balanced reaction $\left(\Delta H_{\text{balanced rxn}}\right)$ is calculated by the formula,

$\Delta H_{\text{balanced rxn}}={\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}-{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ (4)

Where,

• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)$ is standard enthalpy of formation of the products.
• $\Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)$ is of standard enthalpy of formation of the reactants.
• $n$ is number of moles of products.
• $m$ is number of moles of reactants.

In the balanced chemical equation (3) the,

• Number of moles of product ${\text{H}}_2\left(g\right)$ is $2$.
• Number of moles of product ${\text{NH}}_{\text{3}}\left(g\right)$ is $1$.
• Number of moles of product ${\text{BH}}_{\text{3}}\left(g\right)$ is $1$.
• Number of moles of product $\text{HNBH}\left(g\right)$ is $1$.
• Number of moles of reactant ${\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}\left(g\right)$ is $2$.

The ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ for the balanced chemical reaction is calculated by the formula,

${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=2\text{mol}\times \Delta H_{\text{f,}{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}+1\text{mol}\times \Delta H_{\text{f,}{\text{NH}}_3}^{\text{ο}}+1\text{mol}\times \Delta H_{\text{f,}{\text{BH}}_{\text{3}}}^{\text{ο}}+1\text{mol}\times \Delta H_{\text{f,}\text{HNBH}}^{\text{ο}}$

Substitute the value of $\Delta H_{\text{f,}{\text{H}}_{\text{2}}\left(g\right)}^{\text{ο}}$, $\Delta H_{\text{f,}{\text{NH}}_3}^{\text{ο}}$, $\Delta H_{\text{f,}{\text{BH}}_{\text{3}}}^{\text{ο}}$ and $\Delta H_{\text{f,}\text{HNBH}}^{\text{ο}}$ in above expression.

$\begin{array}{c}{\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}=\left(\begin{array}{c}2\text{mol}\times 0\text{kJ}/\text{mol}+1\text{mol}\times -46.1\text{kJ}/\text{mol}+1\text{mol}\times \left(+110.2\text{kJ}/\text{mol}\right)\\ +1\text{mol}\times \left(+56.9\text{kJ}/\text{mol}\right)\end{array}\right)\\ =-46.1\text{kJ}+110.2\text{kJ}+56.9\text{kJ}\\ =121\text{kJ}\end{array}$

The ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ for the balanced chemical reaction (1) is calculated by the formula,

${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times \Delta H_{\text{f,}{\text{H}}_3{\text{NBH}}_{\text{3}}}^{\text{ο}}$ (5)

Substitute the value of $\Delta H_{\text{f,}{\text{H}}_3{\text{NBH}}_{\text{3}}}^{\text{ο}}$ in equation (5).

$\begin{array}{c}{\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}=2\text{mol}\times -38.1\text{kJ}/\text{mol}\\ =-76.2\text{kJ}\end{array}$

Substitute the values of ${\displaystyle \sum n\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Products}\right)}$ and ${\displaystyle \sum m\times \Delta H_{\text{f}}^{\text{ο}}\left(\text{Reactants}\right)}$ in equation (4).

$\begin{array}{c}\Delta H_{\text{balanced rxn}}=121\text{kJ}-\left(-76.2\text{kJ}\right)\\ =121\text{kJ}+76.2\text{kJ}\\ =\underset{\_}{197.2kJ}\end{array}$

Thus, the enthalpy change $\left(\Delta H\right)$ for the given reactions is $\underset{\_}{197.2kJ}$.

(c)

Interpretation Introduction

To determine: The amount of ammonia borane needed to supply $10.0\text{kg}$ of hydrogen.

Answer to Problem 5.139AP

Solution

The amount of ammonia borane needed to supply $10.0\text{kg}$ of hydrogen is $\underset{\_}{153.100kg}$.

Explanation of Solution

Explanation

In the above balanced chemical equation, $2$ moles of ammonia borane are utilized to produce $2$ moles of hydrogen. That means number of moles of ammonia borane and hydrogen are same.

The amount of hydrogen needed is $10.0\text{kg}$. To change this in to grams use conversion factor.

$\begin{array}{l}1\text{kg}={10}^3\text{g}\\ \text{10}\text{kg}={10}^4\text{g}\end{array}$

Thus, in grams the amount of hydrogen is ${10}^4\text{g}$.

The amount of ammonia borane $\left(m_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}\right)$ needed is calculated by the formula,

$m_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}=\frac{m_{{\text{H}}_{\text{2}}}\times M_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}}{M_{{\text{H}}_{\text{2}}}}$ (6)

Where,

• $m_{{\text{H}}_{\text{2}}}$ is the amount of hydrogen in grams.
• $M_{{\text{H}}_{\text{2}}}$ is the molar mass of hydrogen.
• $M_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}$ is the molar mass of ammonia borane.

The molar mass of hydrogen is $\begin{array}{l}=2\times \text{H}\\ =\text{2}\times 1.008\text{g}/\text{mol}\\ =2.016\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}$ is $\begin{array}{l}=1\times \text{N}+1\times \text{B}+6\times \text{H}\\ =1\times 14.007\text{g}/\text{mol}+1\times 10.81\text{g}/\text{mol}+6\times 1.008\text{g}/\text{mol}\\ =14.007\text{g}/\text{mol}+10.81\text{g}/\text{mol}+6.048\\ =30.865\text{g}/\text{mol}\end{array}$

Substitute the values of $m_{{\text{H}}_{\text{2}}}$, $M_{{\text{H}}_{\text{2}}}$ and $M_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}$ in equation (6).

$\begin{array}{c}{m}_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}=\frac{{10}^4\text{g}\times 30.865\text{g}/\text{mol}}{2.016\text{g}/\text{mol}}\\ =153,100.19\text{g}\end{array}$

To, change the value of $m_{{\text{H}}_{\text{3}}{\text{NBH}}_{\text{3}}}$ in $\text{kg}$ use conversion factor.

$\begin{array}{c}1\text{g}={10}^{-3}\text{kg}\\ \text{153,100}\text{.19}\text{g}=\text{153,100}\text{.19}\times {10}^{-3}\text{kg}\\ =\underset{\_}{153.100kg}\end{array}$

Thus, the amount of ammonia borane needed to supply $10.0\text{kg}$ of hydrogen is $\underset{\_}{153.100kg}$.

Conclusion

1. a. The fuel densities of hydrogen in gaseous and liquid phases are $\underset{\_}{1,668.52kJ/L}$ and $\underset{\_}{2.118kJ/L}$.
2. b. The enthalpy change $\left(\Delta H\right)$ for the given reactions is $\underset{\_}{197.2kJ}$.
3. c. The amount of ammonia borane needed to supply $10.0\text{kg}$ of hydrogen is $\underset{\_}{153.100kg}$