Chapter 13, Problem 13.136AP

Interpretation Introduction

Interpretation: The grams of methylamine required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ and grams of carbon dioxide produced in the combustion reaction is to be calculated.

Concept introduction: A reaction in which combustion of a hydrocarbon takes place to form carbon dioxide and water is known as combustion reaction. It is an exothermic reaction. Heat is always liberated in combustion reaction.

To determine: The grams of methylamine required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ and grams of carbon dioxide produced in the combustion reaction.

Answer to Problem 13.136AP

The grams of methylamine required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ is $\underset{\_}{702.04g}$ and grams of carbon dioxide produced in the combustion reaction is $\underset{\_}{993.30g}$.

Explanation of Solution

The heat required to raise the temperature of water ( $Q$) is calculated by the formula,

$Q=cm\Delta T$ (1)

Where,

• $c$ is specific heat of water.
• $m$ is mass of water.
• $\Delta T$ is change in temperature.

The specific heat of water is $\text{4}\text{.186}\text{J/g}{}^{\text{ο}}\text{C}$.

Mass of water is $\text{454}\text{g}$.

The change in temperature is calculated by the formula,

$\Delta T=T_2-T_1$

Where,

• $T_2$ is final temperature.
• $T_1$ is initial temperature.

The value of $T_2$ is $\text{50}\text{.0}{}^{\text{ο}}\text{C}$.

The value of $T_1$ is $\text{20}\text{.0}{}^{\text{ο}}\text{C}$.

Substitute the value of $T_2$ and $T_1$ in the above equation.

$\begin{array}{c}\Delta T=\text{50}\text{.0}{}^{\text{ο}}\text{C}-\text{20}\text{.0}{}^{\text{ο}}\text{C}\\ =\text{30}\text{.0}{}^{\text{ο}}\text{C}\end{array}$

Substitute the value of $\Delta T$, $m$ and $c$ in equation (1).

$\begin{array}{c}Q=\text{4}\text{.186}{\text{J/g}}^{\text{ο}}\text{C}\times \text{454}\text{g}\times \text{30}\text{.0}{}^{\text{ο}}\text{C}\\ =57013.32\text{J}\end{array}$

The heat transfer to water is $100\%$ efficient. Therefore, the heat required for the combustion is $57013.32\text{J}$.

The mass of methylamine required for combustion is calculated by the formula,

$Q=cm\Delta T$

Where,

• $c$ is standard enthalpy of formation of methylamine.
• $m$ is mass of methylamine.
• $\Delta T$ is change in temperature.
• $Q$ is heat required for combustion.

The value of $c$ is $-\text{23}\text{.0}\text{kJ/mol}$.

The conversion of kJ to J is done as,

$\text{1}\text{kJ}={10}^3\text{J}$

Therefore, the conversion of $-\text{23}\text{.0}\text{kJ/mol}$ to J is done as,

$\begin{array}{c}\text{23}\text{.0}\text{kJ/mol}=23\times {10}^3\text{J}\\ =\text{23000}\text{J/mol}\end{array}$

The value of $c$ in J/g is calculated by the formula,

$\text{J/g}=\frac{\text{J/mol}}{\text{Molar}\text{mass}\text{of}\text{methylamine}}$

Molar mass of methylamine is $\text{31}\text{.1}\text{g/mol}$.

Substitute the value of molar mass in the above equation.

$\begin{array}{c}\text{J/g}=\frac{\text{23000}\text{J/mol}}{\text{31}\text{.1}\text{g/mol}}\\ =\text{739}\text{.54}\text{J/g}\end{array}$

The conversion of $\text{739}\text{.54}\text{J/g}$ to ${\text{J/g}}^{\text{ο}}\text{C}$ is done as,

$\begin{array}{c}\text{739}\text{.54}\text{J/g}=\frac{\text{739}\text{.54}\text{J/g}}{273.15{}^{\text{ο}}\text{C}}\\ =2.707{\text{J/g}}^{\text{ο}}\text{C}\end{array}$

Substitute the value of $c$, $Q$, $\Delta T$ in the above equation.

$\begin{array}{c}57013.32\text{J}=2.707{\text{J/g}}^{\text{ο}}\text{C}\times m\times \text{30}\text{.0}{}^{\text{ο}}\text{C}\\ m=\frac{57013.32\text{J}}{2.707{\text{J/g}}^{\text{ο}}\text{C}\times \text{30}\text{.0}{}^{\text{ο}}\text{C}}\\ =\underset{\_}{702.04g}\end{array}$

Number of moles of methylamine is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Substitute the value of mass and molar mass of methylamine in the above equation.

$\begin{array}{c}\text{Moles}=\frac{\text{702}\text{.04}\text{g}}{\text{31}\text{.1}\text{g/mol}}\\ =22.57\text{mol}\end{array}$

Methylamine ( ${\text{CH}}_{\text{3}}{\text{NH}}_{\text{2}}$) burns according to the thermochemical equation:

${\text{2CH}}_{\text{3}}{\text{NH}}_{\text{2}}\left(l\right)+\frac{13}{2}{\text{O}}_{\text{2}}\left(g\right)\to 2{\text{CO}}_{\text{2}}\left(g\right)+2{\text{NO}}_{\text{2}}\left(g\right)+5{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Since, $\text{2}\text{mol}$ of methylamine produces $\text{2}\text{mol}$ of carbon dioxide.

Therefore, $22.57\text{mol}$ of methylamine produces $22.57\text{mol}$ of carbon dioxide.

The amount of carbon dioxide produced is calculated by the formula,

$\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$

The molar mass of carbon dioxide is $\text{44}\text{.01}\text{g/mol}$.

Substitute the value of number of moles and molar mass of carbon dioxide in the above equation.

$\begin{array}{c}22.57\text{mol}=\frac{\text{Mass}}{\text{44}\text{.01}\text{g/mol}}\\ \text{Mass}=22.57\text{mol}\times \text{44}\text{.01}\text{g/mol}\\ =\underset{\_}{993.30g}\end{array}$

Conclusion

The grams of methylamine required to raise the temperature of $\text{454}\text{g}$ of water from $\text{20}\text{.0}{}^{\text{ο}}\text{C}$ to $\text{50}\text{.0}{}^{\text{ο}}\text{C}$ is $\underset{\_}{702.04g}$ and grams of carbon dioxide produced in the combustion reaction is $\underset{\_}{993.30g}$.