Chapter 5, Problem 5.123AP

(a)

Interpretation Introduction

Interpretation: The given reactions are to be identified as exothermic and endothermic.

The value of $\Delta H_{\text{rxn}}^{\text{ο}}$ and the amount of energy absorbed or released is to be calculated for the given values. The $\Delta H_{\text{rxn}}^{\text{ο}}$ for the metabolism of $1$ mole of ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ to give $1$ mole of formaldehyde is to be compared with $509.8\text{kJ}$.

Concept introduction: The reactions, in which heat energy is evolved, are termed as an exothermic process and if heat energy is absorbed then they are termed as an endothermic.

The standard enthalpy change $\left(\Delta H_{\text{rxn}}^{\text{ο}}\right)$ is the change in enthalpy when one mole of product is formed at standard condition.

The amount of energy released or absorbed $\left(q\right)$ is calculated by the formula,

$q=n\times \Delta H_{\text{rxn}}^{\text{ο}}$

To determine: The given reaction is to be identified as exothermic and endothermic.

Answer to Problem 5.123AP

Solution

The given reaction is exothermic.

Explanation of Solution

Explanation

The given reaction is stated as,

${\text{O}}_{\text{2}}\left(g\right)+{\text{CH}}_{\text{3}}\text{OH}\left(l\right)\to 2\text{HCOOH}\left(l\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\text{+1019}\text{.6}\text{kJ}$

The reactions, in which heat energy is evolved, are termed as an exothermic process and if heat energy is absorbed then they are termed as an endothermic.

In the above reaction heat energy is evolved. Therefore, it is termed as an exothermic process.

(b)

Interpretation Introduction

To determine: The value of $\Delta H_{\text{rxn}}^{\text{ο}}$.

Answer to Problem 5.123AP

Solution

The value of $\Delta H_{\text{rxn}}^{\text{ο}}$ is $\underset{\_}{-509.8kJ/mol}$.

Explanation of Solution

Explanation

The given reaction is stated as,

${\text{O}}_{\text{2}}\left(g\right)+{\text{CH}}_{\text{3}}\text{OH}\left(l\right)\to 2\text{HCOOH}\left(l\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\text{+1019}\text{.6}\text{kJ}$

In the above reaction two moles of formic acid are formed. The standard enthalpy change $\left(\Delta H_{\text{rxn}}^{\text{ο}}\right)$ is the change in enthalpy when one mole of product is formed at standard condition. The given value of evolved heat energy should be in negative $\left(\Delta H_{\text{rxn,}\text{HCOOH}}=-\text{1019}\text{.6}\text{kJ}\right)$ as the process is exothermic.

Hence, the $\Delta H_{\text{rxn}}^{\text{ο}}$ of reaction is half of mole to the amount of evolved heat energy for the formation of formic acid $\left(\Delta H_{\text{rxn,}\text{HCOOH}}\right)$ and calculated by the formula,

$\Delta H_{\text{rxn}}^{\text{ο}}=\frac{\Delta H_{\text{rxn,}\text{HCOOH}}}{2\text{mol}}$ (1)

Where,

• $\Delta H_{\text{rxn}}^{\text{ο}}$ is the standard enthalpy change for the formation of one mole of formic acid.
• $\Delta H_{\text{rxn,}\text{HCOOH}}$ is the given amount of evolved heat energy for the formation of formic acid.

Substitute the value of $\Delta H_{\text{rxn,}\text{HCOOH}}$ in equation (1).

$\begin{array}{c}\Delta H_{\text{rxn}}^{\text{ο}}=\frac{-\text{1019}\text{.6}\text{kJ}}{2\text{mol}}\\ =-509.8\text{kJ}/\text{mol}\end{array}$

Hence, the value of $\Delta H_{\text{rxn}}^{\text{ο}}$ is $\underset{\_}{-509.8kJ/mol}$.

(c)

Interpretation Introduction

To determine: The amount of energy absorbed or released.

Answer to Problem 5.123AP

Solution

The amount of energy released is $\underset{\_}{-953.326kJ}$.

Explanation of Solution

Explanation

The mass $\left(m\right)$ of methanol is $60.0\text{g}$.

The given reaction is stated as,

${\text{O}}_{\text{2}}\left(g\right)+{\text{CH}}_{\text{3}}\text{OH}\left(l\right)\to 2\text{HCOOH}\left(l\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\text{+1019}\text{.6}\text{kJ}$

The above reaction is metabolism of methanol liquid $\left({\text{CH}}_{\text{3}}\text{OH}\left(l\right)\right)$. The number of moles $\left(n\right)$ used for metabolism is calculated by the formula,

$n=\frac{m}{M}$ (2)

Where,

• $m$ is the mass of methanol liquid.
• $M$ is the molar mass of methanol.

The molar mass of methanol is $\begin{array}{l}=1\times \text{C}+\text{4}\times \text{H}+1\times \text{O}\\ \text{=1}\times \text{12}\text{.011}\text{g}/\text{mol}+\text{4}\times \text{1}\text{.008}\text{g}/\text{mol}+1\times 15.999\text{g}/\text{mol}\\ \text{=12}\text{.011}\text{g}/\text{mol}+4.032\text{g}/\text{mol}+15.999\text{g}/\text{mol}\\ =32.042\text{g}/\text{mol}\end{array}$

Substitute the values of $M$ and $m$ in equation (2).

$\begin{array}{c}n=\frac{60.0\text{g}}{32.042\text{g}/\text{mol}}\\ =1.87\text{mol}\end{array}$

The amount of energy released or absorbed $\left(q\right)$ is calculated by the formula,

$q=n\times \Delta H_{\text{rxn}}^{\text{ο}}$ (3)

Where,

• $\Delta H_{\text{rxn}}^{\text{ο}}$ is the standard enthalpy change for the formation of one mole of formic acid.

Substitute the values of $\Delta H_{\text{rxn}}^{\text{ο}}$ and $n$  in equation (3).

$\begin{array}{c}q=1.87\text{mol}\times -509.8\text{kJ}/\text{mol}\\ \text{=}-953.326\text{kJ}\end{array}$

The negative sign indicates that the energy is released during the process. Hence, the amount of energy released is $\underset{\_}{-953.326kJ}$.

(d)

Interpretation Introduction

To determine: The comparison of $\Delta H_{\text{rxn}}^{\text{ο}}$ for the metabolism of $1$ mole of ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ to give $1$ mole of formaldehyde with $509.8\text{kJ}$.

Answer to Problem 5.123AP

Solution

The $\Delta H_{\text{rxn}}^{\text{ο}}$ for the metabolism of $1$ mole of ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ to give $1$ mole of formaldehyde is expected to be smaller than $509.8\text{kJ}$.

Explanation of Solution

Explanation

The amount of energy released in two step for the formation of formic acid is $-953.326\text{kJ}$. The formation of formaldehyde is the first step and metabolized by one mole of methanol. Therefore, the amount of energy released would be half of the $-953.326\text{kJ}$.

$\begin{array}{c}=\frac{953.326\text{kJ}}{2}\\ =476.663\text{kJ}\end{array}$

The negative sign of $-953.326\text{kJ}$ is eliminated as it only indicates that heat is evolved.

Hence, the $\Delta H_{\text{rxn}}^{\text{ο}}$ for the metabolism of $1$ mole of ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ to give $1$ mole of formaldehyde is expected to be smaller than $509.8\text{kJ}$.

Conclusion

1. a. The given reaction is exothermic.
2. b. The value of $\Delta H_{\text{rxn}}^{\text{ο}}$ is $\underset{\_}{-509.8kJ/mol}$.
3. c. The amount of energy released is $\underset{\_}{-953.326kJ}$.
4. d. The $\Delta H_{\text{rxn}}^{\text{ο}}$ for the metabolism of $1$ mole of ${\text{CH}}_{\text{3}}\text{OH}\left(l\right)$ to give $1$ mole of formaldehyde is expected to be smaller than $509.8\text{kJ}$