Chapter 8, Problem 73GQ

The equation for the combustion of gaseous methanol is

2 CH₃OH(g) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(g)

1. (a) Using the bond dissociation enthalpies in Table 8.8, estimate the enthalpy change for this reaction. What is the enthalpy of combustion of one mole of gaseous methanol?
2. (b) Compare your answer in part (a) with the value of ${\Delta }_t{H}^{\circ }$calculated using enthalpies of formation data.

Interpretation Introduction

Interpretation:

Enthalpy change for the given reaction and the enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ${\Delta }_{r}H$ for bond breaking is always positive and ${\Delta }_{r}H$ value for formation of bond is always negative.

${\Delta }_{r}H=\Sigma \Delta H\left(bondsbrocken\right)-\Sigma \Delta H\left(bondsformed\right)$

Explanation of Solution

Equation for the given reaction is,

$2C{H}_{3}O{H}_{\left(g\right)}+3O_{2\left(g\right)}\to 2C{O}_{2\left(g\right)}+4H_2{O}_{\left(g\right)}$

The enthalpy change for this reaction can be determined as follows,

Bonds Brocken:

$\begin{array}{l}6molofC-H\\ 2molofO-H\\ 2molofC-O\\ 3molofO=O\end{array}$

$\begin{array}{l}\Sigma \Delta H\left(bondsbrocken\right)=6\times 413kJforC-Hbonds+2\times 463kJforO-H+\\ 2\times 358kJforC-Obonds+3\times 498kJforO=O\\ =5614kJ/mol\\ \end{array}$

Bonds formed:

$\begin{array}{l}4molC=O\\ 8molO-H\end{array}$

$\begin{array}{l}\Sigma \Delta H\left(bondsformed\right)=4\times 745kJforC=Obonds+8\times 463kJforO-H+\\ =6684kJ/mol\end{array}$

$\begin{array}{c}{\text{Δ}}_{\text{r}}\text{H}\text{=}\text{ΣΔH}\left(\text{bonds}\text{brocken}\right)\text{-}\text{ΣΔH}\left(\text{bonds}\text{formed}\right)\\ \text{=}\text{5614}\text{kJ}\text{-}\text{6684}\text{kJ}\\ \text{=}-1070kJ/mol\end{array}$

Enthalpy of combustion of $2mole$ methanol is found to be $\text{-1070}\text{kJ/mol}$ and so the enthalpy of combustion of $1mol$ methanol is $\frac{\text{-1070}\text{kJ/mol}}{\text{2}}=-535kJ/mol$

Interpretation Introduction

Interpretation:

Standard enthalpy change for the given reaction and the standard enthalpy of combustion of one mole of gaseous methanol has to be determined.

Concept Introduction:

Bond dissociation enthalpy is the energy required to break one mole of gaseous bonds to form gaseous atoms. The process of breaking bonds in a molecule is always endothermic therefore ${\Delta }_{r}H$ for bond breaking is always positive and ${\Delta }_{r}H$ value for formation of bond is always negative.

${\Delta }_{r}H=\Sigma \Delta H\left(bondsbrocken\right)-\Sigma \Delta H\left(bondsformed\right)$

Enthalpy of formation:

${\Delta }_{r}H=\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(products\right)-\sum {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)$

Explanation of Solution

Equation for the given reaction is,

$2C{H}_{3}O{H}_{\left(g\right)}+3O_{2\left(g\right)}\to 2C{O}_{2\left(g\right)}+4H_2{O}_{\left(g\right)}$

The standard enthalpy change for this reaction can be determined as follows,

$\begin{array}{c}{\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)\text{=}\text{2}\text{×}\text{-201}\text{kJ/mol}\text{for}{\text{CH}}_{\text{3}}\text{OH}\text{+}\text{3}\text{×}\text{0}\text{kJ/mol}\text{for}{\text{O}}_{\text{2}}\\ \text{=}-402kJ/mole\end{array}$

$\begin{array}{c}{\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{products}\right)\text{=}\text{2}\text{×}\text{-393}\text{.509}\text{kJ/mol}\text{for}{\text{CO}}_2\text{+}4\text{×}-241.83\text{kJ/mol}\text{for}{H}_{2}O\\ \text{=}-1754.3kJ/mole\end{array}$

$\begin{array}{l}{\text{Δ}}_{\text{r}}\text{H=}\Sigma {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{products}\right)\text{-}\Sigma {\text{Δ}}_{\text{f}}{\text{H}}^{\text{0}}\left(\text{reactants}\right)\\ \text{=}\text{-1754}\text{.3}\text{kJ/mole}\text{-}\text{402}\text{kJ/mol}\\ \text{=}-1352.3kJ/mole\end{array}$

Enthalpy of combustion of $2mole$ methanol is found to be $-1352.3kJ/mole$ and so the enthalpy of combustion of $1mol$ methanol is $\frac{\text{-1352}\text{.3}\text{kJ/mole}}{\text{2}}\text{=-676}\text{kJ/mol}$