Chapter 8, Problem 124AP

Interpretation Introduction

Interpretation:

The Lewis structure of vinyl chloride, repeating units of vinyl chloride, and the enthalpy change are to be calculated.

Concept Introduction:

The standard enthalpy for a reaction is the amount of enthalpy change that occurs to occur understandard conditions(room temperature and pressure)

The enthalpy of reaction can be calculated as:

$\Delta H{°}_{\text{rxn}}={\displaystyle \sum BE}\left(\text{reactants}\right)-{\displaystyle \sum BE}\left(\text{products}\right).$

Here, BE represents the average value of bond enthalpy.

Answer to Problem 124AP

Solution:

(a)

(b)

(c) $-1.2\times {10}^6\text{ kJ}$

Explanation of Solution

a)The Lewis structure of vinyl chloride

There are a total of 18 electrons in ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$. Subtract 12 electrons to account for the six bonds in the skeleton, leaving six electrons to distribute as three lone pairs on $\text{Cl}$ atom.

The skeletal structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ is:

The Lewis structure for ${\text{C}}_{\text{2}}{\text{H}}_{\text{3}}\text{Cl}$ is:

b)Draw a portion of the molecule showing three $-{\text{CH}}_2-\text{CHCl}-$ repeating units.

The poly vinyl chloride has the repeating chain of $-{\text{CH}}_2-\text{CHCl}-$.

The portion of poly vinyl chloride for three repeating chain of $-{\text{CH}}_2-\text{CHCl}-$ is as follows:

Given information:

Mass, $m_{\text{vinyl chloride}}=1\times {10}^3\text{ kg}\text{.}$

c) The enthalpy change when $1\times {10}^3\text{ kg}$ of vinyl chloride forms poly(vinyl chloride).

The molar mass of vinyl chloride is $62.5\text{ kg}/\text{mol}\text{.}$

The moles of vinyl chloride can be calculated as:

$n_{\text{vinyl chloride}}=\frac{m_{\text{vinyl chloride}}}{M_{\text{vinyl chloride}}}.$

Substitute $1\times {10}^3\text{ kg}$ for $m_{\text{vinyl chloride}}$ and $62.5\text{ kg}/\text{mol}$ for $M_{\text{vinyl chloride}}$ in the above equation.

$\begin{array}{c}{n}_{\text{vinyl chloride}}=\frac{1\times {10}^3\text{ kg}}{62.5\text{ kg}/\text{mol}}\\ =\frac{1\times {10}^3\times {10}^3\text{ g}}{62.5\text{ kg}/\text{mol}}\\ =0.016\times {10}^6\text{ kg}/\text{mol}\text{.}\end{array}$

The enthalpy of reaction can be calculated as:

$\Delta H{\text{°}}_{\text{rxn}}={\displaystyle \sum BE}\left(\text{reactants}\right)-{\displaystyle \sum BE}\left(\text{products}\right).$

From table 8.6, the enthalpy of formation values are:

$\begin{array}{l}BE\left[\text{C}=\text{C}\right]=620\text{ kJ}/\text{mol;}\\ BE\left[\text{C}-\text{C}\right]=347\text{ kJ}/\text{mol}\text{.}\end{array}$

Now, the standard enthalpy of a given reaction is:

$\Delta H{°}_{\text{rxn}}=\left\{BE\left[\text{C}=\text{C}\right]\right\}-\left\{\text{2}BE\left[\text{C}-\text{C}\right]\right\}.$

Substitute, $620\text{ kJ}/\text{mol}$ for $BE\left[\text{C}=\text{C}\right]$ and $347\text{ kJ}/\text{mol}$ for $BE\left[\text{C}-\text{C}\right]$ in the above equation.

$\begin{array}{c}\Delta H{°}_{\text{rxn}}=\left[620\text{ kJ}/\text{mol}\right]-\left[\text{2}\left(347\text{ kJ}/\text{mol}\right)\right]\\ =620\text{ kJ}/\text{mol}-694\text{ kJ}/\text{mol}\\ =-74\text{ kJ}/\text{mol}\text{.}\end{array}$

The heat released is

$q=n\times \Delta H{°}_{\text{rxn}}.$

Substitute $0.016\times {10}^6\text{ kg}/\text{mol}$ for $n_{\text{vinyl chloride}}$ and $-74\text{ kJ}/\text{mol}$ for $\Delta H{°}_{\text{rxn}}$ in the above reaction.

$\begin{array}{c}q=\left(0.016\times {10}^6\right)\times \left(-74\right)\\ =-1.2\times {10}^6\text{ kJ}\text{.}\end{array}$

Hence, the enthalpy change is $-1.2\times {10}^6\text{ kJ}$.