Chapter 9, Problem 9.138QP

Nitroglycerin, one of the most commonly used explosives, has the structure

The decomposition reaction is

$\begin{array}{l}4{\text{C}}_3{\text{H}}_{\text{5}}{\text{N}}_{\text{3}}{\text{O}}_{\text{9}}(l)\stackrel{}{\to }\\ 12{\text{CO}}_{\text{2}}(g)+10{\text{H}}_2\text{O}(g)+6{\text{N}}_2(g)+{\text{O}}_2(g)\end{array}$

The explosive action is the result of the heat released and the large increase in gaseous volume. (a) Calculate the ΔH° for the decomposition of one mole of nitroglycerin using both standard enthalpy of formation values and bond enthalpies. Assume that the two O atoms in the NO₂ groups are attached to N with one single bond and one double bond. (b) Calculate the combined volume of the gases at STP. (c) Assuming an initial explosion temperature of 3000 K, estimate the pressure exerted by the gases using the result from (b). (The standard enthalpy of formation of nitroglycerin is −371.1 kJ/mol.)

Interpretation Introduction

Interpretation:

$\Delta H°$ for the decomposition of one mole of nitroglycerin has to be calculated using both bond enthalpy and standard enthalpy of formation values.

Concept Introduction:

$\Delta H°$ refers to change in enthalpy. Change in enthalpy in a reaction and bond energy (BE) are related as,

    $\text{ΔH°}\text{=}\text{ΣBE(reactants)-ΣBE(products)}$

Change in enthalpy in a reaction and enthalpy of formation are related by the formula,

    $\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)$

Where,

    $\begin{array}{l}\Delta H_{rxn}^{°}=\text{ standard}\text{enthalpy change in reaction}\\ \Delta H_f^{°}(product)=\text{ standard}\text{enthalpy of formation of product}\\ \Delta H_f^{°}(reac\tan t)=\text{ standard}\text{enthalpy of formation of }reac\tan t\end{array}$

Answer to Problem 9.138QP

$\Delta H°$ Calculated using enthalpy of formation values is $-1413.9kJ/mol.$

$\Delta H°$ Calculated using bond energy values is $-1937kJ/mol.$

Explanation of Solution

The decomposition reaction is,

    $4C_3{H}_5{N}_3{O}_{9(l)}\stackrel{}{\to }12C{O}_{2(g)}+10H_2{O}_{(g)}+6{\text{N}}_{\text{2(g)}}+{\text{O}}_{\text{2(g)}}$

Reduce the equation to show the decomposition of one mole of nitroglycerin.

Dividing by 4,

    $C_3{H}_5{N}_3{O}_{9(l)}\stackrel{}{\to }3C{O}_{2(g)}+\frac{5}{2}{H}_2{O}_{(g)}+\frac{3}{2}{\text{N}}_{\text{2(g)}}+\frac{1}{4}{\text{O}}_{\text{2(g)}}$

Calculation of $\Delta H°$ using enthalpy of formation values from appendix 3 of the text book:

    $\begin{array}{l}\Delta H_{rxn}^{°}=\Sigma \Delta H_f^{°}(product)-\Sigma \Delta H_f^{°}(reac\tan t)\\ =\left(3\right)\left(-395.5kJ/mol\right)+\left(\frac{5}{2}\right)\left(-241.8kJ/mol\right)-\left(1\right)\left(-371.1\text{kJ/mol}\right)\\ =-1413.9\text{kJ/mol}\end{array}$

Calculation of $\Delta H°$ using bond enthalpy values from table 9.4 of the text book:

Bonds broken in reactants	Number of bonds broken	Bond enthalpy $\left(kJ/mol\right)$	Enthalpy change $\left(kJ/mol\right)$
$N=O$	$3$	$607$	$1821$
$C-H$	$5$	$414$	$2070$
$C-C$	$2$	$347$	$694$
$N-O$	$6$	$176$	$1056$
$C-O$	$3$	$351$	$1053$
Bonds formed in product	Number of bonds formed	Bond enthalpy $\left(kJ/mol\right)$	Enthalpy change $\left(kJ/mol\right)$
$C=O$	$6$	$799$	$4794$
$O-H$	$5$	$460$	$2300$
$N\equiv N$	$1.5$	$941.4$	$1412.1$
$O=O$	$0.25$	$498.7$	$124.7$

Using the above data $\Delta H°$ of the reaction is calculated.

    $\begin{array}{l}\text{ΔH° }\text{= ΣBE(reactants)-ΣBE(products)}\\ \text{ΔH° }\text{= }\left(6694-8630.8\right)kJ/mol=-1937kJ/mol\end{array}$

Interpretation Introduction

Interpretation:

Combined volume of gases in the given reaction has to be calculated at STP.

Concept Introduction:

STP refers to standard temperature and pressure conditions which are $25°C$ and $1atm$ pressure.

Answer to Problem 9.138QP

Combined volume of gases in the given reaction at STP is calculated as $162L.$

Explanation of Solution

The decomposition reaction is,

    $4C_3{H}_5{N}_3{O}_{9(l)}\stackrel{}{\to }12C{O}_{2(g)}+10H_2{O}_{(g)}+6{\text{N}}_{\text{2(g)}}+{\text{O}}_{\text{2(g)}}$

Reduce the equation to show the decomposition of one mole of nitroglycerin.

Dividing by 4,

    $C_3{H}_5{N}_3{O}_{9(l)}\stackrel{}{\to }3C{O}_{2(g)}+\frac{5}{2}{H}_2{O}_{(g)}+\frac{3}{2}{\text{N}}_{\text{2(g)}}+\frac{1}{4}{\text{O}}_{\text{2(g)}}$

One mole of nitroglycerin forms $\left(3+2.5+1.5+0.25\right)=7.25molofgases.$

At STP, one mole of ideal gas occupies a volume of $22.4L.$

Therefore total volume of gases in the given reaction at STP is calculated as,

    $7.25mol\times \frac{22.4L}{1mol}=162L$

Interpretation Introduction

Interpretation:

The pressure exerted by the gases at temperature of $3000K$ has to be estimated.

Concept Introduction:

Pressure and temperature are related by ideal gas equation is,

    $PV=nRT$

Where,

    $\begin{array}{l}\text{P}\text{= pressure}\\ \text{V}\text{= volume}\\ \text{T}\text{= temperature}\\ \text{n}\text{= no}\text{.of moles}\\ \text{R}\text{= gas constant}\end{array}$

Answer to Problem 9.138QP

The pressure exerted by the gases at temperature of $3000K$ is calculated as $11.0atm.$

Explanation of Solution

Given data:

    $\begin{array}{l}V=162L\\ T=3000K\\ n=7.25mol\end{array}$

As we know,

    $PV=nRT$

Rewrite the above equation for P,

    $P=\frac{nRT}{V}$

Substitute the known values.

    $\begin{array}{l}P=\frac{\left(7.25mol\right)\times \left(\frac{0.0821L.atm}{mol.K}\right)\times \left(3000K\right)}{162L}\\ =11.0atm\end{array}$