Chapter 6, Problem 114AE

Consider the two space shuttle fuel reactions in Exercises 81 and 82. Which reaction produces more energy per kilogram of reactant mixture (stoichiometric amounts)?

81. The reusable booster rockets of the space shuttle use a mixture of aluminum and ammonium perchlorate as fuel. A possible reaction is

$3\text{Al}\left(s\right)+3{\text{NH}}_4{\text{ClO}}_4\left(s\right)\to {\text{Al}}_2{\text{O}}_3\left(s\right)+{\text{AlCl}}_3\left(s\right)+3\text{NO}\left(g\right)+6{\text{H}}_2\text{O}\left(g\right)$

Calculate ∆H° for this reaction

82. The space shuttle Orbiter utilizes the oxidation of methylhydrazine by dinitrogen tetroxide for propulsion:

$4{\text{N}}_2{\text{H}}_3{\text{CH}}_3\left(l\right)+5{\text{N}}_2{\text{O}}_4\left(l\right)\to 12{\text{H}}_2\text{O}\left(g\right)+9{\text{N}}_2\left(g\right)+4{\text{CO}}_2\left(g\right)$

Calculate ∆H° for this reaction

Interpretation Introduction

Interpretation: One kilogram of reactant mixture produced energy of given reactions should be calculated and the high energy produced reaction should be determined.

Concept Introduction:

Standard enthalpy of formation:

• The change in enthalpy that associate with the formation of one mole of a product from its pure elements, with all substances in its standard states is called as a standard enthalpy of formation.
• The enthalpy of reaction is equal to the subtraction of standard enthalpy of formation of reactants from product.
• Formula:

               ${\text{ΔH}}_{\text{f}}\text{= }{\displaystyle \sum {\text{n}}_{\text{p}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(products)-}{\displaystyle \sum {\text{n}}_{\text{r}}\text{ΔH}}{\text{°}}_{\text{f}}\text{(reactants)}\mathrm{......}\text{(1)}$

                                n is number of moles

• The heat (energy) change of the reaction at constant pressure is equal to enthalpy change.
• The kilogram mass of the reactants produced energy is equal to the ratio of the energy released and mass of the substance $\left(kg\right).$

  $\text{Energy}\text{(heat)}\text{per}\text{kilogram}\text{=}\frac{\text{Energy}\text{(kJ)}}{\text{Mass}\text{(kg)}}\mathrm{......}(2)$

Answer to Problem 114AE

Answer

The ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{/N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}$ mixture released energy is -7129 kJ and this is the more energy released than to ${\text{Al /NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ mixture.

Explanation of Solution

Explanation

Calculate the enthalpy change of given reactions.

Given data:

$\begin{array}{l}{\text{3Al}}_{\text{(s)}}{\text{+3NH}}_{\text{4}}{\text{ClO}}_{\text{4(s)}}\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{+AlCl}}_{\text{3(s)}}{\text{+3NO}}_{\text{(g)}}\uparrow {\text{+6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\mathrm{......}\text{(1)}\\ {\text{4N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3(l)}}{\text{+5N}}_{\text{2}}{\text{O}}_{\text{4(l)}}\to {\text{12H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{+9N}}_{\text{2(g)}}{\text{+4CO}}_{\text{2(g)}}\mathrm{......}\text{(2)}\end{array}$

                       

Enthalpy change of given reaction 1 is,

• The fuel reaction is:

                         ${\text{3Al}}_{\text{(s)}}{\text{+3NH}}_{\text{4}}{\text{ClO}}_{\text{4(s)}}\to {\text{Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}{\text{+AlCl}}_{\text{3(s)}}{\text{+3NO}}_{\text{(g)}}\uparrow {\text{+6H}}_{\text{2}}{\text{O}}_{\text{(g)}}\mathrm{......}\text{(1)}$

$\begin{array}{l}{\text{ΔH}}_{\text{reaction}\text{1}}\text{=}\left[\left\{\text{6×}\left(\frac{\text{-242kJ}}{\text{mol}}\right)\right\}\text{+}\left\{\text{3×}\left(\frac{\text{90kJ}}{\text{mol}}\right)\right\}\text{+}\left\{\text{1×}\left(\frac{\text{-704kJ}}{\text{mol}}\right)\right\}\text{+}\left\{\text{1×}\left(\frac{\text{-1676kJ}}{\text{mol}}\right)\right\}\right]\text{-}\left[\text{3×}\left(\frac{\text{-295kJ}}{\text{mol}}\right)\right]\\ \text{=}\text{-2677}\text{KJ}\end{array}$

                           

Enthalpy change of given reaction 2 is,

  $\begin{array}{l}{\text{ΔΗ}}_{\text{reaction}\text{2}}\text{=}\left[\left\{\text{12×}\left(\frac{\text{-242kJ}}{\text{mol}}\right)\right\}\text{+}\left\{\text{4×}\left(\frac{\text{-393}\text{.5 kJ}}{\text{mol}}\right)\right\}\right]\text{-}\left[\left\{\text{5×}\left(\frac{\text{-20}\text{kJ}}{\text{mol}}\right)\right\}\text{+}\left\{\text{4×}\left(\frac{\text{54}\text{kJ}}{\text{mol}}\right)\right\}\right]\\ \text{=}\text{-4594}\text{KJ}\end{array}$

• The standard enthalpy of formation of reactants and products are taken from the standard data and this are plugging in to equation (1) to get the enthalpy change of the reactions.
• The enthalpy change of reaction (1) is $\text{-2677}\text{KJ}$ and reaction (2) is $\text{-4594}\text{KJ}$.

To calculate the per kilogram of reactant mixture produced energy in reaction 1

The molecular weight of reactants is:

                                ${\text{NH}}_{\text{4}}\text{ClO = 117}\text{.49g,}\text{Al = 26}\text{.98 g}$

                     $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{reaction1}}\text{=}\left[\text{3mol}\left\{\frac{\text{26}\text{.98}\text{g}}{\text{mol}}\right\}\text{+3mol}\left\{\frac{\text{117}\text{.49}\text{g}}{\text{mol}}\right\}\right]\\ \text{=}\text{433}\text{.41g}\\ \text{ = 0}\text{.43341}\text{kg}\\ \text{For}\text{the}\text{1}\text{.0}\text{kg}\text{of}\text{fule}\text{released}\text{energy}\text{is}\text{:}\\ \text{=}\frac{\text{released}\text{energy}\text{(kJ)}}{\text{mass}\text{(kg)}}\\ \text{=}\frac{\text{-2677}\text{kJ}}{\text{0}\text{.43341}\text{kg}}\\ \text{=}\text{-6177KJ}\end{array}$

• The molecular weights of the reactants are multiplied by its coefficients and sum of this give the 1 mole of reactants mass. The given released energy is divided by total mass of reactants to given kilogram of reactants produced energy.
• The per kilogram of reactant mixture produced energy is $\text{-6177KJ}$.

To calculate per kilogram of reactant mixture produced energy in reaction 2.

The molecular weight of reactants is:

                                ${\text{N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}\text{=}\text{46}\text{.08}\text{g, }{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{=}\text{92}\text{.02}\text{g}$

                       $\begin{array}{l}{\text{ΔH}}^{\text{°}}{}_{\text{reaction}\text{2}}\text{=}\left[\text{5mol}\left\{\frac{\text{92}\text{.02}\text{g}}{\text{mol}}\right\}\text{+4mol}\left\{\frac{\text{46}\text{.08}\text{g}}{\text{mol}}\right\}\right]\\ \text{=}\text{644}\text{.42}\text{g}\\ \text{ = 0}\text{.64412 Kg}\\ \text{For}\text{the}\text{1}\text{.0}\text{kg}\text{of}\text{fule}\text{released}\text{energy}\text{is}\text{:}\\ \text{=}\frac{\text{released}\text{energy}\text{(kJ)}}{\text{mass}\text{(kg)}}\\ \text{=}\frac{\text{-4594}\text{kJ}}{\text{0}\text{.64412 Kg}}\\ \text{=}\text{-7129KJ}\end{array}$

The fuel reaction is:

                       ${\text{4N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3(l)}}{\text{+5N}}_{\text{2}}{\text{O}}_{\text{4(l)}}\to {\text{12H}}_{\text{2}}{\text{O}}_{\text{(g)}}{\text{+9N}}_{\text{2(g)}}{\text{+4CO}}_{\text{2(g)}}\mathrm{......}\text{(2)}$

• The molecular weights of the reactants are multiplied by its coefficients and sum of this give the 1 mole of reactants mass.
• The given released energy is divided by total mass of reactants (kg) to get  produced energy per kilogram of reactants.
• The produced energy per kilogram of reactant mixture is $\text{-7129KJ}$.

Conclusion

Conclusion

• The produced energy per kilogram of reactant mixtures are calculated by using enthalpy of formation and the energy per kilogram of reactant mixtures was ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{/N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}$ mixture was -7129 kJ,  ${\text{Al /NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ mixture was $\text{-6177KJ}$.

From the result the produced energy per kilogram of reactant mixture ${\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{/N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}$ was higher than the ${\text{Al /NH}}_{\text{4}}{\text{ClO}}_{\text{4}}$ mixture.