Chapter 6, Problem 6.76QP

Hydrazine, N₂H₄, decomposes according to the following reaction:

$3{\text{N}}_2{\text{H}}_4(l)\stackrel{}{\to }4{\text{NH}}_3(g)+{\text{N}}_2(g)$

(a) Given that the standard enthalpy of formation of hydrazine is 50.42 kJ/mol, calculate ΔH° for its decomposition. (b) Both hydrazine and ammonia burn in oxygen to produce H₂O(l) and N₂(g). Write balanced equations for each of these processes and calculate ΔH° for each of them. On a mass basis (per kg), would hydrazine or ammonia be the better fuel?

Interpretation Introduction

Interpretation:

The standard enthalpy of decomposition of Hydrazine has to be calculated. The equation has to be balanced and $\text{ΔH°}$ has to be calculated in mass per kilogram for each of them and better fuel has to be identified.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.76QP

The standard enthalpy $\text{ΔH°}$ for Hydrazine decomposition is $\text{-336}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Explanation of Solution

Standard enthalpy of formation of ${\text{NH}}_{\text{3}}\text{=-46}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{N}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{N}}_{\text{2}}{\text{H}}_4\text{=50}\text{.45}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of decomposition = $\left[{\text{4ΔH°}}_{\text{f}}{\text{(NH}}_{\text{3}}{\text{)+ΔH°}}_{\text{f}}{\text{(N}}_{\text{2}}\text{)}\right]{\text{-3ΔH°}}_{\text{f}}{\text{(N}}_{\text{2}}{\text{H}}_{\text{4}}\text{)}$

=$\left[\text{(4)(-46}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(0)}\right]\text{-(3)(50}\text{.42}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}$

=$\text{-336}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of decomposition of Hydrazine = $\text{-336}\text{.5}\text{kJ}{\text{mol}}^{\text{-1}}$

Interpretation Introduction

Interpretation:

The standard enthalpy of decomposition of Hydrazine has to be calculated. The equation has to be balanced and $\text{ΔH°}$ has to be calculated in mass per kilogram for each of them and better fuel has to be identified.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.76QP

The standard enthalpy $\text{ΔH°}$ for ${\text{N}}_{\text{2}}{\text{H}}_{\text{4(l)}}\text{=-1}{\text{.941×10}}^{\text{4}}\text{kJ}{\text{kg}}^{\text{-1}}$ and ${\text{NH}}_{\text{3}}$ is $\text{-2}{\text{.245×10}}^{\text{4}}\text{kJ}{\text{kg}}^{\text{-1}}$

The standard enthalpy of Ammonia is found to be $\text{-2}{\text{.245×10}}^{\text{4}}\text{kJ}{\text{kg}}^{\text{-1}}$ .

Balanced equation are given as,

1. a) ${\text{N}}_{\text{2}}{\text{H}}_{\text{4(l)}}{\text{+O}}_{\text{2(g)}}\to {\text{N}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$
2. b) ${\text{4NH}}_{\text{3(g)}}{\text{+3O}}_{\text{2(g)}}\to {\text{2N}}_{\text{2(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Explanation of Solution

The equations are balance by multiplying the equal numbers on product and reactant side. The balanced equations can be given as,

1. a) ${\text{N}}_{\text{2}}{\text{H}}_{\text{4(l)}}{\text{+O}}_{\text{2(g)}}\to {\text{N}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$
2. b) ${\text{4NH}}_{\text{3(g)}}{\text{+3O}}_{\text{2(g)}}\to {\text{2N}}_{\text{2(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

To calculate: the $\text{ΔH°}$ for ${\text{N}}_{\text{2}}{\text{H}}_{\text{4(l)}}{\text{+O}}_{\text{2(g)}}\to {\text{N}}_{\text{2(g)}}{\text{+2H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Standard enthalpy of formation of ${\text{N}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{N}}_{\text{2}}{\text{H}}_4\text{=50}\text{.45}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{O}}_2\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}\text{=}\left[{\text{ΔH°}}_{\text{f}}{\text{(N}}_{\text{2}}{\text{)+2ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{)}\right]\text{-}\left[{\text{ΔH°}}_{\text{f}}{\text{(N}}_{\text{2}}{\text{H}}_{\text{4}}{}_{\text{(l)}}{\text{)+ΔH°}}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=}\left[\text{(1)(0)+(2)(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-}\left[\text{(1)(50}\text{.42}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(1)(0)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-622}\text{.0}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

In terms of mass, $\text{ΔH°}$can be given as,

$\begin{array}{l}{\text{N}}_{\text{2}}{\text{H}}_{\text{4(l)}}{\text{=ΔH°}}_{\text{reaction}}\text{=}\frac{\text{-622}\text{.0}\text{kJ}}{\text{1}\text{mol}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{×}\frac{\text{1}\text{mol}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{32}\text{.05}\text{g}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{×}\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-1}{\text{.941×10}}^{\text{4}}\text{kJ}{\text{kg}}^{\text{-1}}\end{array}$

To calculate the $\text{ΔH°}$ for ${\text{4NH}}_{\text{3(g)}}{\text{+3O}}_{\text{2(g)}}\to {\text{2N}}_{\text{2(g)}}{\text{+6H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}$

Standard enthalpy of formation of ${\text{N}}_{\text{2}}\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{NH}}_{\text{3}}\text{=-46}\text{.3}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{H}}_{\text{2}}\text{O=-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}$

Standard enthalpy of formation of ${\text{O}}_2\text{=0}\text{kJ}{\text{mol}}^{\text{-1}}$

$\begin{array}{l}{\text{ΔH°}}_{\text{reaction}}\text{=}\left[{\text{2ΔH°}}_{\text{f}}{\text{(N}}_{\text{2}}{\text{)+6ΔH°}}_{\text{f}}{\text{(H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\text{)}\right]\text{-}\left[{\text{4ΔH°}}_{\text{f}}{\text{(NH}}_3{}_{\text{(l)}}{\text{)+3ΔH°}}_{\text{f}}{\text{(O}}_{\text{2}}\text{)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=}\left[\text{(2)(0)+(6)(-285}\text{.8}\text{kJ}{\text{mol}}^{\text{-1}}\text{)}\right]\text{-}\left[\text{(4)(-46}\text{.3)}\text{kJ}{\text{mol}}^{\text{-1}}\text{)+(3)(0)}\right]\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=1529}\text{.6}\text{kJ}{\text{mol}}^{\text{-1}}\end{array}$

In terms of mass, $\text{ΔH°}$can be given as,

$\begin{array}{l}{\text{NH}}_{\text{3(g)}}{\text{=ΔH°}}_{\text{reaction}}\text{=}\frac{\text{-1529}\text{.6}\text{kJ}}{\text{4}\text{mol}{\text{NH}}_{\text{3}}}\text{×}\frac{\text{1}\text{mol}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}{\text{17}\text{.03 g}{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}}\text{×}\frac{\text{1000}\text{g}}{\text{1}\text{kg}}\\ \\ {\text{ΔH°}}_{\text{reaction}}\text{=-2}{\text{.245×10}}^{\text{4}}\text{kJ}{\text{kg}}^{\text{-1}}\end{array}$

Interpretation Introduction

Interpretation:

The standard enthalpy of decomposition of Hydrazine has to be calculated. The equation has to be balanced and $\text{ΔH°}$ has to be calculated in mass per kilogram for each of them and better fuel has to be identified.

Concept Introduction:

The change in enthalpy that is associated with the formation of one mole of a substance from its related elements being in standard state is called standard enthalpy of formation (${\text{ΔH}}_{\text{f}}\text{°}$).  The standard enthalpy of formation is used to determine the standard enthalpies of compound and element.

The standard enthalpy of reaction is the enthalpy of reaction that takes place under standard conditions.

The equation for determining the standard enthalpies of compound and element can be given by,

${\text{ΔH°}}_{\text{reaction}}\text{=}{\displaystyle \sum {\text{nΔH°}}_{\text{f}}\text{(products)}}\text{-}{\displaystyle \sum {\text{mΔH°}}_{\text{f}}\text{(reactants)}}$

Answer to Problem 6.76QP

Ammonia is better fuel than Hydrazine.

Explanation of Solution

Ammonia acts a better fuel when compared to Hydrazine because it releases more energy per kilogram of substance.