Chapter 5, Problem 137AP

Hydrazine $\left({\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\right)$ decomposes to form ammonia and nitrogen gases. (a) Write a balanced chemical equation for this process. (b) Given that the standard enthalpy of formation of hydrazine is 50.42 kJ/mol, calculate $\Delta \text{Hrxnº}$ for its decomposition. (c) Both hydrazine and ammonia will burn in oxygen to produce $\Delta \text{Hrxnº}$ Write balanced equations for these processes and determine ${\text{H}}_{\text{2}}\text{O}\left(l\right){\text{ and N}}_{\text{2}}\left(g\right).$ for each process. (d) If equal masses of hydrazine and ammonia were burned in separate bomb calorimeter experiments, which would cause the greater increase in temperature?

Interpretation Introduction

Interpretation:

The balanced reaction, the enthalpy of formation, the enthalpy of reaction, for the given reaction is to be determined and the balanced reaction for burning of hydrazine and ammonia is to be written for each processes and their enthalpy of reaction is to be determined. The compound that causes more increase in temperature is to be identified.

Concept Introduction:

The standard enthalpy change for a reaction is the amount of enthalpy change that occurs at the standard conditions.

The standard enthalpy of reaction is to be determined using the equation as follows:

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

Here, the stoichiometric coefficients are represented by m for reactants and n for products and the enthalpies of formation at standard conditions are represented by $\Delta H_f^{°}$.

The standard enthalpy of formation is the amount of heat change when one mole of compound is formed from its integral elements that are present in their standard states.

The heat is related to the change in temperature of a material, which is as follows:

$q=sm\Delta T.$

Here, q represents heat, s signifies specific heat, m shows mass, and $\Delta T$ is the change in the temperature of the material.

Answer to Problem 137AP

Solution:

a)

${\text{3N}}_{\text{2}}{\text{H}}_4\left(g\right)\to 4{\text{NH}}_3\left(g\right)+{\text{N}}_2\left(g\right).$

b) The reaction for hydrazine and oxygen is as follows:

${\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right).$

and the reaction for ammonia and oxygen is as follows:

${\text{4NH}}_3\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{N}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right).$

$-336.46\text{ kJ}/\text{mol}$.

c)

The enthalpy of reaction of hydrazine with oxygen is $-622.02\text{ kJ}/\text{mol}$

and for reaction of ammonia with oxygen is $-1529.6\text{ kJ}/\text{mol}$.

d)

The heat is more for ammonia, so it will cause more increase in temperature.

Explanation of Solution

a) A balanced chemical equation for the given process.

The reaction for decomposition of hydrazine is as follows:

${\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\to {\text{NH}}_3\left(g\right)+{\text{N}}_2\left(g\right).$

The balancing of a chemical reaction is done by the law of conservation of mass. According to this law, the mass of a species in the chemical reaction is conserved. It means that the mass of an atom on the reactant side is equal to that on the product side.

The reaction after balancing is given as follows:

${\text{3N}}_{\text{2}}{\text{H}}_4\left(g\right)\to 4{\text{NH}}_3\left(g\right)+{\text{N}}_2\left(g\right).$

b) $\Delta H{°}_{rxn}$ for its decomposition

Standard enthalpy of formation of hydrazine is $\Delta H{°}_f\left[{\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\right]=50.42\text{ KJ}/\text{mol}$

The reaction for decomposition of hydrazine is as follows:

${\text{3N}}_{\text{2}}{\text{H}}_4\left(g\right)\to 4{\text{NH}}_3\left(g\right)+{\text{N}}_2\left(g\right)$.

From appendix 2, the enthalpy of formation values is as follows:

$\Delta H{°}_f\left[{\text{NH}}_3\left(g\right)\right]=-46.3\text{ kJ}/\text{mol}\text{.}$

The nitrogen gas is in its most stable form, so its enthalpy of formation is zero.

The standard enthalpy of reaction is calculated by the expression, which is as follows:

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

Now, the standard enthalpy of given reaction is calculated as follows:

$\Delta H{°}_{rxn}=\left\{4\Delta H{°}_f\left[{\text{NH}}_3\left(g\right)\right]+\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right]\right\}-\left\{3\Delta H{°}_f\left[{\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\right]\right\}.$

Substitute $-46.3\text{ kJ}/\text{mol}$

for $\Delta H{°}_f\left[{\text{NH}}_3\left(g\right)\right]$, 0 for $\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right],$

and $50.42\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\right]$

in the above equation.

$\begin{array}{c}\Delta H{°}_{rxn}=\left\{4\times -46.3\text{ kJ}/\text{mol}+0\text{ kJ}/\text{mol}\right\}-\left\{3\times 50.42\text{ kJ}/\text{mol}\right\}\\ =-185.2\text{ kJ}/\text{mol}-151.26\text{ kJ}/\text{mol}\\ =-336.46\text{ kJ}/\text{mol}\text{.}\end{array}$

Hence, the enthalpy of reaction is $-336.46\text{ kJ}/\text{mol}$.

c) Balance equations for the processes and $\Delta H{°}_{rxn}$

for the process.

The reaction of hydrazine with oxygen is as follows:

${\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)+{\text{O}}_2\left(g\right)\to {\text{N}}_2\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The reaction of ammonia with oxygen is as follows:

${\text{4NH}}_3\left(g\right)+3{\text{O}}_2\left(g\right)\to 2{\text{N}}_2\left(g\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right).$

Now, the standard enthalpy of the given reaction is calculated as follows:

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

Now, the standard enthalpy of reaction of hydrazine with oxygen is given as follows:

$\Delta H{°}_{rxn}=\left\{\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right]+2\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{\Delta H{°}_f\left[{\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\right]+\Delta H{°}_f\left[{\text{O}}_2\left(g\right) \right]\right\}$.

Substitute $-285.8\text{ kJ}/\text{mol}$

for $\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$, 0 for $\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right]$, 0 for $\Delta H{°}_f\left[{\text{H}}_2\left(g\right)\right],$

and $50.42\text{ kJ}/\text{mol}$ for $\Delta H{°}_f\left[{\text{N}}_{\text{2}}{\text{H}}_4\left(g\right)\right]$

in the above equation.

$\begin{array}{c}\Delta H{°}_{rxn}=\left\{0+2\left(-285.8\text{ kJ}/\text{mol}\right)\right\}-\left\{\left(50.42\text{ kJ}/\text{mol}\right)+0\right\}\\ =-571.6\text{ kJ}/\text{mol}-50.42\text{ kJ}/\text{mol}\\ =-622.02\text{ kJ}/\text{mol}\text{.}\end{array}$

Now, the standard enthalpy of reaction of ammonia with oxygen is given as follows:

$\Delta H{°}_{rxn}=\left\{2\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right]+6\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]\right\}-\left\{4\Delta H{°}_f\left[{\text{NH}}_3\left(g\right)\right]+3\Delta H{°}_f\left[{\text{O}}_2\left(g\right) \right]\right\}$.

Substitute $-285.8\text{ kJ}/\text{mol}$

for $\Delta H{°}_f\left[{\text{H}}_{\text{2}}\text{O}\left(l\right)\right]$, 0 for $\Delta H{°}_f\left[{\text{N}}_2\left(g\right)\right]$, 0 for $\Delta H{°}_f\left[{\text{H}}_2\left(g\right)\right]$

and $-46.3\text{ kJ}/\text{mol}$

for $\Delta H{°}_f\left[{\text{NH}}_3\left(g\right)\right]$

in the above equation.

$\begin{array}{c}\Delta H{°}_{rxn}=\left\{2\left(0\right)+6\left(-285.8\text{ kJ}/\text{mol}\right)\right\}-\left\{4\left(-46.3\text{ kJ}/\text{mol}\right)+3\left(0\right)\right\}\\ =-1714.8\text{ kJ}/\text{mol}-\left(-185.2\text{ kJ}/\text{mol}\right)\\ =-1714.8\text{ kJ}/\text{mol}+185.2\text{ kJ}/\text{mol}\\ =-1529.6\text{ kJ}/\text{mol}\text{.}\end{array}$

Hence, the enthalpy of reaction of hydrazine with oxygen is $-622.02\text{ kJ}/\text{mol}$

and for reaction of ammonia with oxygen is $-1529.6\text{ kJ}/\text{mol}$.

d) Equal masses of hydrazine and ammonia burn in separate bomb calorimeter, the one which will cause the greater increase in temperature.

The heat is calculated by the expression, which is as follows:

$\begin{array}{c}q=n\times \Delta H\\ =\frac{m}{M}\times \Delta H.\end{array}$ …… (1)

For ammonia:

The molar mass is $M=17\text{ g}/\text{mol}$

The enthalpy is $-1529.6\text{ kJ}/\text{mol}$.

The mass is $m=1\text{ g}$

Substitute $1\text{ g}$

for $m$, $17\text{ g}/\text{mol}$ for $M,$

and $-1529.6\text{ kJ}/\text{mol}$

for $\Delta H$ in the equation (1).

$\begin{array}{c}{q}_{\text{ammonia}}=\frac{1\cancel{\text{ g}}}{17\cancel{\text{g}}/\cancel{\text{mol}}}\times \left(-1529.6\text{ kJ}/\cancel{\text{mol}}\right)\\ =-89.97\text{ kJ}\text{.}\end{array}$

For hydrazine:

The molar mass is $M=32\text{ g}/\text{mol}\text{.}$

The enthalpy is $-622.02\text{ kJ}/\text{mol}\text{.}$

The mass is $m=1\text{ g}\text{.}$

Substitute $1\text{ g}$

for $m$, $32\text{ g}/\text{mol}$ for $M,$

and $-622.02\text{ KJ}/\text{mol}$

for $\Delta H$ in the above equation.

$\begin{array}{c}{q}_{\text{hydrazine}}=\frac{1\cancel{\text{ g}}}{32\cancel{\text{g}}/\cancel{\text{mol}}}\times \left(-622.02\text{ KJ}/\cancel{\text{mol}}\right)\\ =-19.40\text{ kJ}\text{.}\end{array}$

As the heat is more for ammonia, so it will cause more increase in temperature.