Question

The space shuttle orbiter utilizes the oxidation of methylhydrazine by dinitrogen tetroxide for propulsion:

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

Calculate$\mathbf{∆}\mathit{H}\mathbf{°}$ for this reaction using data in Appendix $\mathbf{4}$.

Short answer

The enthalpy for the reaction ${\text{4N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}{\text{(l) + 5N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(l)}\to {\text{12H}}_{\text{2}}{\text{O(g) + 9N}}_{\text{2}}{\text{(g) + 4CO}}_{\text{2}}\text{(g)}$ is obtained as $∆H_{reaction}^{*}=-4694kJ/mol$.

Step-by-step solution

Concept Introduction

The loss of electrons by a molecule, atom, or ion during a process is known as oxidation. When the oxidation state of a molecule, atom, or ion is enhanced, it is called oxidation.

Step 2: ∆H° for different substances

The reaction provided is –

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

Based on the appendix$4$ , the given values for the$∆H°$of the given substances are –

$$\begin{gathered} {\text{N}}_{\text{2}}{\text{H}}_{\text{3}}{\text{CH}}_{\text{3}}\text{(l) = 54} \\ {\text{5N}}_{\text{2}}{\text{O}}_{\text{4}}\text{(l) = - 20} \\ {\text{N}}_{\text{2}}\text{(g) = 0} \\ {\text{CO}}_{\text{2}}\text{(G) = - 393.5} \\ {\text{H}}_{\text{2}}\text{O(g) = - 242} \end{gathered}$$

Step 3:  ∆H°for the reaction

To calculate for the, use the formula below, plug in thevalues, then solve.

$$\begin{gathered} ∆H_{reaction}^{*}=\sum H_{product}^{*}=\sum H_{reac\tan t}^{*} \\ ∆H_{reaction}^{*}=\left[12\times ∆H_{H_{2}O}^{*}+9\times ∆H_{N_2}^{*}+4\times ∆H_{C{O}_2}^{*}\right]-\left[4\times ∆H_{N_2{H}_{3}C{H}_3}^{\times }+5\times ∆H_{N_2{O}_4}^{\times }\right] \\ ∆H_{reaction}^{*}=[12\times (-242)+(9\times 0)+4\times (-393.5\left)\right]-[4\times 54+5\times (-20\left)\right] \\ ∆H_{reaction}^{*}=-4694KJ/mol \end{gathered}$$

Therefore, the enthalpy of reaction is equal to $-4694kJ/mol$.