Chapter 5, Problem 86GQ

Interpretation Introduction

Interpretation:

The enthalpy change of combustion per gram of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}$ has to be calculated.

Concept Introduction:

The standard enthalpy change of combustion of a compound is the enthalpy change which occurs when one gram of the compound is burned completely in oxygen under standard conditions, and with everything in its standard state.

                    ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  $\text{ΔH}\text{=}{\text{Δ}}_{\text{r}}\text{H×number of moles}$

Answer to Problem 86GQ

The enthalpy change of combustion per gram of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is $\text{-8}\text{.919 J/gK}$ and the enthalpy change of combustion per gram of ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}$ is $\text{-32}\text{.93J/gK}$

Explanation of Solution

For ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ the balanced equation is:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}{\text{+ O}}_{\text{2}}\stackrel{}{\to }{\text{N}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\left[\left(\text{1mol×-393}\text{.5kJ/mol}\right)\text{+}\left(\text{0kJ/mol}\right)\right]$-$\left[\left(\text{1mol×50kJ/mol}\text{+0}\right)\right]$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-446}\text{.5kJ}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  ${\text{Δ}}_{\text{r}}\text{H}$=$\text{-446}\text{.5kJ}$$\text{/mol}\text{×}\frac{\text{1mol}}{\text{50}\text{.0604 g}}$=$\text{-8}\text{.919 J/gK}$

So, the molar enthalpy of formation of hydrazine is $\text{-8}\text{.919 J/gK}$

For ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}$ the balanced equation is:

  ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}{\text{ + 4O}}_{\text{2}}\to {\text{2CO}}_{\text{2}}{\text{+ 4H}}_{\text{2}}\text{O}\text{+}{\text{N}}_{\text{2}}$

  ${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=${\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(products)}\text{-}{\text{ΣnΔ}}_{\text{f}}{\text{H}}^{\text{0}}\text{(reactants)}$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\left[\left(\text{2mol×-393}\text{.5kJ/mol}\right)\text{+}\left(\text{4mol×-285}\text{.83kJ/mol}\right)+0\right]$-$\left[\left(\text{1mol×48}\text{.9kJ/mol}\text{+0}\right)\right]$

${\text{Δ}}_{\text{r}}{\text{H}}^{\text{0}}$=$\text{-1979}\text{.2kJ}$

The change in enthalpy, $\text{ΔH}$ in kJ per mole of a given reactant for the reaction can be calculated as:

  ${\text{Δ}}_{\text{r}}\text{H=}\frac{\text{enthalpy change}}{\text{number of moles }}$

  $\text{ΔH}\text{=}{\text{Δ}}_{\text{r}}\text{H×number of moles}$

  ${\text{Δ}}_{\text{r}}\text{H}$=$\frac{\text{-1979}\text{.2kJ}}{\text{1mol}}\text{×}\frac{\text{1mol}}{\text{60}\text{.1g}}$=$\text{-32}\text{.93J/gK}$

So, the molar enthalpy of formation of $\text{1,1-dimethylhydrazine}$ is $\text{-32}\text{.93J/gK}$

Conclusion

The enthalpy change of combustion per gram of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ and ${\text{N}}_{\text{2}}{\text{H}}_{\text{2}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}$ was calculated.