Chapter 4, Problem 90QRT

Interpretation Introduction

Interpretation:

Reaction that provides more heat has to be identified.

Answer to Problem 90QRT

The heat evolved from one gram of 1,1-dimethylhydrazine is more than enthalpy evolved from one gram of hydrazine.

Explanation of Solution

The given equations for combustion are as follows,

    $\begin{array}{l}{N}_2{H}_4\left(l\right)+O_2\left(g\right)\to N_2\left(g\right)+2H_{2}O\left(g\right)\\ \\ N_2{H}_2{\left(C{H}_3\right)}_2\left(l\right)+4O_2\left(g\right)\to 2C{O}_2\left(g\right)+N_2\left(g\right)+4H_{2}O\left(g\right)\end{array}$

The combustion enthalpy value is as follows,

    $\begin{array}{c}{\Delta }_c{H}^{\circ }{}_{hydrazine}\text{ = }\left[{\Delta }_f{H}^{\circ }{N}_2\left(g\right)+2{\Delta }_f{H}^{\circ }\left(H_{2}O\left(g\right)\right)\right]-\left[{\Delta }_f{H}^{\circ }{N}_2{H}_4\left(l\right)-{\Delta }_f{H}^{\circ }\left(O_2\left(g\right)\right)\right]\\ \\ \text{= }\left(\text{0 kJ/mol}\right)\text{+2}\left(\text{-241}\text{.818 kJ/mol}\right)\text{-}\left(\text{50}\text{.6 kJ/mol}\right)\text{-}\left(\text{0 kJ/mol}\right)\\ \\ =\text{ -534}\text{.2 kJ}\end{array}$

$\begin{array}{c}{\Delta }_c{H}^{\circ }{}_{DMH}\text{ = }\left[2\times {\Delta }_f{H}^{\circ }C{O}_2\left(g\right)+4{\Delta }_f{H}^{\circ }\left(H_{2}O\left(g\right)\right)+{\Delta }_f{H}^{\circ }\left(N_2\left(g\right)\right)\right]-\left[{\Delta }_f{H}^{\circ }DMH-4{\Delta }_f{H}^{\circ }\left(O_2\left(g\right)\right)\right]\\ \\ \text{= }\left(\text{2×-393}\text{.509 kJ/mol}\right)\text{+2}\left(\text{-241}\text{.818 kJ/mol}\right)\text{-}\left(\text{50}\text{.6 kJ/mol}\right)\text{-}\left(\text{0 kJ/mol}\right)-\left(\text{49}\text{.2kJ/mol}\right)-4\left(\text{-0kJ/mol}\right)\\ \\ =\text{ -1803}\text{.5 kJ}\end{array}$

The enthalpy evolved for one gram of the two given substances are calculated as shown below,

    $\begin{array}{l}For{\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}\Rightarrow \frac{534.2\text{ kJ produced}}{1{\text{ mole N}}_{\text{2}}{\text{H}}_{\text{4}}}\times \frac{1{\text{ mole N}}_{\text{2}}{\text{H}}_{\text{4}}}{32.0450{\text{ g N}}_{\text{2}}{\text{H}}_{\text{4}}}\\ \\ =\text{ 16}\text{.67 }\frac{kJ\text{ produced}}{g{\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}}\left[{\text{Molar mass of N}}_{\text{2}}{\text{H}}_{\text{4}}\text{=32}\text{.0450g/mol}\right]\\ \\ For\text{ DMH }\Rightarrow \frac{1803.5\text{ kJ produced}}{1\text{ mole DMH}}\times \frac{1\text{ mole DMH}}{60.0980\text{ g DMH}}\\ \\ =\text{ 30}\text{.009 }\frac{kJ\text{ produced}}{g\text{ DMH}}\left[\text{Molar mass of DMH=}60.0980\text{g/mol}\right]\end{array}$

From above calculations it is clear that enthalpy evolved from one gram of 1,1-dimethylhydrazine is more than enthalpy evolved from one gram of hydrazine.