Hydrazine (N₂H₄) is used as a rocket fuel. The combustion products are nitrogen (N₂) and water (H₂O). Use the following information about bond energies to estimate the change in enthalpy associated with the combustion of 1 mol of hydrazine. $$\begin{array} { | c | c | } \hline \text { Bond } & \begin{array} { c } \text { Bond Energy } \ ( \mathrm { kJ } / \mathrm { mol } ) \end{array} \ \hline \mathrm { N } - \mathrm { N } & 163 \ \hline \mathrm { N } = \mathrm { N } & 418 \ \hline \mathrm { N } & 941 \ \equiv & \ \mathrm { N } & \ \hline \mathrm { N } - \mathrm { H } & 388 \ \hline \mathrm { O } - \mathrm { O } & 146 \ \hline \mathrm { O } = \mathrm { O } & 495 \ \hline \mathrm { O } - \mathrm { H } & 463 \ \hline \end{array}$$

Question

Quizplus · Accepted Answer

The combustion of hydrazine can be represented by the following equation:N2H4 + O2 -> N2 + 2H2OThe change in enthalpy associated with this reaction can be estimated using the bond energies of the reactants and products. The bond energies of the reactants are:N-N: 163 kJ/molN=N: 418 kJ/molN-H: 388 kJ/molThe bond energies of the products are:N-N: 163 kJ/molO=O: 495 kJ/molO-H: 463 kJ/molThe change in enthalpy associated with the reaction is:ΔH = Σ(bond energies of reactants) - Σ(bond energies of products)ΔH = (163 kJ/mol + 418 kJ/mol + 388 kJ/mol) - (163 kJ/mol + 495 kJ/mol + 463 kJ/mol)ΔH = -583 kJ/molTherefore, the change in enthalpy associated with the combustion of 1 mol of hydrazine is -583 kJ/mol.

Hydrazine (N2H4) Is Used as a Rocket Fuel A) -583 K J

Hydrazine (N₂H₄) Is Used as a Rocket Fuel A) -583 K J