Question

Given the following reactions,$$\begin{array}{lr}\mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ}=-534.2 \mathrm{~kJ} \\\\\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ}=-241.8 \mathrm{~kJ}\end{array}$$ Calculate the heat of formation of hydrazine.

Short answer

Answer: The heat of formation of hydrazine (N2H4) is -433.0 kJ/mol.

Step-by-step solution

Write the formation reaction of hydrazine

First, we need to write the formation reaction for hydrazine (N2H4). The formation reaction is the reaction in which 1 mole of a compound is formed from its elements in their standard states. The formation reaction for hydrazine is: \[ \frac{1}{2}\mathrm{N}_{2}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l) \] Our goal is to find the enthalpy change (\(\Delta H^{\circ}\)) of this reaction.

Manipulate the given reactions using Hess's Law

To use Hess's Law, we want to manipulate the given reactions so that when they are added together, they result in the formation reaction of hydrazine. The given reactions are: 1. $$\mathrm{N}_{2}\mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g)~~~~~~\Delta H^{\circ}_1=-534.2~~\mathrm{kJ}$$ 2. $$\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(g)~~~~~~\Delta H^{\circ}_2=-241.8~~\mathrm{kJ}$$ To cancel out the oxygen gas, we need to multiply reaction 2 by 4 (so that the oxygen gas produces and consumes 2 moles): $$4(\mathrm{H}_{2}(g)+\frac{1}{2}\mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}\mathrm{O}(g))$$ $$4\mathrm{H}_{2}(g)+2\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{H}_{2}\mathrm{O}(g)~~~~~~4\Delta H^{\circ}_2=-4(241.8)~~\mathrm{kJ}$$ Now, the first given reaction should be reversed: $$\mathrm{N}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l)+\mathrm{O}_{2}(g)~~~~~~\Delta H^{\circ}_1 = 534.2~~\mathrm{kJ}$$

Add the manipulated reactions to get the formation reaction

Now we can add the manipulated reactions to get the formation reaction of hydrazine: $$\mathrm{N}_{2}(g)+2\mathrm{H}_{2}\mathrm{O}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l)+\mathrm{O}_{2}(g)$$ $$4\mathrm{H}_{2}(g)+2\mathrm{O}_{2}(g) \longrightarrow 4\mathrm{H}_{2}\mathrm{O}(g)$$ Summation: $$\frac{1}{2}\mathrm{N}_{2}(g) + 2\mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}\mathrm{H}_{4}(l)$$ Add the enthalpy changes for the manipulated reactions to calculate the heat of formation for hydrazine: $$\Delta H^{\circ}_{\text{formation}} = 534.2 \mathrm{kJ} - 4(241.8 \mathrm{kJ})$$

Calculate the heat of formation

Finally, calculate the heat of formation for hydrazine: $$\Delta H^{\circ}_{\text{formation}} = 534.2 \mathrm{kJ} - 967.2 \mathrm{kJ} = -433.0 \mathrm{kJ}$$ Therefore, the heat of formation of hydrazine (N2H4) is -433.0 kJ/mol.