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Solved: Hydrazine, N2H4, decomposes according to the
Chapter 6, Problem 76P(choose chapter or problem)
Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), decomposes according to the following reaction:
\(3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\)
(a) Given that the standard enthalpy of formation of hydrazine is \(50.42 \mathrm{~kJ} / \mathrm{mol}\), calculate \(\Delta H^{\circ}\) for its decomposition. (b) Both hydrazine and ammonia burn in oxygen to produce \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{N}_{2}(g)\). Write balanced equations for each of these processes and calculate \(\Delta H^{\circ}\) for each of them. On a mass basis (per \(\mathrm{kg}\) ), would hydrazine or ammonia be the better fuel?
Questions & Answers
QUESTION:
Hydrazine, \(\mathrm{N}_{2} \mathrm{H}_{4}\), decomposes according to the following reaction:
\(3 \mathrm{~N}_{2} \mathrm{H}_{4}(l) \longrightarrow 4 \mathrm{NH}_{3}(g)+\mathrm{N}_{2}(g)\)
(a) Given that the standard enthalpy of formation of hydrazine is \(50.42 \mathrm{~kJ} / \mathrm{mol}\), calculate \(\Delta H^{\circ}\) for its decomposition. (b) Both hydrazine and ammonia burn in oxygen to produce \(\mathrm{H}_{2} \mathrm{O}(l)\) and \(\mathrm{N}_{2}(g)\). Write balanced equations for each of these processes and calculate \(\Delta H^{\circ}\) for each of them. On a mass basis (per \(\mathrm{kg}\) ), would hydrazine or ammonia be the better fuel?
ANSWER:
Step 1 of 4
(a)
Here, we are going to calculate the standard enthalpy change for the decomposition reaction. Given that,
\(\mathrm{~N}_{2} \mathrm{H}_{4}(\mathrm{l}) \rightarrow 4 \mathrm{NH}_{3}(\mathrm{~g})+\mathrm{N}_{2}(\mathrm{~g})-(1)\)
\(\Delta H_{f}^{\circ}\left(\mathrm{NH}_{3}, g\right)=-46.3 \mathrm{~kJ} / \mathrm{mol}\)
\(\Delta H_{f}^{\circ}\left(\mathrm{N}_{2} H_{4}, l\right)=50.42 \mathrm{~kJ} / \mathrm{mol}\)
We know that the standard enthalpy change of a reaction is calculated by using the following equation.
\(\Delta H_{r x n}^{0}=\sum n \Delta H_{f}^{o} \text { (product) }-\sum n \Delta H_{f}^{o} \text { (reactant) }\)
Therefore, from equation (1), we get
\(\Delta H_{r x n}^{o} =\left[4 \times \Delta H_{f}^{o}\left(N H_{3}, g\right)+\Delta H_{f}^{o}\left(N_{2}, g\right)\right]-\left[3 \times \Delta H_{f}^{o}\left(N_{2} H_{4}, l\right)\right]\)
\(=4 \times(-46.3 \mathrm{~kJ} / \mathrm{mol})+0-3 \times 50.42 \mathrm{~kJ} / \mathrm{mol}\)
\(=-185.2 \mathrm{~kJ} / \mathrm{mol}-151.26 \mathrm{~kJ} / \mathrm{mol}\)
\(=-336.46 \mathrm{~kJ} / \mathrm{mol}\)
\(=-336.5 \mathrm{~kJ} . \mathrm{mol}\)
Therefore, the standard enthalpy change for the decomposition reaction is \(-336.5 \mathrm{~kJ} / \mathrm{mol}\).