Question

Methyl hydrazine (N2H3CH3) is a common liquid propellant used in rocket fuels. Look for the standard molar enthalpies of formation of the reactants and products. Calculate the ∆H˚ for the reaction per mole of N2H3CH3.

4N2H3CH3(l) + 5N2O4(l) → 12H2O(g) + 9N2(g) + 4CO(g)

Answer 1

The standard enthalpy of reaction of the given reaction is -865.71 kJ per mole of N₂H₃CH₃.

What is the standard molar enthalpy of formation?

The standard molar enthalpy of formation of a compound is defined as the enthalpy of formation of 1.0 mol of the pure compound in its stable state from the pure elements in their stable states at P = 1.0 bar at a constant temperature.

Let's consider the following equation.

4 N₂H₃CH₃(l) + 5 N₂O₄(l) → 12 H₂O(g) + 9 N₂(g) + 4 CO(g)

We can calculate the standard enthalpy of the reaction using the following expression.

ΔH° = Σnp × ΔH°f(p) - Σnr × ΔH°f(r)

where,

• ΔH° is the standard enthalpy of the reaction.
• n is stoichiometric coefficient.
• ΔH°f is the standard molar enthalpy of formation.
• p are the products.
• r are the reactants.

ΔH° = 12 mol × ΔH°f(H₂O(g)) + 9 mol × ΔH°f(N₂(g)) + 4 mol × ΔH°f(CO(g)) - 4 mol × ΔH°f(N₂H₃CH₃(l)) - 5 mol × ΔH°f(N₂O₄(l))

ΔH° = 12 mol × (-241.81 kJ/mol) + 9 mol × (0 kJ/mol) + 4 mol × (-110.53 kJ/mol) - 4 mol × (54.20 kJ/mol) - 5 mol × (-19.56 kJ/mol)

ΔH° = -3462.84 kJ

In the balanced equation, there are 4 moles of N₂H₃CH₃. The standard enthalpy of reaction per mole of N₂H₃CH₃ is:

-3462.84 kJ / 4 mol = -865.71 kJ/mol

The standard enthalpy of reaction of the given reaction is -865.71 kJ per mole of N₂H₃CH₃.

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