2. Hydrazine (N2H4) is continued to be used as a liquid propellant and for termi

Question

2. Hydrazine (N2H4) is continued to be used as a liquid propellant and for terminal descent of spacecraft. All the following 3 reactions (left unbalanced for you to solve) have to happen to produce the gases for the propulsion: NHs + N2H4--> N2 + H2 (3) Reactions 1 and 2 are extremely exothermic and they produce a large volume of hot gas from a small volume of liquid, making hydrazine a fairly efficient thruster propellant with a vacuum specific impulse of about 220 seconds Determine the volume of hydrogen (H2) produced from the combustion of 2 L of hydrazine at the edge of the troposphere where the temperature is -68 °C and the atmospheric pressure is 0.1 atm Assume ideal gas low is applicable for this problem.

Explanation / Answer

Write the balanced reactions first.

3 N2H4 --------> 4 NH3 + N2 ……..(1)

N2H4 ----------> N2 + 2 H2 ………(2)

2 NH3 + N2H4 ---------> 2 N2 + 5 H2 ………(3)

Multiply (3) by 2 and add to equations (1) and (2) to obtain

3 N2H4 + N2H4 + 4 NH3 + 2 N2H4 --------> 4 NH3 + N2 + N2 + 2 H2 + 4 N2 + 10 H2

Cancel the common terms and obtain

6 N2H4 --------> 6 N2 + 12 H2

or, N2H4 ---------> N2 + 2 H2

As per the balanced chemical equation,

1 mole N2H4 = 2 moles H2

Since the ideal gas law is valid, we may assume for hydrazine that 1 mole of N2H4 occupies 22.4 L at STP; therefore, mole(s) of N2H4 corresponding to 2 L N2H4 = (2 L)*(1 mole/22.4 L) = 0.08928 mole.

As per the stoichiometry of the equation,

0.08928 mole N2H4 = (0.08928 mole N2H4)*(2 mole H2/1 mole N2H4) = 0.17856 mole H2.

Use the ideal gas law; the temperature is T = -68°C = (-68 + 273) K = 205 K; we have

P*V = n*R*T

====> V = n*R*T/P = (0.17856 mole)*(0.082 L-atm/mol.K)*(205 K)/(0.1 atm) = 30.015936 L 30.016 L (ans).

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