Hydrazine, N2H4 can be made (on paper) by reaction of molecular nitrogen and molecular hydrogen,
2 H 2 ( g ) + N 2 ( g ) --> N 2 H 4 ( g ) and suppose that the equilibrium constant Kp has the value 1×10^–3.
(a) A stoicheometric mixture of H2 and N2 is placed in a container with 1.00 atm total pressure, and the container is sealed. Compute the equilibrium concentration of hydrazine and the final pressure in the container.
(b) If N2H4 were prepared and put into a cylinder at 1 atm pressure, and then subsequently allowed to come to equilibrium at this same total pressure, how much hydrazine would remain (what partial pressure)?
(c) Hydrazine is used as rocket fuel, so it can be made in large amounts. How do you suppose this is done? Why doesn’t it just decompose to nitrogen and hydrogen?
(a) Well, let's take a look at this equilibrium reaction. We have 2 moles of hydrogen gas reacting with 1 mole of nitrogen gas to form 1 mole of hydrazine. So, at equilibrium, if we assume x moles of hydrazine are formed, we will have 2x moles of hydrogen gas and x moles of nitrogen gas consumed.
Now, the total pressure in the container is given as 1.00 atm. Since we have a stoichiometric mixture, the initial pressure of hydrogen gas is equal to the initial pressure of nitrogen gas, which we can call "P".
Using the ideal gas law, we can relate the partial pressure of each gas to its moles:
P(H2) = (2x / x + 2x) * P
P(N2) = (x / x + 2x) * P
The total pressure at equilibrium is given by the sum of the pressure of each component:
P(total) = P(H2) + P(N2) + P(N2H4)
The equilibrium constant Kp is defined as:
Kp = (P(N2H4) / P(N2))^b * (P(H2))^c
Here, b is the coefficient of N2H4 and c is the coefficient of H2.
Since Kp = 1×10^-3, we can rearrange the equation to solve for P(N2H4):
P(N2H4) = Kp * (P(N2))^(-b) * (P(H2))^(-c)
Substituting the expressions for P(N2) and P(H2) in terms of x into the equation above, we can solve for P(N2H4). Given that b = 1 and c = 2, you can plug in the values and solve for x. Then, you can calculate the equilibrium concentration of hydrazine using the equation x/volume, where volume is the volume of the container.
(b) If we start with 1 atm of hydrazine in the cylinder and let it come to equilibrium at the same total pressure, we need to determine how much of the hydrazine is left.
Using the same equation as in part (a), we can set up the equation:
P(N2H4) = Kp * (P(N2))^(-b) * (P(H2))^(-c)
Since we're given that the total pressure is 1 atm, we substitute P(total) = 1 atm into the equation above and solve for P(N2H4), which gives us the partial pressure of hydrazine remaining in the cylinder.
(c) Hydrazine is indeed used as rocket fuel because it has high energy content and releases a large amount of energy when it decomposes. However, to prevent the spontaneous decomposition of hydrazine, it is typically stored with stabilizers and inhibitors. These additives help to slow down or prevent the decomposition reaction. Additionally, the storage conditions, such as temperature and pressure, are carefully controlled to minimize the decomposition rate.
When hydrazine is used as rocket fuel, it is typically combined with an oxidizer, such as dinitrogen tetroxide (N2O4), which enhances the combustion process. The hydrazine and oxidizer are stored separately and are mixed in the rocket engine just before ignition to avoid premature reactions.
So, in summary, the stability of hydrazine is maintained by careful storage conditions, the use of stabilizers and inhibitors, and the separation from oxidizers until the moment of combustion.
(a) To compute the equilibrium concentration of hydrazine and the final pressure in the container, we can use the equation for the equilibrium constant:
Kp = (P[N2H4]) / (P^2[H2] * P^2[N2])
Since a stoichiometric mixture of H2 and N2 is placed in the container, we can assume the initial concentrations of H2 and N2 are the same. Let's assume the initial concentration of H2 and N2 is x. Therefore, the initial pressure of each gas is x atm.
According to the balanced equation, the mole ratio between N2H4, H2, and N2 is 1:2:1, which means the equilibrium concentration of N2H4 is 2x.
Substituting these values into the equation for Kp:
1×10^–3 = (2x) / (x^2 * x^2)
Simplifying the equation:
1×10^–3 = 2 / (x^3)
Cross-multiplying and rearranging:
2x^3 = 1×10^–3
x^3 = (1×10^–3) / 2
x^3 = 5×10^–4
Taking the cube root of both sides:
x = ∛(5×10^–4)
x ≈ 0.082
Therefore, the equilibrium concentration of hydrazine is 2x ≈ 2 * 0.082 ≈ 0.164 atm.
The final pressure in the container is the sum of the partial pressures of H2, N2, and N2H4:
Pfinal = P[H2] + P[N2] + P[N2H4]
Pfinal = x + x + 2x
Pfinal = 4x
Pfinal = 4 * 0.082
Pfinal ≈ 0.328 atm
So, the equilibrium concentration of hydrazine is approximately 0.164 atm and the final pressure in the container is approximately 0.328 atm.
(b) If N2H4 were prepared and put into a cylinder at 1 atm pressure and then allowed to come to equilibrium at the same total pressure, we can find the partial pressure of hydrazine (N2H4) remaining.
Let's assume the initial concentration of N2H4 is y atm. Since the initial total pressure is also 1 atm, the initial partial pressure of N2H4 is also y atm.
Using the equation for the equilibrium constant:
Kp = (P[N2H4]) / (P^2[H2] * P^2[N2])
Substituting the known values:
1×10^–3 = y / (1^2 * 1^2)
1×10^–3 = y
Therefore, the partial pressure of hydrazine remaining is 1×10^–3 atm.
(c) Hydrazine can be made in large amounts by various industrial processes, such as the Raschig Process or the Peroxide Process. These processes involve reacting ammonia (NH3) with an oxidizing agent, such as sodium hypochlorite or hydrogen peroxide, to produce hydrazine.
Hydrazine does not readily decompose to nitrogen and hydrogen because the decomposition reaction requires a significant input of energy. Additionally, the decomposition reaction is endothermic, which means it absorbs heat from the surroundings. Therefore, unless the hydrazine is exposed to high temperatures or catalysts, it remains relatively stable and does not decompose easily. This stability allows it to be used as rocket fuel without rapidly decomposing during combustion.
(a) To find the equilibrium concentration of hydrazine and the final pressure in the container, we need to use the equilibrium constant expression and the given information.
1. Start by writing the balanced chemical equation:
2 H2(g) + N2(g) ⇌ N2H4(g)
2. Let's assume the initial concentration of H2 and N2 is "x" M. Since the stoichiometry of the reaction is 2:1, the initial concentration of N2H4 would be 0 M.
3. At equilibrium, let's assume the concentration of N2H4 is "y" M.
4. Using the given equilibrium constant Kp = 1×10^–3, we can write the equilibrium expression as:
Kp = (y^2) / (x^2 * (1 - y))
5. Since the total pressure is 1.00 atm and the initial pressure of each gas is the same, the partial pressure of H2 and N2 at equilibrium would be (1.00 atm - y).
6. According to the ideal gas law, the partial pressures can be related to the molar concentrations:
Partial Pressure = (Concentration * R * Temperature) / Volume
7. Rearranging the equation to solve for concentration:
Concentration = (Partial Pressure * Volume) / (R * Temperature)
8. The final pressure is the sum of the partial pressures:
Final Pressure = Partial Pressure of H2 + Partial Pressure of N2 + Partial Pressure of N2H4
9. Substitute the relevant variables and solve the equations to find the equilibrium concentration of hydrazine and the final pressure in the container.
(b) If N2H4 is prepared and put into a cylinder at 1 atm pressure, and then allowed to reach equilibrium, the amount of hydrazine remaining can be determined by using the equilibrium constant expression.
1. Use the equilibrium constant expression from part (a):
Kp = (y^2) / (x^2 * (1 - y))
2. Since the total pressure is 1 atm, the partial pressure of N2H4 at equilibrium is also y atm.
3. Set up the equation using the given equilibrium constant and the partial pressure of N2H4:
1×10^–3 = (y^2) / ((1 atm - y)^2)
4. Solve the equation to find the value of y, which represents the partial pressure of N2H4 at equilibrium.
(c) Hydrazine can be made in large amounts for use as rocket fuel through various methods, including the following:
1. The most common method involves the catalytic reduction of nitrogen gas (N2) using hydrogen gas (H2) over a catalyst, such as a transition metal catalyst like Raney nickel.
2. In this process, nitrogen and hydrogen are reacted under carefully controlled conditions, typically at high temperatures and pressures. The reaction produces hydrazine and water as byproducts.
3. The presence of a catalyst helps facilitate the reaction by providing an alternative pathway with lower activation energy, allowing the reaction to proceed at a faster rate.
4. To prevent decomposition of hydrazine into nitrogen and hydrogen, the reaction is usually performed under inert conditions to avoid the presence of other reactive species that could cause unwanted side reactions.
5. Additionally, the hydrazine produced is often stored and handled in a way that minimizes its decomposition. It is usually stored in tanks or containers that are resistant to degradation and under controlled temperature and pressure conditions.