Dinitrogen tetroxide decomposes according to N2O4 (g) = 2NO2 (g)
In a certain experiment, N2O4(g) at an initial pressure of 0.554 bar is introduced into an empty reaction container; after equilibrium is established, the total pressure is 0.770 bar. A quantity of NO2(g) is injected into the container and the total pressure jumps to 0.906 bar. Calculate the total pressure after equilibrium is reestablished.
I need help setting up this problem. Please help.
I would do this.
............N2O4 ==> 2NO2
initial..0.554.........0
change....-p..........+2p
equil....0.554-p.......2p
Total P @ equil = 0.770; therefore,
0.554-p+2p = 0.770 and this allows you to calculate Kp.
NO2 is injected to make Ptotal = 0.906 bar so that makes 0.906-0.770 = 0.136 bar injected.
I would make a new ICE chart with NO2 start = 0.136 + old equil pressure and N2O4 = old equil and recalculate new pressures using the Kp from the first part. You know, of course that the reaction will be shifted to the left.
Sure, I'd be happy to help. But first, let me tell you a chemistry joke to lighten the mood:
Why do chemists like nitrates so much?
Because they're cheaper than day rates!
Now, let's set up the problem. We're given the chemical equation for the decomposition of dinitrogen tetroxide (N2O4) into nitrogen dioxide (NO2):
N2O4 (g) ⇌ 2NO2 (g)
We're also given the initial pressure of N2O4 and the total pressure after equilibrium is established. Let's label these values:
Initial pressure of N2O4 (Pi) = 0.554 bar
Total pressure at equilibrium (Ptotal) = 0.770 bar
Now, we're told that a quantity of NO2 gas is injected, causing the total pressure to jump to 0.906 bar. Let's label this change in pressure as ΔP:
Change in pressure (ΔP) = 0.906 bar - 0.770 bar = 0.136 bar
Since the reaction is N2O4 (g) ⇌ 2NO2 (g), we can see that the number of moles of N2O4 decreases by 1 mole for every 2 moles of NO2 produced. Therefore, we can conclude that the change in pressure is also proportional to the change in the number of moles of N2O4:
ΔP ∝ Δn(N2O4)
Now, to calculate the total pressure after equilibrium is reestablished, we need to find the change in the number of moles of N2O4. We can use the ideal gas law to relate pressure, volume, and number of moles:
PV = nRT
Where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Since the temperature and volume are constant, we can simplify the equation to:
P ∝ n
So, if the initial pressure of N2O4 decreases by ΔP, the number of moles of N2O4 also decreases by Δn(N2O4):
ΔP ∝ Δn(N2O4)
Now, with this information, we can calculate the total pressure after equilibrium is reestablished. But before we do that, let me tell you another joke:
Why don't scientists trust atoms?
Because they make up everything!
Now, back to the problem. We know that the initial pressure of N2O4 (Pi) is 0.554 bar, and the change in pressure (ΔP) is 0.136 bar. Therefore, the change in the number of moles of N2O4 can be calculated using the relationship we established:
ΔP ∝ Δn(N2O4)
0.136 bar ∝ Δn(N2O4)
Since the stoichiometric coefficient of N2O4 is 1, we can say that:
Δn(N2O4) = 0.136 bar
Now, let's calculate the total pressure after equilibrium is reestablished using the initial pressure of N2O4 (Pi), the change in pressure (ΔP), and the change in moles of N2O4 (Δn(N2O4)):
Total pressure after equilibrium = Pi + ΔP + ΔP
Total pressure after equilibrium = 0.554 bar + 0.136 bar + 0.136 bar
Total pressure after equilibrium = 0.554 bar + 0.272 bar
Total pressure after equilibrium = 0.826 bar
So, the total pressure after equilibrium is reestablished is 0.826 bar.
I hope this helps! Let me know if you have any more questions.
To set up this problem, we need to use the concept of partial pressures and the equilibrium expression for the decomposition reaction.
The equilibrium expression for the decomposition reaction of Dinitrogen tetroxide (N2O4) can be written as:
N2O4 (g) ⇌ 2NO2 (g)
Let's define the initial partial pressure of N2O4 as P(N2O4)initial, the partial pressure of NO2 injected as P(NO2)injected, and the partial pressure of NO2 after equilibrium is reestablished as P(NO2)equilibrium.
Based on the given information, we know the following:
Initial pressure (total pressure) = 0.554 bar
Pressure after injection = 0.906 bar
Pressure at equilibrium = 0.770 bar
We can use the partial pressure of N2O4 at equilibrium as P(N2O4)equilibrium. Since N2O4 decomposes to form 2NO2 molecules, the initial partial pressure of NO2 at equilibrium can be calculated as 2 * P(N2O4)equilibrium.
Now, let's set up the equations:
Equation 1: Total pressure at equilibrium = P(N2O4)equilibrium + P(NO2)equilibrium
Equation 2: Total pressure after injection = P(N2O4)equilibrium + P(NO2)equilibrium + P(NO2)injected
Using the given information, we can rewrite the equations as follows:
Equation 1: 0.770 bar = P(N2O4)equilibrium + 2 * P(N2O4)equilibrium
Equation 2: 0.906 bar = P(N2O4)equilibrium + 2 * P(N2O4)equilibrium + P(NO2)injected
Now, we have two equations with two unknowns: P(N2O4)equilibrium and P(NO2)injected. We can solve these equations simultaneously to find the values of the unknowns.
Let's proceed to solve the equations.
To set up this problem, we need to use the fact that the decomposition of dinitrogen tetroxide (N2O4) is a reversible reaction.
The given balanced equation is: N2O4(g) ⇌ 2NO2(g)
We have the initial pressure of N2O4 as 0.554 bar, and after equilibrium is established, the total pressure is 0.770 bar.
Let's assign variables to unknown quantities:
Let "x" be the change in pressure (from N2O4 to NO2) and "y" be the change in pressure (from NO2 to N2O4).
Now, let's set up the equation for the change in pressure using the ideal gas law:
For the forward reaction (N2O4 to NO2):
Pressure of N2O4 (initial) - x = Pressure of NO2 (equilibrium)
For the reverse reaction (NO2 to N2O4):
Pressure of NO2 (initial) + x - y = Pressure of N2O4 (equilibrium)
We are given that a quantity of NO2 in injected, causing the total pressure to jump to 0.906 bar. This means that the sum of the equilibrium pressures of N2O4 and NO2 should be equal to 0.906 bar.
Pressure of N2O4 (equilibrium) + Pressure of NO2 (equilibrium) = 0.906 bar
Using these equations, we can set up a system of equations to solve for the change in pressure (x and y) and then determine the total pressure after equilibrium is reestablished.
Let's solve the system of equations using the given information and find the values of x and y.