5.0 moles of nitrogen gas is reacted with 2.5 moles of oxygen gas in an 10. L container. At equilibrium, 1.5 moles of nitrogen dioxide is collected. What is the equilibrium constant for the formation of NO2 gas? A. 0.568 b. 0.756 c. 5.29 d. 1.76
(N2) = mols/L = 5.0 mols/10. L = 0.5 M initial
(O2) = 2.5 mols/10. L = 0.25 M initial
(NO2) = 1.5 mols/10. L = 0.15 M at equilibrium
.............................N2 + 2O2 ==> 2NO2
Initial...................0.5......0.25.........0
change.................-x.........-2x........2x
equilibrium .......0.5-x.....0.25-2x....2x
The problem tells you that 2x = 0.15 M
Knowing x you can calculate 0.5-x and 0.25-2x so now you know the equilibrium concentrations of all three components. Plug those values into Kc expression and solve for Kc. Post your work if you get stuck.
To find the equilibrium constant (Kc) for the formation of NO2 gas, we can use the balanced chemical equation for the reaction:
2 NO(g) + O2(g) ⇌ 2 NO2(g)
The equilibrium constant expression is given by:
Kc = [NO2]^2 / ([NO]^2 × [O2])
Given the moles of nitrogen gas (N2), oxygen gas (O2), and nitrogen dioxide (NO2) at equilibrium, we can calculate the concentrations:
[NO2] = 1.5 moles / 10 L = 0.15 M
[NO] = (5.0 moles - 1.5 moles) / 10 L = 0.35 M
[O2] = (2.5 moles) / 10 L = 0.25 M
Plugging these values into the equilibrium constant expression:
Kc = (0.15)^2 / ((0.35)^2 × 0.25) = 0.567
So, the equilibrium constant for the formation of NO2 gas is approximately 0.567.
Therefore, the answer is A. 0.568.
To determine the equilibrium constant for the formation of NO2 gas, we can use the balanced chemical equation for the reaction:
N2(g) + 2O2(g) ⇌ 2NO2(g)
The equilibrium constant expression for this reaction is given by:
Kc = ([NO2]^2) / ([N2][O2]^2)
Where [NO2], [N2], and [O2] represent the molar concentrations of NO2, N2, and O2 at equilibrium, respectively.
We are given the initial moles of N2 and O2, as well as the moles of NO2 at equilibrium. From this information, we can determine the concentrations of NO2, N2, and O2 at equilibrium using the ideal gas law:
PV = nRT
Since we have a 10.0 L container, we can assume that the pressure (P), temperature (T), and gas constant (R) are constant, allowing us to use the equation:
n = PV / RT
First, let's calculate the concentrations of the gases at equilibrium:
For NO2:
We are given that 1.5 moles of NO2 is collected in a 10.0 L container.
So the concentration of NO2 at equilibrium, [NO2], is:
[NO2] = (moles of NO2) / (total volume) = 1.5 moles / 10.0 L = 0.15 M
For N2:
We are given that we started with 5.0 moles of N2 in a 10.0 L container.
So the concentration of N2 at equilibrium, [N2], is:
[N2] = (moles of N2) / (total volume) = 5.0 moles / 10.0 L = 0.50 M
For O2:
We are given that we started with 2.5 moles of O2 in a 10.0 L container.
So the concentration of O2 at equilibrium, [O2], is:
[O2] = (moles of O2) / (total volume) = 2.5 moles / 10.0 L = 0.25 M
Now let's substitute these values into the equilibrium constant expression:
Kc = ([NO2]^2) / ([N2][O2]^2)
= (0.15 M)^2 / ((0.50 M)(0.25 M)^2)
= 0.0225 M^2 / 0.03125 M^3
= 0.72
Therefore, the equilibrium constant for the formation of NO2 gas is approximately 0.72.
Since none of the given answer choices match exactly, it seems that the correct answer might not be among the options provided.