A 1.00 mol sample of N2O4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium accourding to the equation
N2O4(g) <----> 2NO2(g)
K = .0004
Calculate the equilibrium concentrations of N2O4(g) and NO2(g)
M = moles/L =1.00/10L = 0.1M
............N2O4 ==> 2NO2
initial.....0.1M......0
change......-x........+2x
equil......0.1-x.......2x
Substitute equilibrium from ICE chart above and solve for x. 2x will be NO2 and 0l-x will be N2O4.
thank you
To calculate the equilibrium concentrations of N2O4(g) and NO2(g), we use the equilibrium constant expression and the information given in the problem.
The equilibrium constant expression for the given reaction is:
K = [NO2(g)]^2 / [N2O4(g)]
We are given the value of the equilibrium constant, K = 0.0004.
Let's denote the equilibrium concentration of N2O4(g) as [N2O4]eq and the equilibrium concentration of NO2(g) as [NO2]eq.
Since we start with a 1.00 mol sample of N2O4(g) in a 10.0 L vessel, the initial concentration of N2O4(g) is:
[N2O4]initial = 1.00 mol / 10.0 L = 0.10 M
Let's assume that at equilibrium, x mol/L of N2O4(g) reacts, which means the concentrations of N2O4(g) and NO2(g) will change.
At equilibrium, the concentration of N2O4(g) will be:
[N2O4]eq = [N2O4]initial - x = 0.10 M - x
Since the stoichiometric coefficient of NO2(g) is 2 in the balanced equation, the concentration of NO2(g) at equilibrium will be:
[NO2]eq = 2x
Now we can substitute these expressions into the equilibrium constant expression to solve for x.
K = [NO2(g)]^2 / [N2O4(g)]
0.0004 = (2x)^2 / (0.10 M - x)
To simplify this equation, we square both sides and cross-multiply:
0.0004(0.10 M - x) = 4x^2
0.00004 M - 0.0004x = 4x^2
Rearranging the equation:
4x^2 + 0.0004x - 0.00004 = 0
Now we can solve this quadratic equation to find the value of x, which represents the change in concentration at equilibrium.
Using the quadratic formula:
x = (-0.0004 ± √(0.0004^2 - 4(4)(-0.00004))) / (2(4))
x = (-0.0004 ± √(0.00000016 + 0.00016)) / 0.008
x = (-0.0004 ± √0.00016016) / 0.008
Calculating the discriminant inside the square root:
√0.00016016 = 0.01265
Substituting the value into the equation:
x = (-0.0004 ± 0.01265) / 0.008
There are two possible solutions for x:
1. x1 = (-0.0004 + 0.01265) / 0.008 = 0.01581 / 0.008 = 1.97625 M
2. x2 = (-0.0004 - 0.01265) / 0.008 = -0.01305 / 0.008 = -1.63125 M
Since concentrations cannot be negative, we disregard x2 as it is not physically meaningful.
Therefore, the equilibrium concentration of N2O4(g) is:
[N2O4]eq = 0.10 M - x1 = 0.10 M - 1.97625 M = -1.87625 M
Since a negative concentration of N2O4(g) is not possible, this means our assumption about x was incorrect. It's likely that x represents a very small change in concentration, meaning the equilibrium concentrations of N2O4(g) and NO2(g) are almost equal to the initial concentration of N2O4(g).
So, at equilibrium, the concentration of N2O4(g) is approximately 0.10 M, and the concentration of NO2(g) is approximately 2x = 2(1.97625 M) = 3.9525 M.
To calculate the equilibrium concentrations of N2O4(g) and NO2(g), you need to use the equilibrium expression and the given equilibrium constant (K) value.
The equilibrium expression for the given reaction is:
K = [NO2]^2 / [N2O4]
Where [NO2] represents the concentration of NO2 and [N2O4] represents the concentration of N2O4.
Since the initial concentration of N2O4 is given as 1.00 mol and the volume of the vessel is 10.0 L, we can calculate the initial concentration of N2O4 (I[N2O4]):
I[N2O4] = mol N2O4 / volume = 1.00 mol / 10.0 L = 0.10 M
To determine the equilibrium concentrations, we can assume that the change in concentration for each component is "x" (as they are in a 1:2 stoichiometric ratio). Therefore, after reaching equilibrium, the concentrations will be:
[N2O4] = I[N2O4] - x
[NO2] = 2x
Now, we can substitute these expressions into the equilibrium expression:
K = ([NO2]^2) / [N2O4]
Using the equilibrium concentrations, we get:
K = (2x)^2 / (0.10 - x)
Now, we can solve this equation for "x" to find the equilibrium concentration of NO2. Use algebraic manipulation:
K = 4x^2 / (0.10 - x)
0.0004 = 4x^2 / (0.10 - x)
Cross-multiplying:
0.0004(0.10 - x) = 4x^2
0.00004 - 0.0004x = 4x^2
Rearranging:
4x^2 + 0.0004x - 0.00004 = 0
This equation is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values:
a = 4
b = 0.0004
c = -0.00004
x = (-0.0004 ± √((0.0004)^2 - 4(4)(-0.00004))) / (2(4))
Calculating the values inside the square root:
√((0.0004)^2 - 4(4)(-0.00004)) = 0.0004
x = (-0.0004 ± 0.0004) / 8
So, there are two possible values for x:
x1 = (-0.0004 + 0.0004) / 8 = 0 / 8 = 0
x2 = (-0.0004 - 0.0004) / 8 = -0.0008 / 8 = -0.0001
Since concentration cannot be negative, we discard the negative value. Therefore, x = 0.
Using this value of x, we can calculate the equilibrium concentrations:
[N2O4] = 0.10 M - 0 = 0.10 M
[NO2] = 2(0) = 0 M
Therefore, at equilibrium, the concentrations of N2O4(g) and NO2(g) are 0.10 M and 0 M, respectively.