The equilibrium constant for the gas phase
reaction
N2O4 ⇀↽ 2 NO2
at a certain temperature is K = 0.0466. If
the initial concentrations are [N2O4] = 1.0 M,
[NO2] = 0.0 M, what are the final concentrations
of [N2O4] and [NO2], respectively?
1. 1.0 M ; 0.22 M
2. 0.8 M ; 0.2 M
3. 1.2 M ; 0.4 M
4. 0.8 M ; 0.1 M
5. 0.9 M ; 0.2 M
let [NO₂] equal 2x
(2x)^2 = 0.0466 * (1.0 - x)
4 x^2 + 0.0466 x - 0.0466 = 0
you can use the quadratic formula to find x
but answer 5 is the only one that makes sense
DRBOB222 WHERE ARE U WHEN SOMEONE NEEDS HELP IN CHEMISTRY U FOOL I THOUGHT YOU WERE THE CHEMISTRY ANSWERER
To find the final concentrations of [N2O4] and [NO2], we can use the equilibrium expression and the given equilibrium constant. According to the reaction:
N2O4 ⇀↽ 2 NO2
The equilibrium expression is:
K = [NO2]^2 / [N2O4]
Given that the initial concentration of [N2O4] is 1.0 M and [NO2] is 0.0 M, we can substitute these values into the equilibrium expression and solve for the final concentrations.
K = [NO2]^2 / [N2O4]
0.0466 = [NO2]^2 / 1.0
Solving for [NO2]:
0.0466 = [NO2]^2
[NO2]^2 = 0.0466
[NO2] = sqrt(0.0466)
[NO2] ≈ 0.22 M
So, the final concentration of [NO2] is approximately 0.22 M.
To find the final concentration of [N2O4], we can use the stoichiometry of the reaction. Since the ratio of [NO2] to [N2O4] is 2:1, the final concentration of [N2O4] will be half the concentration of [NO2]:
[N2O4] = 0.22 / 2
[N2O4] ≈ 0.11 M
Therefore, the final concentrations of [N2O4] and [NO2] are approximately 0.11 M and 0.22 M, respectively.
None of the provided options match these values, so the correct answer is not among the given choices.
To solve this question, we need to use the equilibrium constant expression and the initial concentrations of the reactants to calculate the final concentrations.
The equilibrium constant expression for the given reaction is:
K = [NO2]^2 / [N2O4]
Given that the equilibrium constant (K) is 0.0466, the equation becomes:
0.0466 = [NO2]^2 / [N2O4]
Initially, [N2O4] = 1.0 M and [NO2] = 0.0 M. Let's assume the concentration of [NO2] at equilibrium be 'x'. Therefore, the concentration of [N2O4] at equilibrium will be '1.0 - x'.
Now, substitute these equilibrium concentrations into the equilibrium constant expression:
0.0466 = (x^2) / (1.0 - x)
Rearranging the equation, we get:
0.0466 * (1.0 - x) = x^2
0.0466 - 0.0466x = x^2
Rearranging, we have a quadratic equation:
x^2 + 0.0466x - 0.0466 = 0
To solve the quadratic equation, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = 1, b = 0.0466, and c = -0.0466. Plugging in these values, we get:
x = (-0.0466 ± √((0.0466)^2 - 4(1)(-0.0466))) / (2(1))
Performing the calculations, we find two possible values for 'x': x ≈ 0.2177 or x ≈ -0.2643. Since the concentration cannot be negative, we discard the negative value.
Therefore, the final concentration of [NO2] is approximately 0.2177 M. To find the final concentration of [N2O4], we subtract this value from the initial concentration:
Final concentration of [N2O4] = 1.0 M - 0.2177 M ≈ 0.7823 M
Thus, the final concentrations of [N2O4] and [NO2], respectively, are approximately 0.7823 M and 0.2177 M.
Therefore, the answer is not among the options provided.