At #25^@ "C"#, for the reaction #"N"_2"O"_4(g) rightleftharpoons 2"NO"_2(g)#, the equilibrium constant is #K_c = 5.85xx10^(3)#. #"20.0 g"# of #"N"_2"O"_4# were added to a #"5.00L"# container. What is the molar concentration of #"NO"_2# at equilibrium?
#N_2O_4(g) ⇌ 2NO_2(g)#
You can determine the initial molar concentration of dinitrogen tetroxide by calculating its molar mass, which can be done using the chemical formula.
Therefore, by assuming that it fills the entire container as a gas (i.e., using the volume of the container as your volume), you can determine the concentration that it was at initially:
You now have enough knowledge to solve the remaining steps of the problem.
The response that was in balance was:
additionally:
In the case of the small x approximation
WITHOUT THE APPROXIMATION OF THE SMALL X
which the quadratic formula is used to solve, yielding:
which, if you accept the response that makes sense physically, gives you
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To solve this problem, we need to start by calculating the initial concentration of N2O4 in the container. We do this by converting the given mass of N2O4 to moles and then dividing by the volume of the container.
Given: Mass of N2O4 = 20.0 g Volume of container = 5.00 L
Next, we use the stoichiometry of the reaction to determine the change in concentration of N2O4 and NO2 at equilibrium. Since 1 mole of N2O4 produces 2 moles of NO2, the change in concentration of N2O4 is 2x and the change in concentration of NO2 is +4x, where x is the change in concentration.
We then use the equilibrium constant expression, Kc, to set up an equation:
Kc = [NO2]^2 / [N2O4]
Substituting the equilibrium concentrations into the equation and solving for x gives us the change in concentration of N2O4 and NO2 at equilibrium.
Finally, we use the initial concentration of N2O4 and the calculated change in concentration to find the equilibrium concentration of NO2.
Calculations:

Calculate initial concentration of N2O4: moles of N2O4 = mass / molar mass = 20.0 g / 92.02 g/mol = 0.217 mol initial concentration of N2O4 = moles / volume = 0.217 mol / 5.00 L = 0.0434 M

Set up equilibrium expression: Kc = (4x)^2 / (0.0434  2x)

Substitute Kc and initial concentration into the equation: 5.85x10^(3) = (4x)^2 / (0.0434  2x)

Solve for x:
5.85x10^(3) = (16x^2) / (0.0434  2x) 5.85x10^(3)(0.0434  2x) = 16x^2 0.0002539  0.0117x = 16x^2 16x^2 + 0.0117x  0.0002539 = 0
Using the quadratic formula, we find: x ≈ 0.012 M

Calculate equilibrium concentration of NO2: [NO2] = initial concentration + change in concentration [NO2] = 0 + (4 * 0.012) = 0.048 M
Therefore, the molar concentration of NO2 at equilibrium is approximately 0.048 M.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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