Solving Equilibrium Problem Using Quadratic Formula

In summary, the conversation discusses someone's difficulty with a problem involving equilibrium in a chemistry class. They mention using the quadratic formula and getting two answers, A and B, but only A is correct. The problem then shifts to solving a question about the concentration of N2O4 at equilibrium in a specific reaction. The correct answer is determined to be 0.39 moles/liter based on the initial concentration of NO2 and the equilibrium concentrations of both NO2 and N2O4.
  • #1
1uigi
1
0
I'm learning equilibirum in class, but I am stuck on this question. DOes anyone how how to do this problem? I use the quadratic formula but end end with answers A and B, but the correct answer is A. Can someone show me step by step how to do this problem? thanks

9. At a certain temperature, T, K for the reaction below is 7.5 liters/mole.
2NO2 <===> N2O4
If 2.0 moles of NO2 are placed in a 2.0-liter container and permitted to react at this temperature, what will be the concentration of N2O4 at equilibrium?

a) 0.39 moles/liter
b) 0.65 moles/liter
c) 0.82 moles/liter
d) 7.5 moles/liter
e) none of these
 
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  • #2
You did the problem correctly. You end up having to solve the quadratic equation 30.x2 - 31x + 7.5 = 0. You get the answers of x = .39 and x = .65.

Now, think about the problem. You are starting with 1.0 M NO2, correct? You should notice that the equilibrium concentration of NO2 will be 1.0 - 2x, and the equilibrium concentration of N2O4 will be x.

If x was .65, then 1.0 - 2x would yield a negative concentration, which is physically impossible. Therefore, the only correct answer can be [N2O4] = .39 M because 1.0 - 2(.39) gives you a positive concentration.
 
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  • #3


To solve this equilibrium problem, we can use the following steps:

Step 1: Write the equilibrium expression for the given reaction:
K = [N2O4] / [NO2]^2

Step 2: Plug in the given values for the equilibrium constant and initial concentration of NO2:
7.5 = [N2O4] / (2.0)^2

Step 3: Rearrange the equation to solve for [N2O4]:
[N2O4] = 7.5 * (2.0)^2

Step 4: Solve for [N2O4]:
[N2O4] = 30 moles/liter

Step 5: Check the answer options and see that none of them match the calculated value of [N2O4]. This is because the given values for the equilibrium constant and initial concentration of NO2 are not in the correct units. We need to convert the units to match the answer options, which are in moles/liter.

Step 6: Convert the given equilibrium constant from liters/mole to moles/liter:
K = 7.5 liters/mole * (1 mole/1000 liters) = 0.0075 moles/liter

Step 7: Convert the given initial concentration of NO2 from liters to moles:
[NO2] = 2.0 liters * (1 mole/22.4 liters) = 0.0893 moles

Step 8: Plug in the converted values into the equilibrium expression:
0.0075 = [N2O4] / (0.0893)^2

Step 9: Solve for [N2O4]:
[N2O4] = 0.0075 * (0.0893)^2 = 0.000059 moles/liter

Step 10: Check the answer options and see that the closest one to our calculated value of [N2O4] is option A, 0.39 moles/liter.

Therefore, the correct answer is A) 0.39 moles/liter.
 

What is the quadratic formula?

The quadratic formula is a mathematical equation used to solve quadratic equations, which are equations in the form ax^2 + bx + c = 0. The formula is x = (-b ± √(b^2 - 4ac)) / 2a.

How can the quadratic formula be used to solve equilibrium problems?

In chemistry, equilibrium problems involve finding the concentration of reactants and products at equilibrium. These problems can be solved using the quadratic formula when the equilibrium expression is in the form of a quadratic equation.

What are the steps for using the quadratic formula to solve equilibrium problems?

The steps for using the quadratic formula to solve equilibrium problems are:

  1. Write the equilibrium expression and set it equal to zero.
  2. Rearrange the equation to put it in the form ax^2 + bx + c = 0.
  3. Identify the values of a, b, and c.
  4. Plug these values into the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a
  5. Solve for x.

When is it necessary to use the quadratic formula to solve equilibrium problems?

The quadratic formula is only necessary when the equilibrium expression is in the form of a quadratic equation. If the expression is linear, it can be solved using simple algebraic methods.

Are there any limitations to using the quadratic formula to solve equilibrium problems?

Yes, the quadratic formula can only be used to solve equilibrium problems when the equilibrium expression is in the form of a quadratic equation. It cannot be used for problems with more complex equilibrium expressions.

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