Question

At 35 degrees Celcius,

K = 1.6x10^-5
for the reaction:
2NOCl<--> 2NO + Cl2

If 2.0mol NOCl and 1.0mol Cl2 are placed into a 1.0-L flask, calculate the equilibrium concentrations of all species.

Answer:
[NOCl] = 2.0M
[NO] = 8.0x10^-3
[Cl2] = 1.0M

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What I did:
___2NOCl<--> 2NO + Cl2
I__2.0_______0_____1.0
C__-2x_______+2x___+x
E__2-2x______2x_____1+x

K = 1+x * 2x^2 / (2-2x)^2
x = -0.99987 or x = 5.609x10^-3 or x = -5.705x10^-3

I am obviously doing something wrong.
Please help - I at least tried my homework.

Answer 1

K = 1+x * 2x^2 / (2-2x)^2

One quick thing I see is the 2x^2 came from (2x)^2 so that would be 4x^2.

Answer 2

Here is a cubic equation solver if you need one. My answer was x = 0.003976 which rounds to 0.004 so (NO) = 0.008M

http://www.1728.com/cubic.htm

Answer 3

[AI-generated answer — Step-by-Step Bot]

You did a good job setting up the initial reaction and the changes in the concentrations. However, it seems that you made a mistake in calculating the equilibrium concentrations.

Let's go through the calculation again step-by-step:

Step 1: Set up the initial concentrations:
[NOCl] = 2.0 M (initial)
[NO] = 0 M (initial)
[Cl2] = 1.0 M (initial)

Step 2: Set up the changes in concentration:
2NOCl <--> 2NO + Cl2
I: 2.0 0 1.0
C: -2x +2x +x
E: 2-2x 2x 1+x

Step 3: Substitute the expressions for equilibrium concentrations into the equilibrium expression:
K = [NO]^2[Cl2] / [NOCl]^2
= (2x)^2(1+x) / (2-2x)^2

Step 4: Solve for x (the change in concentration):
K = (2x)^2(1+x) / (2-2x)^2
1.6x10^-5 = (2x)^2(1+x) / (2-2x)^2

To simplify the calculation, we can assume that the value of x is small, so we can ignore the term (1+x) in the numerator and approximate 2-2x as 2:

1.6x10^-5 = (2x)^2 / 2^2
1.6x10^-5 = 4x^2 / 4
1.6x10^-5 = x^2 / 1

x^2 = 1.6x10^-5
x = √(1.6x10^-5)
x = 4.0x10^-3

Since x represents the change in concentration at equilibrium, we can use it to calculate the equilibrium concentrations:

[NOCl] = 2.0 - 2x
= 2.0 - 2(4.0x10^-3)
= 2.0 - 8.0x10^-3
= 1.992 M

[NO] = 2x
= 2(4.0x10^-3)
= 8.0x10^-3 M

[Cl2] = 1 + x
= 1 + 4.0x10^-3
= 1.004 M

Therefore, the equilibrium concentrations are:
[NOCl] = 1.992 M
[NO] = 8.0x10^-3 M
[Cl2] = 1.004 M

I hope this clarifies the calculation for you! If you have any further questions, feel free to ask.

Answer 4

[AI-generated answer — Explain Bot]

It looks like you're on the right track with your initial setup of the ICE (Initial, Change, Equilibrium) table. However, there seems to be a mistake in your equation for calculating the equilibrium constant (K) expression.

The correct expression for K in this case should be:

K = ([NO]^2 * [Cl2]) / [NOCl]^2

where [NO], [Cl2], and [NOCl] represent the equilibrium concentrations of each species, respectively.

Now let's go through the calculations step by step:

1. Start with the given initial concentrations:
[NOCl]_initial = 2.0 M
[NO]_initial = 0 M
[Cl2]_initial = 1.0 M

2. Set up the ICE table and write the expressions for the change in concentration ([x]) for each species:
___2NOCl <--> 2NO + Cl2
I__2.0________0_______1.0
C__-2x________+2x______+x
E__(2-2x)____2x_______(1+x)

3. Substitute the expressions for equilibrium concentrations into the equation for K:
K = ([NO]^2 * [Cl2]) / [NOCl]^2

So, K = ((2x)^2 * (1+x)) / ((2-2x)^2)

4. Solve for x by substituting the given equilibrium constant value:
1.6 x 10^(-5) = ((2x)^2 * (1+x)) / ((2-2x)^2)

Now you can solve this equation to find the correct value of x, which represents the change in concentration of NOCl. This will allow you to calculate the equilibrium concentrations ([NO], [Cl2], and [NOCl]).

Note that the given equilibrium constant (K = 1.6 x 10^(-5)) can also be used to check your final concentrations of [NO], [Cl2], and [NOCl].