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
---------
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.
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.
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
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.
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].