Well I almost understand this question but still got stuck, I will show you the question and my answer so I need someone to tell me if I did it correctly or not and how I should fix it.

Question

Well I almost understand this question but still got stuck, I will show you the question and my answer so I need someone to tell me if I did it correctly or not and how I should fix it.

at 55degrees the K for the reaction: 2NO2(g)<=>N2O4 is 1.15
calc the concentration of N2O4 present in equilibrium with 0.50 mole of NO2

I first used the ICE chart then calculated
1.15 = x/(0.5-2x)^2
O.5-2x~ ).5
1.15 = x/(0.5)^2
1.15(0.25) = x
x= 0.2875
2(0.2875) - 0.575
Therefore N2O4 = 0.575M?

I don't agree that you can assume .5-2x is approximately .5. Your final answer shows that it is not.

1.15 = x/(0.5-2x)^2

(0.5-2x)^2= x/1.15

.25 -2x +4x^2= .696x

put the terms together, and use the quadratic equation for x.

then would my answer look like this

0.25-2.696x + 4x^2 = 0

and I would also like to understand how did you get 0.696?

I don't know how I got .696 . It should have been 1/1.15=.869

Now that you have the quadratic,

0.25-2.896x + 4x^2 = 0
Use the quadratic equation

x= (2.896 +- sqrt (8.23-4) )/8

x= .618
or x=.104

Now obviously, the first is impossible (why?), so the concentration the product is .104, and the concentration of the reactant is .5-.208

check my calc work

Answer 1

1.15=x/[2{0.5-x}]^2 u get x=0.9546 and x=0.2619 corrt anxw ix last 1

Answer 2

To solve the quadratic equation, you need to use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

In this case, your quadratic equation is:
4x^2 - 2.896x + 0.25 = 0

Comparing this to the standard quadratic equation (ax^2 + bx + c = 0), you have:
a = 4, b = -2.896, c = 0.25

Plugging these values into the quadratic formula, you get:

x = (-(-2.896) ± √((-2.896)^2 - 4 * 4 * 0.25)) / (2 * 4)
x = (2.896 ± √(8.383616 - 4)) / 8
x = (2.896 ± √4.383616) / 8

Calculating the square root:
x = (2.896 ± 2.0946) / 8

This gives two possible solutions:
x = (2.896 + 2.0946) / 8 = 4.9906 / 8 = 0.6238
x = (2.896 - 2.0946) / 8 = 0.8014 / 8 = 0.10018

So, the possible values for x are approximately 0.6238 and 0.10018.

Now, let's check your calculation work:
To find the concentration of N2O4, we subtract the value of x from the initial concentration of NO2.
Concentration of N2O4 = 0.5 mol - 2 * 0.10018 mol ≈ 0.2996 mol

Therefore, the correct concentration of N2O4 at equilibrium is approximately 0.2996 mol.