Question

Equilibrium equation
Br2+Cl2−><−2BrCl
all gases
at 400 K Kc= 7

If .25 mol Br2 and . 55 mol Cl2 are introduced into a 3.0 L container at 400 K, what will be the equilibrium concentration of Br2? Cl2? and BrCl2?

So i made an ICE Table

Br2+Cl2−><−2BrCl

I .083 M .183 M 0M
C -x -x +2x
E .083-x .183-x 2x

But i can't figure the math out for some reason...i know you have to use Kc= [Brcl] ^2/ [Br2] [Cl2]


Can someone please show me the actual math and the process of how to figure this out...i'm having alot of trouble with it.

Answer 1

The Math Equation would be... \[7= \frac{ 2x ^{2} }{ (.083-x)(.183-x) }\]

Answer 2

7 = (2x)^2/(.183-x)(.083-x)
Simplify and you will have to use the quadratic equation to solve for x.

Answer 3

ok, but im sorry that doens't help...

Answer 4

i understand that you have to solve for x

Answer 5

cross multiply to get:
7(.083-x)(.183-x) = 4x^2
can you continue from here?

Answer 6

do you foil to 7(.0152-.083x-.183x-x^2)= 2x^2 ?

Answer 7

Yes...but note on the right it is(2x)^2 which is 4x^2

Answer 8

wouldn't it ust be 4x then though

Answer 9

and on the left it should be +x^2 for the last term

Answer 10

no, it you are squaring 2x...2x*2x = 4x^2

Answer 11

oh ok

Answer 12

Please continue and I can check when I get back or perhaps someone else can check your work.

Answer 13

ok i'll try! thank you so much

Answer 14

no problem...

Answer 15

hint: next step is: 7(.0152 - .266x +x^2) = 4x^2

Answer 16

awesome, that's what i thought

Answer 17

now would you * 7 by everything in the parentheses?

Answer 18

so it would then be .1064- 1.0862x +7x^2 = 4x^2
then get it to =o by .1064 - 1.0862 + 3x^2 = 0
then solve with ax^2+bx+c=0
3x^2 - 1.0826x + .1064 =0
square root 3x^2 = 1.732x
1.732x - 1.862x = -.1064
then -.13x = -.1064
x= .8185 ???

Answer 19

Ughh this didn't work in the chem problem

Answer 20

I found out that x should =.064 but i don't understand how the math is wrong

Answer 21

@IloveHW do you know?

Answer 22

This is what I have:
7(.0152 - .266x +x^2) = 4x^2
.1064 -1.86x +7x^2 = 4x^2

so you have 3x^2 - 1.86x + .1064 = 0
ax^2 + bx + c = 0
so you have:
a =3, b =-1.86 , c = .1064

now to solve for the two roots:
x1,2 = (-b +- (b^2-4ac)^1/2)/2a

you need to sub in your values for a, b and c into the above equation to solve for x1 and x2.
Note: +- means plus/minus so:
x1 = (-b + (b^2-4ac)^1/2)/2a
x2 = (-b - (b^2-4ac)^1/2)/2a

give it a try.

http://www.math.com/students/calculators/source/quadratic.htm

Answer 23

that has already been corrected....look above