# For the following reaction: 2NO(g)+Br2(g)<—>2NOBr(g) Kc given: 750?

If 2.5 moles of NO and 1.0 moles of Br2 are placed in a 1.0 liter container. Calculate the equilibrium concentrations of all the components.

### 2 Answers

- davidLv 71 week agoFavourite answer
2NO(g)+Br2(g)<—>2NOBr(g) Kc given: 750?

If 2.5 moles of NO and 1.0 moles of Br2 are placed in a 1.0 liter container. Calculate the equilibrium concentrations of all the components.

... because it is a 1L container moles will equal molar concentration making this a little easier.

Kc = [NOBr]^2 / ([NO]^2 X [BR2])

init [NO] = 2.5 M

init [Br2] = 1.0 M

init [NOBr] = 0

reactants are used .. products are formed

is x of Br2 is used the 2x of NO are used and 2x NOBr are formed

at equilib

[NO] = 2.5 - 2x

[Br2] = 1.0 - x

[NOBr] = 2x

Kc = [NOBr]^2 / ( [NO]^2 * [Br2] )

750 = (2x)^2 / [ (2.5 - 2x)^2 * (1.0 - x) << assuming x is small .. solve

750[ (2.5)^2 * (1.0) = (2x)^2

x = 1171.88 << not possible .. subtraction create negative conc. values which are not possible

so solving would involve using a calculator .. maybe graphing an looking for the zeros

750 = (2x)^2 / [ (2.5 - 2x)^2 * (1.0 - x)

x = 0.982 M

[NO] = 2.5 - 2x = 0.536 M

[Br2] = 1.0 - x = 0.018M

[NOBr] = 2x = 1.964M

- Anonymous1 week ago
I don't know friend

I'm sorry