? asked in Science & MathematicsChemistry · 1 week ago

# 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.

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• david
Lv 7
1 week ago

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

• Anonymous
1 week ago

I don't know friend

I'm sorry

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