? 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
    Favourite 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

       

  • Anonymous
    1 week ago

    I don't know friend

    I'm sorry

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