Given the reaction: N2 + O2 = 2NO for which the Keq at 2273 K is 1.2 x 10-4 a. Write the equilibrium constant expression for the react

Question

Given the reaction: N2 + O2 = 2NO for which the Keq at 2273 K is 1.2 x 10-4
a. Write the equilibrium constant expression for the reaction.
b. Write the equation that would allow you solve for the concentration of NO.
c. What is the concentration of NO if [NZ] = 0.166M and [02] = 0.145M?

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Sigridomena 2 months 2021-07-30T10:37:29+00:00 1 Answers 3 views 0

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    2021-07-30T10:38:39+00:00

    Answer:

    (a): The expression of equilibrium constant is K_{eq}=\frac{[NO]^2}{[N_2][O_2]}

    (b): The equation to solve the concentration of NO is [NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}

    (c): The concentration of NO is 0.0017 M.

    Explanation:

    The equilibrium constant is defined as the ratio of the concentration of products to the concentration of reactants raised to the power of the stoichiometric coefficient of each. It is represented by the term K_{eq}

    (a):

    The given chemical equation follows:

    N_2+O_2\rightarrow 2NO

    The expression for equilbrium constant will be:

    K_{eq}=\frac{[NO]^2}{[N_2][O_2]}

    (b):

    The equation to solve the concentration of NO follows:

    [NO]=\sqrt{K_{eq}\times [N_2]\times [O_2]}            ……(1)

    (c):

    Given values:

    K_{eq}=1.2\times 10^{-4}

    [N_2]_{eq}=0.166M

    [O_2]_{eq}=0.145M

    Plugging values in equation 1, we get:

    [NO]=\sqrt{(1.2\times 10^{-4})\times 0.166\times 0.145}

    [NO]=\sqrt{2.88\times 10^{-6}}

    [NO]=0.0017 M

    Hence, the concentration of NO is 0.0017 M.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )