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For the reaction 2NOBr (g) ↔ 2NO (g) + Br2 (g), when the concentrations are [NOBr] = 0.10 m, [NO] = 0.010 M, and [Br2] = 0.0050 M, what is the equilibrium constant of this reaction? A. 5.0 × 10–5 M B. 0.000005 M C. 5.0 × 105 M D. 5.0 M E. None of the Above

Answer :

A. Remember that K=[Reactants]/[products]
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Answer:  B. 0.000005

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.

The given balanced equilibrium reaction is,

                    [tex]NOBr(g)\rightleftharpoons 2NO(g)+Br_2(g)[/tex]

At eqm. conc.            0.10 M      0.010 M      0.0050 M

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[NO]^2[Br_2]}{[NOBr]}[/tex]

Now put all the given values in this expression, we get :

[tex]K_c=\frac{(0.010)^2\times (0.0050)}{(0.10)}[/tex]

[tex]K_c=5\times 10^{-6}=0.000005[/tex]

Thus the value of the equilibrium constant is 0.000005 M .

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