Calculate the equilibrium concentrations of N2O4 and NO2 at 25 C in a vessel tha

Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 C in a vessel that contains an initial N2O4 concentration of 0.0438 M . The equilibrium constant Kc for the reaction N2O4(g)2NO2(g) is 4.64×103 at 25 C. Express your answers using four decimal places separated by a comma. [N2O4], [NO2] =

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The air pollutant NO is produced in automobile engines from the high-temperature reaction N2(g)+O2(g)2NO(g);Kc=1.7×103at 2300 K.

If the initial concentrations of N2 and O2 at 2300 K are both 1.58 M, what are the concentrations of NO when the reaction mixture reaches equilibrium?

Express your answer to two significant figures and include the appropriate units.

If the initial concentrations of N2 and O2 at 2300 K are both 1.58 M, what are the concentrations of N2 when the reaction mixture reaches equilibrium?

Concentrations of O2

If the initial concentrations of N2 and O2 at 2300 K are both 1.58 M, what are the concentrations of O2 when the reaction mixture reaches equilibrium?

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The interconversion of L--lysine and L--lysine, for which at Kc=7.20 at 333 K, is catalyzed by the enzyme lysine 2,3-aminomutase.
At 333 K, a solution of L--lysine at a concentration of 3.60×103 M is placed in contact with lysine 2,3-aminomutase.

What are the equilibrium concentrations of L--lysine?

What are the equilibrium concentrations of L- lysine?

Explanation / Answer

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 C in a vessel that contains an initial N2O4 concentration of 0.0438 M . The equilibrium constant Kc for the reaction N2O4(g)2NO2(g) is 4.64×103 at 25 C. Express your answers using four decimal places separated by a comma. [N2O4], [NO2] =

[N2O4] = 0.0438

K = [NO2]^2/ [N2O4]

K = 4.64*10^-3

solve for all concnetrations

K = [NO2]^2/ [N2O4]

initially:

[N2O4] = 0.0438

[NO2] = 0

in equilibrium

[N2O4] = 0.0438 - x

[NO2] = 0 +2x

then

K = (0 +2x)^2 /(0.0438 - x)

4.64*10^-3 = (4x^2)/(0.0438 - x)

solve for x

(4.64*10^-3) * ((0.0438 - x)) = 4x^2

0.0438 (4.64*10^-3) - (4.64*10^-3) x -4x^2 =

x = 0.0065

substiutte

[N2O4] = 0.0438 - 0.0065 = 0.0373

[NO2] = 0 +2*0.0065 = 0.013

NOTE: Please consider posting multiple questions in multiple set of Q&A-. We are not allowed to answer to multilpe questions in a single set of Q&A.

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