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 4.64×10−3 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=1.7×10−3at 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.
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Concentrations of NO= |
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-α-lysine and
L-β-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-α-lysine at a concentration of
3.60×10−3 M is placed in contact with lysine
2,3-aminomutase.
What are the equilibrium concentrations of L-α-lysine?
What are the equilibrium concentrations of L- β lysine?
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 4.64×10−3 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
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