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For the current reaction, 2NO2 ↔ N2O4, we have:

Kc = [tex]\frac{[N_{2} O_{4}]}{[NO_{2}]^{2} }[/tex]

Concentration of NO2 = 11.95
Concebtration of N2O4 = 6.05
Based on the current concentrations of NO2 and N2O4, what is Kc?

For The Current Reaction 2NO2 N2O4 We Have Kc TexfracN2 O4NO22 Tex Concentration Of NO2 1195 Concebtration Of N2O4 605 Based On The Current Concentrations Of NO class=

Sagot :

Spaceidn

Answer:

[tex]\displaystyle K_c \approx 0.0424[/tex]

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Chemistry

Equilibrium

  • Equilibrium Constant K
  • Concentrations - Denoted in [Brackets]

Step-by-step explanation:

Step 1: Define

[RxN]   2NO₂ ⇆ N₂O₄

[Equilibrium Rate Law]   [tex]\displaystyle K_c = \frac{[N_2O_4]}{[NO_2]^2}[/tex]

NO₂ = 11.95 M

N₂O₄ = 6.05 M

Step 2: Find K

  1. Substitute [ERL]:                    [tex]\displaystyle K_c = \frac{[6.05]}{[11.95]^2}[/tex]
  2. Exponents:                            [tex]\displaystyle K_c = \frac{[6.05]}{[142.803]}[/tex]
  3. Divide:                                   [tex]\displaystyle K_c = 0.042366[/tex]
  4. Round (Sig Figs):                  [tex]\displaystyle K_c \approx 0.0424[/tex]

This value of K tells us that the reactants are favored in the equilibrium reaction (K < 0.1).

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