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

Kc =
([N_(2) O_(4)])/([NO_(2)]^(2) )

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 = ([N_(2) O_(4)])/([NO_(2)]^(2) ) Concentration-example-1
User MikeLim
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1 Answer

11 votes

Answer:


\displaystyle K_c \approx 0.0424

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]

Explanation:

Step 1: Define

[RxN] 2NO₂ ⇆ N₂O₄

[Equilibrium Rate Law]
\displaystyle K_c = ([N_2O_4])/([NO_2]^2)

NO₂ = 11.95 M

N₂O₄ = 6.05 M

Step 2: Find K

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

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

User Nitish Kumar Pal
by
3.9k points