Question

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

Answer 1

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]

Step-by-step explanation:

Step 1: Define

[RxN]   2NO₂ ⇆ N₂O₄

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

NO₂ = 11.95 M

N₂O₄ = 6.05 M

Step 2: Find K

1. Substitute [ERL]:                    $\displaystyle K_c = \frac{[6.05]}{[11.95]^2}$
2. Exponents:                            $\displaystyle K_c = \frac{[6.05]}{[142.803]}$
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).