Question

At a given temperature T in a 10.0 L flask, 2.0 moles of N$_2$O$_4$(g) is heated. At equilibrium, 20% of N$_2$O$_4$(g) dissociates into NO$_2$(g). The value of K$_C$ for the reaction N$_2$O$_4$(g) $\rightleftharpoons$ 2NO$_2$(g) is:

• A. 2 10^-2
• B. 4 10^-2
• C. 3 10^-2
• D. 6 10^-2

Hint:

For dissociation problems, always express equilibrium concentrations in terms of degree of dissociation before substituting into K$_C$.

Answer 1

The Correct Option is B

Step 1: Write the equilibrium expression.
For the dissociation \[ N_2O_4(g) \rightleftharpoons 2NO_2(g) \] the equilibrium constant in terms of concentration is \[ K_C = \frac{[NO_2]^2}{[N_2O_4]} \]

Step 2: Note the starting amounts.
We begin with $2.0$ moles of $N_2O_4$ in a $10.0$ L flask, and no $NO_2$ yet. The degree of dissociation is $20\%$, that is $0.2$.

Step 3: Find how much reacts.
Moles of $N_2O_4$ that break apart: \[ 2.0 \times 0.2 = 0.4 \text{ mol} \]

Step 4: Find equilibrium moles.
Leftover $N_2O_4 = 2.0 - 0.4 = 1.6$ mol. Each mole of $N_2O_4$ gives two of $NO_2$, so \[ NO_2 = 2 \times 0.4 = 0.8 \text{ mol} \]

Step 5: Convert to concentrations.
Dividing by the $10.0$ L volume: \[ [N_2O_4] = \frac{1.6}{10} = 0.16, \quad [NO_2] = \frac{0.8}{10} = 0.08 \]

Step 6: Compute $K_C$.
\[ K_C = \frac{(0.08)^2}{0.16} = \frac{0.0064}{0.16} = 0.04 = 4 \times 10^{-2} \]
\[ \boxed{4 \times 10^{-2}} \]