To solve this problem, we need to determine if the reaction will proceed in the forward direction, reverse direction, or if it is already at equilibrium by comparing the reaction quotient ([tex]\( Q_c \)[/tex]) to the equilibrium constant ([tex]\( K_{eq} \)[/tex]).
### Step-by-Step Solution:
1. Write the balanced chemical equation for the reaction:
[tex]\[
N_2O_4(g) \rightleftharpoons 2 NO_2(g)
\][/tex]
2. Identify the equilibrium constant [tex]\( K_{eq} \)[/tex]:
[tex]\[
K_{eq} = 6.5
\][/tex]
3. Determine the initial concentrations of the reactants and products:
Given:
- [tex]\( 0.250 \)[/tex] mol of [tex]\( N_2O_4 \)[/tex] in a [tex]\( 1.0 \)[/tex]-L container
- [tex]\( 0.250 \)[/tex] mol of [tex]\( NO_2 \)[/tex] in a [tex]\( 1.0 \)[/tex]-L container
Since the volume of the container is [tex]\( 1.0 \)[/tex] L, the molarity (concentration) is the same as the number of moles:
- [tex]\([N_2O_4] = 0.250 \, \text{mol/L}\)[/tex]
- [tex]\([NO_2] = 0.250 \, \text{mol/L}\)[/tex]
4. Calculate the reaction quotient [tex]\( Q_c \)[/tex]:
The expression for the reaction quotient [tex]\( Q_c \)[/tex] for the given reaction is:
[tex]\[
Q_c = \frac{[NO_2]^2}{[N_2O_4]}
\][/tex]
Substitute the initial concentrations:
[tex]\[
Q_c = \frac{(0.250 \, \text{mol/L})^2}{0.250 \, \text{mol/L}} = \frac{0.0625}{0.250} = 0.25
\][/tex]
5. Compare [tex]\( Q_c \)[/tex] with [tex]\( K_{eq} \)[/tex]:
- [tex]\( Q_c = 0.25 \)[/tex]
- [tex]\( K_{eq} = 6.5 \)[/tex]
Since [tex]\( Q_c < K_{eq} \)[/tex]:
[tex]\[
0.25 < 6.5
\][/tex]
6. Determine the direction of the reaction:
- If [tex]\( Q_c < K_{eq} \)[/tex], the reaction will proceed in the forward direction (towards the products) to reach equilibrium.
- If [tex]\( Q_c > K_{eq} \)[/tex], the reaction will proceed in the reverse direction (towards the reactants) to reach equilibrium.
- If [tex]\( Q_c = K_{eq} \)[/tex], the reaction is already at equilibrium.
In this case, since [tex]\( Q_c < K_{eq} \)[/tex], the reaction will proceed in the forward direction.
### Conclusion:
The reaction will proceed in the forward direction.
