`K_c` for the equilibrium `N_2O_4(g) Leftrightarrow 2NO_2 (g)" at "298^@"C is "5.7 xx 10^(-9).` Which species has a higher concentration at equilibrium?

Answer

To determine which species has a higher concentration at equilibrium for the reaction \( N_2O_4(g) \leftrightarrow 2NO_2(g) \) at \( 298^\circ C \) with a given equilibrium constant \( K_c = 5.7 \times 10^{-9} \), we can follow these steps:

### Step 1: Write the expression for the equilibrium constant \( K_c \)

For the reaction \( N_2O_4(g) \leftrightarrow 2NO_2(g) \), the equilibrium constant \( K_c \) is defined as:

\[ K_c = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 2: Substitute the value of \( K_c \)

We know that \( K_c = 5.7 \times 10^{-9} \). Thus, we can write:

\[ 5.7 \times 10^{-9} = \frac{[NO_2]^2}{[N_2O_4]} \]

### Step 3: Analyze the value of \( K_c \)

Since \( K_c \) is much less than 1 (specifically \( 5.7 \times 10^{-9} \)), this indicates that at equilibrium, the concentration of the reactants (in this case, \( N_2O_4 \)) is much higher than that of the products (in this case, \( NO_2 \)).

### Step 4: Conclusion

From the analysis, we can conclude that at equilibrium, the concentration of \( N_2O_4 \) is higher than that of \( NO_2 \).

### Final Answer

**The species with a higher concentration at equilibrium is \( N_2O_4 \).**

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