The given chemical reaction is:

$4 \text{NO}(g) + 3 \text{O}_2(g) \rightleftharpoons 2 \text{N}_2\text{O}_5(g)$

To find the equilibrium constant expression $K_c$, we use the general form of the equilibrium law for a reaction of the type:

$aA + bB \rightleftharpoons cC + dD$

The equilibrium constant expression is given by:

[ K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} ]

Applying to the Given Reaction

For the reaction provided:

• $a = 4$, for NO
• $b = 3$, for $\text{O}_2$
• $c = 2$, for $\text{N}_2\text{O}_5$

The equilibrium constant expression $K_c$ will be:

[ K_c = \frac{[\text{N}_2\text{O}_5]^2}{[\text{NO}]^4 [\text{O}_2]^3} ]

Now, let's check the options in the image:

1. $\frac{[\text{N}_2\text{O}_5]}{[\text{NO}][\text{O}_2]^2}$
2. $\frac{[\text{N}_2\text{O}_5]^2}{[\text{NO}]^4 [\text{O}_2]^3}$
3. $\frac{[\text{NO}]^4}{[\text{N}_2\text{O}_5]^2 [\text{O}_2]}$
4. $\frac{[\text{N}_2\text{O}_5]}{[\text{NO}]^2 [\text{O}_2]^3}$
5. $\frac{[\text{N}_2\text{O}_5]^2}{[\text{NO}]^4 [\text{O}_2]^2}$

The correct option is the second one:

[ K_c = \frac{[\text{N}_2\text{O}_5]^2}{[\text{NO}]^4 [\text{O}_2]^3} ]

Would you like me to explain in more detail or have any other questions?

Related Questions:

1. What is the difference between $K_c$ and $K_p$?
2. How do temperature changes affect the value of $K_c$?
3. What is the significance of the equilibrium constant being greater than 1?
4. How does the reaction quotient $Q$ compare to $K_c$?
5. How can you calculate $K_c$ if given concentrations at equilibrium?

Tip: Always ensure to balance the chemical equation before determining the equilibrium constant expression.