Question

The equilibrium constant for the reaction

Answer 1

The question is incomplete, here is the complete question:

The equilibrium constant for the reaction

N₂O₄(g)⇌2NO₂ at 2°C is Kc = 2.0

If each yellow sphere represents 1 mol of N₂O₄(g) and each gray sphere 1 mol of NO₂ which of the following 1.0 L containers represents the equilibrium mixture at 2°C?

The image is attached below.

Answer: The system which represents the equilibrium having value of $K_c=2.0$ is system (b)

Explanation:

Equilibrium constant in terms of concentration is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

For a general chemical reaction:

$aA+bB\rightarrow cC+dD$

The expression for $K_{c}$ is written as:

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

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$      .......(1)

We are given:

Volume of the container = 1.0 L

Value of $K_c$ = 2.0

Molarity of the substance is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume}}$

For the given images:

• For a:

Number of Gray spheres = 8 moles

Number of yellow spheres = 4 moles

Putting values in expression 1, we get:

$K_c=\frac{(8/1)^2}{(4/1)}\\\\K_c=16$

• For b:

Number of Gray spheres = 4 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(4/1)^2}{(8/1)}\\\\K_c=2$

• For c:

Number of Gray spheres = 6 moles

Number of yellow spheres = 6 moles

Putting values in expression 1, we get:

$K_c=\frac{(6/1)^2}{(6/1)}\\\\K_c=6$

• For d:

Number of Gray spheres = 2 moles

Number of yellow spheres = 8 moles

Putting values in expression 1, we get:

$K_c=\frac{(2/1)^2}{(8/1)}\\\\K_c=\frac{1}{2}$

Hence, the system which represents the equilibrium having value of $K_c=2.0$ is system (b)