The molecular scenes below depict the reaction $\mathbf{\text{Y}}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{\text{Z}}$ at four different times, out of sequence, as it reaches equilibrium. Each sphere (Y is red and Zis green) represents$\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{}\mathbf{\text{mol}}$ and the volume is$\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathbf{\text{L}}$ .

(a) Which scene represents equilibrium?

(b) List the scenes in the correct sequence.

(c) Calculate${\mathit{K}}_{\mathbf{\text{c}}}$ .

Chemical equilibrium is a state in which no net change in the amounts of reactants and products occurs during a reversible chemical reaction.

The given reaction:

$Y\ge 2Z$

Y and Z are in the 1:2 ratio as one molecule of v gives 2 molecules of Z .

In the molecular scenes:

Y - red-coloured spheres

Z - green-coloured spheres

Scene A:

Y-7 molecules

Z-2 molecules

There isn't a 1:2 ratio so this scene isn't at equilibrium

Scene C:

Y- 8 molecules

So this scene isn't at equilibrium

Scene B and : D

Y-4 molecules.

Z-8 molecules.

Y and Z are in the 1:2 ratio so these scenes are at equilibrium.

Y molecules are presented initially and they are in scene C

The number of Y molecules decreases with time and on the other side the number of Z molecules increases

The scene A is presented after the scene C because it has 7Y molecules and 2Z molecules

Then scenes B and D equally represent the next step

Therefore, the order of given scenes is:

$C\to A\to B=D$.

Firstly we will calculate the concentration of Y and Z :

$\left[\text{Y}\right]=\frac{\text{n}}{\text{V}}$

$\left[\text{Y}\right]=\frac{4·0.025\text{mol}}{0.40\text{L}}$

$\left[\text{Y}\right]=0.25\text{mol}/\text{L}$

$\left[\text{Z}\right]=\frac{\text{n}}{\text{V}}$

$\left[\text{Z}\right]=\frac{8·0.025\text{mol}}{0.40\text{L}}$

$\left[\text{Z}\right]=0.50\text{mol}/\text{L}$

.

So now we can calculate $K_c$as:

${\text{K}}_{\text{c}}=\frac{{\left[\text{Z}\right]}^2}{\left[\text{Y}\right]}$

${\text{K}}_{\text{c}}=\frac{{(0.50\text{mol}/\text{L})}^2}{0.25\text{mol}/\text{L}}$

${\text{K}}_{\text{c}}=1.0$

Therefore, the required value is ${\text{K}}_{\text{c}}=1.0$.