Q17.106CP - Chemistry: Molecular Nature Of Matter And Change

Question

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  Z is 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 ${\mathit{K}}_{\mathbf{\text{c}}}$ .

Step-by-Step Solution

Answer

a) Y and Z are represent equilibrium.

b) Scene in the correct order is $C\to A\to B=D$ .

c) The required value is ${\text{K}}_{\text{c}}=1.0$ .

Step 1: Definition of equilibrium

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

Step 2: (a) Which scene represents equilibrium

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.

Step 3: (b) List the scenes in the correct sequence

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:

$C\to A\to B=D$ .

Step 4: (c) Calculate K c

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