Question

Consider the reaction in a 1.0-L rigid flask.

$\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

Answer the following questions for each situation (a–d):

i. Estimate a range (as small as possible) for the requested substance. For example, [A] could be between 95M and 100 M.

ii. Explain how you decided on the limits for the estimated range.

iii. Indicate what other information would enable you to narrow your estimated range.

iv. Compare the estimated concentrations for a through d, and explain any differences.

a. If at equilibrium,$\left[\mathbf{A}\right]=\mathbf{1}\mathbf{M}$ , and then 1 mole of C is added, estimate the value for $\left[\mathbf{A}\right]$once equilibrium is re-established.

b. If at equilibrium,$\left[\mathbf{B}\right]=\mathbf{1}\mathbf{M}$, and then 1 mole of C is added, estimate the value for $\left[\mathbf{B}\right]$once equilibrium is re-established.

c. If at equilibrium$\left[\mathbf{C}\right]=\mathbf{1}\mathbf{M}$,and then 1 mole of C is added,estimate the value for$\left[\mathbf{C}\right]$once equilibrium is re-established.

d. If at equilibrium$\left[\mathbf{D}\right]=\mathbf{1}\mathbf{M}$ , and then 1 mole of C is added, estimate the value for $\left[\mathbf{D}\right]$once equilibrium is re-established.

Short answer

Step-by-Step Solution

Step-by-step solution

Solution of subpart (i)

For condition (a)

Given reaction:

$\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At equilibrium $\left[A\right]$ = 1M, 1mol of C is added. This helps the equilibrium to re-establish itself.

Addition of C at equilibrium increases the concentration of products. Hence, according to La Chatlier’s principle, the equilibrium shifted itself to backward direction that is to the left hand side.

Now, the total no. of moles of reactants = 3mol

Ratio of no. of moles of A to the total no. of moles of reactants = 1:3

Addition of C increases the concentration of reactants.

Now, the ratio of no. of moles of A to the total no. of moles of reactants = 1:4

Hence, the concentration of A at equilibrium is higher than 1M.

Therefore, the range will be from 1.25M to 1.75M.

For condition (b)

Given reaction:

$\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At equilibrium $\left[B\right]$= 1M, 1mol of C is added. This helps the equilibrium to re-establish itself.

Addition of C at equilibrium increases the concentration of products. Hence, according to La Chatlier’s principle, the equilibrium shifted itself to backward direction that is to the left hand side.

Now, the total no. of moles of reactants = 3mol

Ratio of no. of moles of B to the total no. of moles of reactants = 2:3

Addition of C increases the concentration of reactants.

Now, the ratio of no. of moles of B to the total no. of moles of reactants = 2:4

Hence, the concentration of A at equilibrium is higher than 1M.

Therefore, the range will be from 1.5M to 2.5M.

For condition (c)

Given reaction:

$\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At equilibrium $\left[C\right]$ = 1M, 1mol of C is added. This helps the equilibrium to re-establish itself.

Addition of C at equilibrium increases the concentration of products. Hence, according to La Chatlier’s principle, the equilibrium shifted itself to backward direction that is to the left hand side.

Now, the total no. of moles of reactants = 3mol

Ratio of no. of moles of A to the total no. of moles of reactants = 1:3

Addition of C increases the concentration of reactants.

Now, the ratio of no. of moles of A to the total no. of moles of reactants = 1:4

Hence, the concentration of C at equilibrium is lower than 2M.

Therefore, the range will be from 1.25M to 1.75M.

For condition (d)

Given reaction:

$\mathit{A}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{2}\mathit{B}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathit{C}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathit{D}\mathbf{\left(}\mathit{g}\mathbf{\right)}$

At equilibrium = 1M, 1mol of C is added. This helps the equilibrium to re-establish itself.

Addition of C at equilibrium increases the concentration of products. Hence, according to La Chatlier’s principle, the equilibrium shifted itself to backward direction that is to the left hand side.

Now, the total no. of moles of reactants = 3mol

Ratio of no. of moles of A to the total no. of moles of reactants = 1:3

Addition of C increases the concentration of reactants.

Now, the ratio of no. of moles of A to the total no. of moles of reactants = 1:4

Hence, the concentration of D at equilibrium is lower than 2M.

Therefore, the range will be from 1.25M to 1.75M.

Solution of subpart (ii)

Explanation

You can explain the above ranges with the help of mole ratio which is used as a conversion factor between reactants and products.

Explanation of mole ratio

The ratio between the moles of two compounds in a reaction is called mole ratio.

 Step 3: Solution of subpart (iii)

Explanation of the answer

As all the reactants and products are in gaseous form, therefore to explain the above ranges you can use temperature and pressure.

Explanation with help of temperature

As you know, equilibrium constant is a function of temperature. When you increase the temperature, you also increase the energy. Therefore, the equilibrium will shift itself in a direction where there is a possibility of consumption of energy.

Explanation with help of pressure

When you increase the pressure at constant volume, you also increase the no. of moles per unit volume. Therefore, the equilibrium will shift itself in a direction where there is a possibility of decrease in no. of moles.

 Step 4: Solution of subpart (iv)

Explanation of the answer

Addition of C at equilibrium increases the concentration of products. Hence, according to La Chatlier’s principle, the equilibrium shifted itself to backward direction that is to the left hand side.

Therefore, the estimated concentration of reactants will be higher than that of the products.