Question

Given the reaction

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

Consider the following situations:

i. You have 1.3M A and 0.8M B initially.

ii. You have 1.3M A, 0.8M B, and 0.2M C initially.

iii. You have 2.0M A and 0.8M B initially.

After equilibrium has been reached, order i–iii in terms of increasing equilibrium concentrations of D. Explain your sequence. Then give the order in terms of increasing equilibrium concentration of B and explain.

Short answer

You need to calculate the concentration of D and B for three different situations.

Step-by-step solution

Equilibrium constant for the reaction

According to the question the reaction is,

$A\left(g\right)+B\left(g\right)\rightleftharpoons C\left(g\right)+D\left(g\right)$

You can write the equilibrium constant as

$K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}$

Where, K = equilibrium constant;

$\left[A\right]$= concentration of A at equilibrium;

$\left[B\right]$= concentration of B at equilibrium;

$\left[C\right]$= concentration of C at equilibrium;

$\left[D\right]$= concentration of D at equilibrium;

When concentration of D is increased:Step 2: Calculation of concentration of D for situation (i)

According to the question;

$\left[A\right]$= 1.3M

$\left[B\right]$= 0.8M

Now,

$$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[D\right]=\frac{K\left[A\right]\left[B\right]}{\left[C\right]} \\ \left[D\right]=\frac{K\left(1.3\right)\left(0.8\right)}{\left[C\right]} \end{gathered}$$

Hence,${\left[D\right]}_i=\frac{K\left(1.3\right)\left(0.8\right)}{\left[C\right]}$

Calculation of concentration of D for situation (ii)

According to the question;

$\left[A\right]$= 1.3M

$\left[B\right]$= 0.8M

$\left[C\right]$= 0.2M

Now,

$$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[D\right]=\frac{K\left[A\right]\left[B\right]}{\left[C\right]} \\ \left[D\right]=\frac{K\left(1.3\right)\left(0.8\right)}{\left(0.2\right)} \end{gathered}$$

Hence,${\left[D\right]}_{ii}=\frac{K\left(1.3\right)\left(0.8\right)}{\left(0.2\right)}$

Calculation of concentration of D for situation (iii)

According to the question;

$\left[A\right]$= 2M

$\left[B\right]$= 0.8M

Now,

$$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[D\right]=\frac{K\left[A\right]\left[B\right]}{\left[C\right]} \\ \left[D\right]=\frac{K\left(2\right)\left(0.8\right)}{\left[C\right]} \end{gathered}$$

Hence, ${\left[D\right]}_{iii}=\frac{K\left(2\right)\left(0.8\right)}{\left[C\right]}$

Conclusion

Concentration of D is highest in situation (iii).

Hence the order of increasing equilibrium concentration of D is

(ii)<(i)<(iii)

When concentration of B is increased:Step 6: Calculation of concentration of B for situation (i)

According to the question;

$\left[A\right]$= 1.3M

Now,

$$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[B\right]=\frac{\left[C\right]\left[D\right]}{K\left[A\right]} \\ \left[B\right]=\frac{\left[C\right]\left[D\right]}{K\left(1.3\right)} \end{gathered}$$

Hence,${\left[B\right]}_i=\frac{\left[C\right]\left[D\right]}{K\left(1.3\right)}$

Calculation of concentration of B for situation (ii)

According to the question;

$\left[A\right]$= 1.3M

$\left[C\right]$= 0.2M

Now,

localid="1662203461637" $$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[B\right]=\frac{\left[C\right]\left[D\right]}{K\left[A\right]} \\ \left[B\right]=\frac{\left(0.2\right)\left[D\right]}{K\left(1.3\right)} \end{gathered}$$

Hence,${\left[B\right]}_{ii}=\frac{\left(0.2\right)\left[D\right]}{K\left(1.3\right)}$

Calculation of concentration of B for situation (iii)Step 8: Calculation of concentration of B for situation (iii)

According to the question;

$\left[A\right]$= 2M

Now,

$$\begin{gathered} K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]} \\ \left[B\right]=\frac{\left[C\right]\left[D\right]}{K\left[A\right]} \\ \left[B\right]=\frac{\left[C\right]\left[D\right]}{K\left(2\right)} \end{gathered}$$

Hence,${\left[B\right]}_{iii}=\frac{\left[C\right]\left[D\right]}{K\left(2\right)}$

Conclusion

Concentration of B is highest in situation (ii).

Hence the order of increasing equilibrium concentration of B is

(iii)<(i)<(ii)