Chapter 17.3, Problem 2RQ

Interpretation Introduction

Interpretation: Effect of the changes mentioned on the following reaction has to be explained.

Concept introduction: When the rate at which forward and backward reaction occur become equal such that concentration of reactants and products remains to be constant, that state is described as equilibrium. It is indicated by double sided arrow $\rightleftharpoons$ . It can be physical or chemical in nature.

There are various parameters on which equilibrium depends such as temperature, pressure, amount of reactant or product etc. For an equilibrium reaction:

  $a\text{A}+b\text{B}\rightleftharpoons c\text{C}+d\text{D}$

The expression for equilibrium constant is:

  $K=\frac{{\left[\text{C}\right]}^c{\left[\text{D}\right]}^d}{{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b}$

Where,

$K$ is equilibrium constant.

  $\left[\text{A}\right]$ and $\left[\text{B}\right]$ are reactant concentrations for A and B.

  $\left[\text{C}\right]$ and $\left[\text{D}\right]$ are product concentrations for C and D.

a, b, c and d are coefficients of reactants.

Answer to Problem 2RQ

1. Addition of ${\text{UO}}_2$to system will not affect equilibrium.
2. $\text{5}\text{mol}\text{of Xe}\left(g\right)$ when added to system at constant volume, speeds up the process of equilibrium establishment only by accelerating both forward and backward reaction. The point at which equilibrium is achieved remains unaffected.
3. If $\text{HF}$ attacks and reacts with glass, equilibrium shifts to left.
4. Equilibrium shifts to right if water vapor is removed.
5. If the size of reaction vessel is increased, equilibrium shifts to right.

Explanation of Solution

For reaction,

  ${\text{UO}}_2\left(s\right)+4\text{HF}\left(g\right)\rightleftharpoons {\text{UF}}_{\text{4}}\left(g\right)+2{\text{H}}_{\text{2}}\text{O}\left(g\right)$

Since ${\text{UO}}_2\left(s\right)$ is pure solid, addition or removal of it will not affect equilibrium state.

Similarly, increase in volume of vessel or decrease seen in pressure will shift equilibrium to right as there are 2 moles of gaseous product than 1 mole of gaseous reactant in vessel.

Decrease in concentration of $\text{HF}$ (reactant) will shift equilibrium to left to oppose the change as per Le Chatelier’s principle.

Lastly presence of the catalyst only fasten up the equilibrium establishment and do not affect the value of equilibrium constant.

Conclusion

1. Addition of ${\text{UO}}_2$to system will not affect equilibrium.
2. $\text{5}\text{mol}\text{of Xe}\left(g\right)$when added to system at constant volume, speeds up the process of equilibrium establishment only by accelerating both forward and backward reaction. The point at which equilibrium is achieved remains unaffected.
3. If $\text{HF}$ attacks and reacts with glass, equilibrium shifts to left.
4. Equilibrium shifts to right if water vapor is removed.
5. If the size of reaction vessel is increased, equilibrium shifts to right.