Chapter 17, Problem 33A

(a)

Interpretation Introduction

Interpretation: The effect of increasing pressure by injecting oxygen (reactant) on the equilibrium needs to be explained.

Concept introduction: According to Le Chatelier's principle if any stress is given to the equilibrium, it will shift in a direction to reduce the given stress. If change in concentration, pressure or temperature takes place at equilibrium, it will shift in the direction to undo the change or its effect.

Answer to Problem 33A

The reaction will move in the backward direction.

Explanation of Solution

According to Le Chatelier’s principle, if the pressure is increased, the direction of equilibrium will shift so as to reduce the pressure. Pressure is created when gas molecules collides with the sides of their container. The more the number of molecules in the container, the more will be the pressure. The system can decrease the pressure by reacting in such a way as to generate fewer molecules.

The concerned reaction: $4N{H}_3\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+6H_{2}O\left(g\right)$

  $\begin{array}{c}\text{In this reaction change in the number of moles }\Delta n=(4+6)-(4+5)\\ =+1\end{array}$

So, here in the backward reaction moles of gas decreases. If pressure is increased reaction shifts in backward direction.

(b)

Interpretation Introduction

Interpretation: The effect of addition of desiccant in the system needs to be explained.

Concept introduction: According to Le Chatelier's principle if any stress is given to the equilibrium, it will shift in a direction to reduce the given stress. If change in concentration, pressure or temperature takes place at equilibrium, it will shift in the direction to undo the change or its effect.

Answer to Problem 33A

The reaction equilibrium will shift towards right.

Explanation of Solution

The concerned reaction: $4N{H}_3\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+6H_{2}O\left(g\right)$

Equilibrium constant for the reaction is given by the equation, $K=\frac{{\left[NO\right]}^4{\left[H_{2}O\right]}^6}{{\left[N{H}_3\right]}^4{\left[O_2\right]}^5}$

When $H_{2}O$ (product) is absorbed then the value of $K$ will decrease as the concentration of $H_{2}O$ is in nominator. To maintain the value of $K$ concentration of the products ( $N{H}_3$ and $O_2$ ) will decrease. Hence, the reaction will shift towards right.

(c)

Interpretation Introduction

Interpretation: The effect on equilibrium when the system is compressed and ammonia liquefies needs to be explained.

Concept introduction: According to Le Chatelier's principle if any stress is given to the equilibrium, it will shift in a direction to reduce the given stress. If change in concentration, pressure or temperature takes place at equilibrium, it will shift in the direction to undo the change or its effect.

Answer to Problem 33A

The reaction will move in the backward direction.

Explanation of Solution

The concerned reaction: $4N{H}_3\left(g\right)+5O_2\left(g\right)\rightleftharpoons 4NO\left(g\right)+6H_{2}O\left(g\right)$

Equilibrium constant for the reaction is given by the equation, $K=\frac{{\left[NO\right]}^4{\left[H_{2}O\right]}^6}{{\left[N{H}_3\right]}^4{\left[O_2\right]}^5}$

When the system is compressed pressure increases. According to Le Chatelier’s principle, if the pressure is increased, the direction of equilibrium will shift so as to reduce the pressure. Pressure is created when gas molecules collides with the sides of their container. The more the number of molecules in the container, the more will be the pressure. The system can decrease the pressure by reacting in such a way as to generate fewer molecules.

And also when $N{H}_3$ is started to get liquefy concentration of the gaseous $N{H}_3$ decreases. To maintain the value of $K$ reaction equilibrium shifts to backward direction to increase the amount of reactants.