Question

Consider the following exothermic reaction at equilibrium:

${\mathbf{N}}_2\left(\mathbf{g}\right)\mathbf{+}\mathbf{3}{\mathbf{H}}_2\left(\mathbf{g}\right)\rightleftharpoons \mathbf{2}{\mathbf{NH}}_3\left(\mathbf{g}\right)$

Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

Short answer

When${\text{N}}_{\text{2}}\left(\text{g}\right)$is added,$\left[{\text{N}}_{\text{2}}\right]$will increase, $\left[{\text{H}}_{\text{2}}\right]$will decrease, and $\left[{\text{NH}}_{\text{3}}\right]$will increase.

When$\left[{\text{H}}_{\text{2}}\right]\left(\text{g}\right)$is removed,$\left[{\text{N}}_{\text{2}}\right]$will increase,$\left[{\text{H}}_{\text{2}}\right]$will decrease, and$\left[{\text{NH}}_{\text{3}}\right]$will decrease.

When$N{H}_3\left(g\right)$is added,$\left[{\text{N}}_{\text{2}}\right]$will increase,$\left[{\text{H}}_{\text{2}}\right]$will increase, and $\left[{\text{NH}}_{\text{3}}\right]$will decrease.

When $Ne\left(g\right)$is added, ,$\left[{\text{N}}_{\text{2}}\right]$ ,$\left[{\text{H}}_{\text{2}}\right]$and $\left[{\text{NH}}_{\text{3}}\right]$will remain constant

When the temperature is increased, $\left[{\text{N}}_{\text{2}}\right]$will increase,$\left[{\text{H}}_{\text{2}}\right]$will increase, and $\left[{\text{NH}}_{\text{3}}\right]$will decrease.

Step-by-step solution

Define Le Chatelier’s Principle

Le Chatelier’s principle states that when a system at equilibrium is distributed by factors such as temperature, pressure, and concentration, it tends to attain its equilibrium position by shifting its equilibrium position to either the reactant side or the product side by increasing or decreasing the concentrations of reactants and products.

Explanation

The equilibrium reaction is as follows:

${\text{N}}_{\text{2}}\left(\text{g}\right){\text{+3H}}_{\text{2}}\left(\text{g}\right)\to {\text{2NH}}_{\text{3}}\left(\text{g}\right)$

When${\text{N}}_{\text{2}}\left(\text{g}\right)$is added, the reaction will shift to the right side, i.e., the product side in order to reestablish the equilibrium. Thus,$\left[{\text{N}}_{\text{2}}\right]$will increase,$\left[{\text{H}}_{\text{2}}\right]$will decrease, and$\left[{\text{NH}}_{\text{3}}\right]$will increase.

When$\left[{\text{H}}_{\text{2}}\right]\left(\text{g}\right)$is removed, the reactant is decreased;so product formation will also be decreased and the reaction will shift to the left, i.e., the reactant side. Thus,$\left[{\text{N}}_{\text{2}}\right]$will increase,$\left[{\text{H}}_{\text{2}}\right]$will decrease, and$\left[{\text{NH}}_{\text{3}}\right]$will decrease.

When$N{H}_3\left(g\right)$is added, the reactants have to increase and so the reaction will shift to the left, i.e., the product side. Thus$\left[{\text{N}}_{\text{2}}\right]$,will increase,$\left[{\text{H}}_{\text{2}}\right]$ will increase, and $\left[{\text{NH}}_{\text{3}}\right]$will decrease.

When$Ne\left(g\right)$is added, the reaction will not be affected because Ne is a noble gas. ,Thus$\left[{\text{N}}_{\text{2}}\right]$ , $\left[{\text{H}}_{\text{2}}\right]$and$\left[{\text{NH}}_{\text{3}}\right]$will remain constant.

When the temperature is increased, the reaction will shift to the left, i.e., the reactant side, to reestablish the equilibrium because the reaction is an exothermic reaction, and heat is produced on the product. Thus,$\left[{\text{N}}_{\text{2}}\right]$ will increase,$\left[{\text{H}}_{\text{2}}\right]$will increase, and $\left[{\text{NH}}_{\text{3}}\right]$will decrease.