Q17.7P - Chemistry: Molecular Nature Of Matter And Change

Question

In a study of the thermal decomposition of Lithium peroxide, 

$\mathbf{2}{\mathbf{Li}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{Li}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

A chemist finds that, as long as some Lithium peroxide is present at the end of the experiment, the amount of oxygen obtained in a given container at a given temperature is the same. Explain.

Step-by-Step Solution

Answer

Since the system is in equilibrium

Step 1: Equilibrium

The equilibrium of a reaction at a given temperature is the state in which the rate of formation of the product is equal to the rate of decomposition of the reactant.   This is dynamic in nature.

Step 2: Explanation

The reaction at equilibrium for the decomposition of lithium peroxide is as shown below:

 $2L{i}_2{O}_2\left(s\right)\rightleftharpoons 2L{i}_{2}O\left(s\right)+O_2\left(g\right)$

The given reaction is in equilibrium state. The equilibrium state of a chemical reaction is the state in which the reactants and the products are in equilibrium to each other or it can be defined as the rate of formation of product is equal to the rate of backward reaction.

Hence when there is any lithium peroxide present, it will be decomposed to lithium oxide and oxygen. This results in the formation of oxygen when there is lithium peroxide even at the end of the reaction.