Q17.69P - Chemistry: Molecular Nature Of Matter And Change

Question

How would you adjust the volume of the container in order to maximize product yield in each of the following reactions?

(a) $\mathit{F}{\mathit{r}}_{\mathbf{3}}{\mathit{O}}_{\mathbf{4}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{\rightleftharpoons }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{3}\mathit{F}\mathit{e}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$ 

(b) $\mathbf{2}\mathbf{C}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{+}{\mathbf{0}}_{\mathbf{2}}\mathbf{\left(}\mathbf{g}\mathbf{\right)}\mathbf{ }\mathbf{ }\mathbf{\rightleftharpoons }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{2}\mathbf{CO}\mathbf{\left(}\mathbf{g}\mathbf{\right)}$

Step-by-Step Solution

Answer

Answer

(a) Change in volume will have no effect.

(b) Increase in volume.

Step 1: Definition of Concept

Le-Chatelier's principle states that, whenever a system at equilibrium is disturbed, the system will undergo reactions and try to cancel that effect and retain equilibrium. Changes in concentration of any component, temperature, pressure, or volume are all examples of disturbances.

Step 2: Explain how would you adjust the volume of the container in order to maximize product

(a)

Considering the given reaction,

${\mathrm{Fe}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+4{\mathrm{H}}_2\left(\mathrm{g}\right) \rightleftharpoons 3\mathrm{Fe}\left(\mathrm{s}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

4 mol of gas is present on the reactant side, and 4 mol of gas is present on the product side.

There are an equal number of gas molecules on the reactant and product sides. The reaction will be unaffected by a change in container volume.

Step 3: Explain how would you adjust the volume of the container in order to maximize product

(b)

Considering the given reaction,

$2\mathrm{C}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{CO}\left(\mathrm{g}\right)$

1 mol of gas is present on the reactant side, and 2 mol of gas is present on the product side.

There are more gas molecules on the product side than on the reactant side.

The pressure inside the container should be lower to favour the side with the most gas molecules. There is more "space" in a low-pressure container. As a result, increasing the container's volume will help the products form more easily.