Problem 23 - Chemistry and Chemical Reactivity

Question

Calculate \(K\) for the reaction $$ \mathrm{SnO}_{2}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g}) $$ given the following information: \(\operatorname{SnO}_{2}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows \operatorname{Sn}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K=8.12\) $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \quad K=0.771 $$

Step-by-Step Solution

Answer

The value of \( K \) for the reaction is 13.66.

Step 1: Write the Given Reactions

We have two reactions with their equilibrium constants provided. The first reaction is \( \mathrm{SnO}_{2}( ext{s}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}( ext{s}) + 2 \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \) with \( K = 8.12 \). The second reaction is \( \mathrm{H}_{2}(\mathrm{g}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \) with \( K = 0.771 \).

Step 2: Reverse the Second Reaction

To match the intermediate terms with the desired reaction, reverse the second reaction: \( \mathrm{H}_{2}\mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{g}) + \mathrm{CO}_{2}(\mathrm{g}) \). When a reaction is reversed, the equilibrium constant \( K \) becomes \( \frac{1}{K} \), resulting in \( K = \frac{1}{0.771} = 1.297 \).

Step 3: Add the Reactions

Sum the reactions after reversing the second one:1. \( \mathrm{SnO}_{2} + 2 \mathrm{H}_{2} \rightleftharpoons \mathrm{Sn} + 2 \mathrm{H}_{2}\mathrm{O} \) 2. \( 2 \mathrm{(H}_{2}\mathrm{O} + \mathrm{CO}) \rightleftharpoons 2 \mathrm{(H}_{2} + \mathrm{CO}_{2}) \)Adding these gives:\( \mathrm{SnO}_{2} + 2 \mathrm{CO} \rightleftharpoons \mathrm{Sn} + 2 \mathrm{CO}_{2} \)

Step 4: Calculate the Overall Equilibrium Constant

The equilibrium constant for the overall reaction is the product of the constants of the added reactions. Thus: \[ K_{overall} = 8.12 \times (1.297)^2 \] Calculate \( (1.297)^2 = 1.6829 \), then:\[ K_{overall} = 8.12 \times 1.6829 = 13.66 \]

Key Concepts

Chemical Equilibrium

Chemical equilibrium is a state where the rate of the forward reaction equals the rate of the reverse reaction in a chemical process. At equilibrium, the concentrations of reactants and products remain constant over time. This doesn't mean that the reactants and products are equal, but that their ratios do not change. In our example, the equilibrium involves the reaction of tin(IV) oxide with carbon monoxide to produce tin and carbon dioxide. Both of these reactions have reached a state where the conversion of reactants to products happens at the same rate as the conversion of products back to reactants.
Equilibrium can be expressed using an equilibrium constant, denoted as \( K \). It is specific for a given reaction at a certain temperature. The larger the \( K \) value, the more the reaction favors the products at equilibrium. Conversely, a smaller \( K \) value indicates a reaction that favors the reactants. Understanding chemical equilibrium is crucial for predicting the amount of reactants and products present in a reaction at any given time.

Reaction Quotient

The reaction quotient, \( Q \), is similar to the equilibrium constant \( K \), but it applies to the concentrations or pressures of reactants and products at any point before the system reaches equilibrium. By comparing \( Q \) to \( K \), we can predict the direction the reaction will proceed to achieve equilibrium.

• If \( Q = K \), the system is at equilibrium, and no further changes in concentration occur.
• If \( Q < K \), the reaction will proceed in the forward direction, forming more products, to reach equilibrium.
• If \( Q > K \), the reaction will proceed in the reverse direction, forming more reactants, to reach equilibrium.

In the provided exercise, understanding the concept of \( Q \) is essential because it helps to evaluate and predict how the system behaves as changes occur.Analyzing \( Q \) versus \( K \) allows chemists to adjust conditions to achieve the desired products or maximize yields by steering the system towards equilibrium.

Le Chatelier's Principle

Le Chatelier's principle explains how a system at equilibrium responds to external changes. According to this principle, if an external change is applied (like changes in concentration, pressure, or temperature), the equilibrium will shift to counteract this change and restore balance.
When the pressure is increased, the equilibrium will favor the side of the reaction with fewer moles of gas. When the concentration of a reactant is increased, the system shifts towards the products, whereas a decrease in concentration results in the reaction favoring the reactants.
This principle is a handy predictive tool. For instance, if more \( ext{CO}_2 \) is added to our reaction at equilibrium, the system will shift to favor the reverse reaction and form more tin and carbon monoxide to reduce \( ext{CO}_2 \) levels. Likewise, removing a product will shift the equilibrium to produce more of it, thus manipulating yields and optimizing chemical processes.