Problem 23 - Chemistry and Chemical Reactivity

Question

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

Step-by-Step Solution

Answer

The equilibrium constant \(K\) for the target reaction is approximately 13.7.

Step 1: Identify the Target Reaction

The target reaction is given as \(\operatorname{SnO}_{2}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g}) \rightleftarrows\mathrm{Sn}(\mathrm{s})+2 \mathrm{CO}_{2}(\mathrm{g})\), and we need to find its equilibrium constant \(K\).

Step 2: Analyze the Provided Reactions and Constants

We have two provided reactions: \(\mathrm{SnO}_{2}(\mathrm{s})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftarrows\mathrm{Sn}(\mathrm{s})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) with \(K_1 = 8.12\), and \(\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g})\) with \(K_2 = 0.771\).

Step 3: Manipulate the Given Equations to Match the Target Reaction

The goal is to manipulate these reactions algebraically to arrive at the overall target reaction. Notice that if we reverse the second reaction, which is \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) \rightleftarrows\mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})\), the equilibrium constant becomes \(\frac{1}{K_2}\).

Step 4: Combine the Manipulated Reactions

Combine the reversed second reaction with the first reaction to form the target reaction:1. Rewrite the first reaction: \(\mathrm{SnO}_{2}(\mathrm{s}) + 2\mathrm{H}_{2}(\mathrm{g}) \rightleftarrows \mathrm{Sn}(\mathrm{s}) + 2\mathrm{H}_{2}\mathrm{O}(\mathrm{g})\).2. Add doubled reversed reaction: \(2(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \rightleftarrows \mathrm{H}_{2}(\mathrm{g}) + \mathrm{CO}_{2}(\mathrm{g}))\).This results in:- Left-hand side: \(\mathrm{SnO}_{2}(\mathrm{s}) + 2\mathrm{CO}(\mathrm{g}) + 2\mathrm{H}_{2}(\mathrm{g})\)- Right-hand side: \(\mathrm{Sn}(\mathrm{s}) + 2\mathrm{CO}_{2}(\mathrm{g}) + 2\mathrm{H}_{2}(\mathrm{g})\)

Step 5: Calculate the Equilibrium Constant for the Target Reaction

Combining the equilibrium constants, the equilibrium constant for the combination of the two original reactions is:\(K = K_1 \times \left(\frac{1}{K_2}\right)^2 = 8.12 \times \left(\frac{1}{0.771}\right)^2\).Calculate this value to find the equilibrium constant \(K\) for the target reaction.

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, and they can be described using chemical equations.
These equations show the substances that react and form products during a chemical reaction.
In any reaction, you will find reactants on the left side and products on the right side of the equation.

• Reactants are substances that are consumed during the reaction.
• Products are generated as a result of the reaction.

An important rule to remember is that the mass and number of atoms are conserved in chemical reactions. This means that the number of atoms for each element in the reactants equals the number of atoms in the products. Understanding chemical reactions is crucial as it allows scientists to predict the direction and extent of reactions. This prediction can be made by calculating the equilibrium constant, which provides insights into the balance between reactants and products at equilibrium.

Calculating K

The equilibrium constant, denoted as \( K \), is a crucial value in understanding chemical reactions at equilibrium. It quantifies how far a reaction will proceed towards products under a given set of conditions.

The value of \( K \) can be calculated from the concentrations of products and reactants at equilibrium using the following generic equilibrium expression: \[K = \frac{[products]}{[reactants]}\] In practice, you replace "products" and "reactants" with their measurable concentrations or partial pressures. For reactions involving solids or pure liquids, their concentrations are typically not included in the expression because their activities are constant and set to 1.

In our specific example, we are given two reactions with known \( K \) values and asked to find the \( K \) for another reaction. This requires manipulation of the given equations and their \( K \) values.

Reaction Manipulation

Reaction manipulation involves altering the given chemical equations to achieve the desired target equation. This process is necessary for calculating the equilibrium constant of a target reaction when direct data is unavailable.

Here, you either reverse reactions or modify the stoichiometric coefficients and adjust the equilibrium constants accordingly:

• Reversing a reaction inverts its equilibrium constant, turning \( K \) into \( \frac{1}{K} \).
• Multiplying a reaction by a factor alters its \( K \) by raising it to the power of that factor.

In our exercise, to match the target reaction, a given reaction was reversed, and its \( K \) recalculated as reciprocal. This kind of manipulation helps us construct a reaction pathway that arrives at the target equation. Calculating \( K \) for the new pathway yields the equilibrium constant for the new, desired reaction setup.

Equilibrium Expressions

Equilibrium expressions are mathematical representations that describe the ratio of concentrations of products to reactants at equilibrium, aligned with the chemical equation's stoichiometry.
These expressions are essential for calculating the equilibrium constant \( K \), and they guide us in predicting whether the reaction favors the formation of reactants or products. For a generic reaction of the form:\[aA + bB \rightleftharpoons cC + dD\]The equilibrium expression would be:\[K = \frac{[C]^c [D]^d}{[A]^a [B]^b}\] Each concentration is raised to the power of its coefficient in the balanced equation. In our specific example, even though we are not substituting actual concentrations, understanding this framework enables us to assemble reactions algebraically to solve for the target \( K \), ensuring compliance with equilibrium principles. This process ensures that complex reaction dynamics are simplified for practical analysis and calculation.