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

Short Answer

Expert verified
The equilibrium constant \( K \) for the target reaction is approximately 10.53.

Step by step solution

01

Understand the Reaction System

The given reaction system involves two auxiliary reactions with known equilibrium constants and one target reaction for which we need to find the equilibrium constant \( K \). We will use Hess's Law of reaction equilibrium to interconnect these reactions.
02

Write the Target Reaction

Write down the target reaction for which we need to calculate the equilibrium constant:\[ \mathrm{SnO}_{2}(\mathrm{s}) + 2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}(\mathrm{s}) + 2 \mathrm{CO}_{2}(\mathrm{g}) \]
03

Write and Analyze Auxiliary Reactions

List the auxiliary reactions given:1. \[ \mathrm{SnO}_{2}(\mathrm{s}) + 2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}(\mathrm{s}) + 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \quad K_1 = 8.12 \]2. \[ \mathrm{H}_{2}(\mathrm{g}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) + \mathrm{CO}(\mathrm{g}) \quad K_2 = 0.771 \]
04

Adjust the Auxiliary Reactions

To obtain the target reaction, write the second reaction in reverse since it involves \( \mathrm{CO}_{2} \) and \( \mathrm{CO} \) on the opposite sides compared to the target reaction. This means the reverse reaction has an equilibrium constant \( K_2' = \frac{1}{0.771} \).
05

Combine the Reactions

Combine adjusted reactions to get the desired reaction:- Use the first reaction as is.- Reverse the second reaction.Combined, these will yield:\[ \mathrm{SnO}_{2}(\mathrm{s}) + 2 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Sn}(\mathrm{s}) + 2 \mathrm{CO}_{2}(\mathrm{g}) \]
06

Calculate the Equilibrium Constant for the Target Reaction

According to Hess's Law, the equilibrium constant for the target reaction is the product of the equilibrium constants of the adjusted reactions:\[ K = K_1 \times \frac{1}{K_2} = 8.12 \times \frac{1}{0.771} \approx 10.53 \]
07

Verify and Conclude

Verify that the reaction combination summarizes correctly to the target reaction and the equilibrium constants were manipulated properly. Conclude that the equilibrium constant \( K \) for the target reaction is approximately 10.53.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The equilibrium constant, denoted as \( K \), is a crucial aspect of chemical equilibrium and provides insights into the favorability of a reaction towards products or reactants. This constant is derived from the reaction's balanced equation, specifically from the concentrations of the products divided by the concentrations of the reactants, raised to their respective stoichiometric coefficients.
For the given gas-phase reaction, the general form becomes:
  • \[ K = \frac{{[\text{products}]}}{{[\text{reactants}]}} \]
In heterogeneous equilibria like the one we are analyzing, the concentration of solids (like \( \text{SnO}_2 \) and \( \text{Sn} \)) is not included in the \( K \) expression as their concentration remains constant. Consequently, the equilibrium constant for this reaction depends solely on the concentrations of the gaseous reactants and products: \( \text{CO} \) and \( \text{CO}_2 \).
Understanding \( K \) allows chemists to predict which way a reaction might shift under different conditions or whether it has reached equilibrium.
Reaction Equation
A reaction equation provides a quantitative description of the amounts of reactants and products involved in a chemical reaction. In this context, we focus on the specific equation provided:
  • \[ \text{SnO}_2(\text{s}) + 2 \text{CO}(\text{g}) \rightleftharpoons \text{Sn}(\text{s}) + 2 \text{CO}_2(\text{g}) \]
This equation illustrates a reversible change occurring between tin(IV) oxide and carbon monoxide, forming tin and carbon dioxide. Each component is accompanied by its physical state: (s) stands for solid and (g) for gas. In a balanced chemical equation, the number of atoms for each element remains consistent on both sides, ensuring mass conservation. Successfully understanding and manipulating these equations involves recognizing patterns and relationships among the molecules and the conditions under which these transformations occur.
It's essential to express and balance these equations correctly to accurately calculate the equilibrium constant, thereby understanding the extent and direction of the reaction.
Hess's Law
Hess's Law is a fundamental principle in chemistry stating that a chemical reaction's total enthalpy change is the same, regardless of its pathway. It allows for the combination and manipulation of reactions to determine unknown equilibrium constants or enthalpies by summing known quantities.
This principle is effectively applied in the given exercise to compute the equilibrium constant for a reaction not directly ascertainable from elementary data.
  • First, identify auxiliary reactions. These show known equilibrium constants:
    • \( \text{SnO}_2(\text{s}) + 2 \text{H}_2(\text{g}) \rightleftharpoons \text{Sn}(\text{s}) + 2 \text{H}_2\text{O}(\text{g}), \quad K_1 = 8.12 \)
    • \( \text{H}_2(\text{g}) + \text{CO}_2(\text{g}) \rightleftharpoons \text{H}_2\text{O}(\text{g}) + \text{CO}(\text{g}), \quad K_2 = 0.771 \)
  • Next, adjust and reverse reactions as needed. This step involves reversing the second reaction to match the target reaction components:
    • \( K_2' = \frac{1}{0.771} \)
  • Then, apply Hess's Law by multiplying the equilibrium constants from the adjusted reactions to find the equilibrium constant for the desired reaction:
    • \( K = K_1 \times K_2' = 8.12 \times \frac{1}{0.771} \approx 10.53 \)
    By leveraging Hess's Law, one can systematically deduce equilibrium constants of composite reactions through logical rearrangement and evaluation of known reactions.

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Most popular questions from this chapter

Hydrogen and carbon dioxide react at a high temperature to give water and carbon monoxide. $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CO}(\mathrm{g}) $$ (a) Laboratory measurements at \(986^{\circ} \mathrm{C}\) show that there are 0.11 mol each of \(\mathrm{CO}\) and \(\mathrm{H}_{2} \mathrm{O}\) vapor and 0.087 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) at equilibrium in a 50.0 -L container. Calculate the equilibrium constant for the reaction at \(986^{\circ} \mathrm{C}\) (b) Suppose 0.010 mol each of \(\mathrm{H}_{2}\) and \(\mathrm{CO}_{2}\) are placed in a 200.0 -L. container. When equilibrium is achieved at \(986^{\circ} \mathrm{C},\) what amounts of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) in moles, would be present? [Use the value of \(K_{c}\) from part (a).]

The reaction $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$ has an equilibrium constant, \(K_{c}\) of 170 at \(25^{\circ} \mathrm{C}\). If \(2.0 \times 10^{-3}\) mol of \(\mathrm{NO}_{2}\) is present in a \(10 .\) -L. flask along with \(1.5 \times 10^{-3}\) mol of \(\mathrm{N}_{2} \mathrm{O}_{4}\), is the system at equilibrium? If it is not at equilibrium, does the concentration of \(\mathrm{NO}_{2}\) increase or decrease as the system proceeds to equilibrium?

A \(\mathrm{COCl}_{2}\) decomposes to \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) at high temperatures. \(K_{c}\) at \(600 \mathrm{K}\) for the reaction is \(0.0071 .\) $$ \mathrm{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If 0.050 mol of \(\mathrm{COCl}_{2}\) is placed in a 12.5 -L. flask, what is the total pressure at equilibrium at \(600 \mathrm{K} ?\)

Sulfuryl chloride, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\), is used as a reagent in the synthesis of organic compounds. When heated to a sufficiently high temperature, it decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=0.045\) at \(375^{\circ} \mathrm{C}\) (a) \(A\) 10.0-L. flask containing 6.70 g of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is heated to \(375^{\circ} \mathrm{C}\). What is the concentration of each of the compounds in the system when equilibrium is achieved? What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (b) What are the concentrations of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) at equilibrium in the 10.0 -L flask at \(375^{\circ} \mathrm{C}\) if you begin with a mixture of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) \((6.70 \mathrm{g})\) and \(\mathrm{Cl}_{2}(0.10 \mathrm{atm}) ?\) What fraction of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) has dissociated? (c) Compare the fractions of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in parts (a) and (b). Do they agree with your expectations based on Le Chatelier's principle?

The equilibrium constant for the dissociation of iodine molecules to iodine atoms $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}) $$ is \(3.76 \times 10^{-3}\) at 1000 K. Suppose 0.105 mol of \(\mathrm{I}_{2}\) is placed in a 12.3 -L flask at 1000 K. What are the concentrations of \(\mathrm{I}_{2}\) and I when the system comes to equilibrium?

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