Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equilibrium constant for the reaction $$ \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g}) $$ is \(1.7 \times 10^{-3}\) at \(2300 K\) (a) What is \(K\) for the reaction when written as follows? $$ 1 / 2 \mathrm{N}_{2}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g}) $$ (b) What is \(K\) for the following reaction? $$ 2 \mathrm{NO}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$

The reaction $$ \mathrm{PCl}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{PC}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ was examined at \(250^{\circ} \mathrm{C}\). At equilibrium, \(\left|\mathrm{PCl}_{5}\right|=\) \(4.2 \times 10^{-5} \mathrm{mol} / \mathrm{L},\left[\mathrm{PCl}_{3}\right]=1.3 \times 10^{-2} \mathrm{mol} / \mathrm{L}_{-}\) and \(\left[\mathrm{Cl}_{2}\right]=3.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate \(K_{\epsilon}\) for the reaction.

The dissociation of calcium carbonate has an equilibrium constant of \(K_{\mathrm{p}}=1.16\) at \(800^{\circ} \mathrm{C}\) $$ \mathrm{CaCO}_{3}(\mathrm{s}) \rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) $$ (a) What is \(K_{c}\) for the reaction? (b) If you place \(22.5 \mathrm{g}\) of \(\mathrm{CaCO}_{3}\) in a 9.56 -L container at \(800^{\circ} \mathrm{C},\) what is the pressure of \(\mathrm{CO}_{2}\) in the container? (c) What percentage of the original 22.5 -g sample of \(\mathrm{CaCO}_{3}\) remains undecomposed at equilibrium?

The total pressure for a mixture of \(\mathrm{N}_{2} \mathrm{O}_{4}\) and \(\mathrm{NO}_{2}\) is 0.15 atm. If \(K_{p}=7.1\) (at \(25^{\circ} \mathrm{C}\) ), calculate the partial pressure of each gas in the mixture. $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$

Equal numbers of moles of \(\mathrm{H}_{2}\) gas and \(\mathrm{I}_{2}\) vapor are mixed in a flask and heated to \(700^{\circ} \mathrm{C}\). The initial concentration of each gas is \(0.0088 \mathrm{mol} / \mathrm{L},\) and \(78.6 \%\) of the \(\mathrm{I}_{2}\) is consumed when equilibrium is achieved according to the equation $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ Calculate \(K_{c}\) for this reaction.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free