/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 A mixture containing 3.9 moles o... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation: $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftarrows \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

Short Answer

Expert verified
The equilibrium constant, \( K_{c} \), is 1.722.

Step by step solution

01

Write the balanced equation and define changes

The balanced equation for the reaction is \( \mathrm{NO}(g) + \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g) + \mathrm{CO}(g) \). Initially, we have 3.9 moles of \( \mathrm{NO} \) and 0.88 moles of \( \mathrm{CO}_{2} \). If \( x \) moles react, the change in moles will be: \( -x \) for both \( \mathrm{NO} \) and \( \mathrm{CO}_{2} \), and \( +x \) for both \( \mathrm{NO}_{2} \) and \( \mathrm{CO} \).
02

Determine the equilibrium moles

At equilibrium, the number of moles of \( \mathrm{CO}_{2} \) is given as 0.11 moles. Therefore, \( 0.88 - x = 0.11 \), solving this gives \( x = 0.77 \). Thus, the equilibrium moles are: \( \mathrm{NO} = 3.9 - 0.77 = 3.13 \), \( \mathrm{CO}_{2} = 0.11 \), \( \mathrm{NO}_{2} = 0.77 \), and \( \mathrm{CO} = 0.77 \).
03

Calculate equilibrium concentrations

To find the concentration, assume the reaction occurs in a 1 L flask, hence moles equal concentrations (Molarity), giving: \( [\mathrm{NO}] = 3.13 \), \( [\mathrm{CO}_{2}] = 0.11 \), \( [\mathrm{NO}_{2}] = 0.77 \), and \( [\mathrm{CO}] = 0.77 \).
04

Write the equilibrium expression

The equilibrium constant expression is \( K_{c} = \frac{[\mathrm{NO}_{2}][\mathrm{CO}]}{[\mathrm{NO}][\mathrm{CO}_{2}]} \). Insert the equilibrium concentrations into this equation.
05

Calculate the equilibrium constant

Substitute the values into the expression to find \( K_{c} \): \[ K_{c} = \frac{(0.77)(0.77)}{(3.13)(0.11)} \]Calculate to get:\[ K_{c} = \frac{0.5929}{0.3443} = 1.722 \]

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 91Ó°ÊÓ!

Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a state in which the rates of the forward and reverse reactions are equal in a chemical reaction. This means there's no net change in the concentrations of reactants and products over time. In our example, the reaction between nitric oxide (\( \mathrm{NO} \)) and carbon dioxide (\( \mathrm{CO}_2 \)) forms nitrogen dioxide (\( \mathrm{NO}_2 \)) and carbon monoxide (\( \mathrm{CO} \)).
Equilibrium signifies balance in the reaction process. Although the amounts of \( \mathrm{NO} \), \( \mathrm{CO}_2 \), \( \mathrm{NO}_2 \), and \( \mathrm{CO} \) don't change at equilibrium, the chemicals continue to react. This continuous reaction is called dynamic equilibrium.
Understanding chemical equilibrium is crucial because it determines the concentrations of reactants and products, leading to the calculation of the equilibrium constant, \( K_{c} \). This constant helps predict the extent of the reaction and the potential yield of products.
Reaction Stoichiometry
In the world of chemistry, stoichiometry refers to the quantitative relationship between elements in a chemical reaction. The balanced equation for a chemical reaction provides this stoichiometric relationship and allows chemists to predict the amounts of substances consumed and produced.
In our balanced reaction \( \mathrm{NO}(g) + \mathrm{CO}_2(g) \rightleftharpoons \mathrm{NO}_2(g) + \mathrm{CO}(g) \), each molecule of \( \mathrm{NO} \) reacts with one molecule of \( \mathrm{CO}_2 \) to produce one molecule each of \( \mathrm{NO}_2 \) and \( \mathrm{CO} \). This is a 1:1:1:1 stoichiometric relationship.
Stoichiometry helps in calculating the changes in the number of moles of each substance as the reaction proceeds toward equilibrium, which further aids in determining the equilibrium concentrations.
Balanced Equation
A balanced chemical equation is essential for understanding the chemical reaction's stoichiometry. It represents the number of units of each reactant and product involved in the reaction.
For our example, the balanced equation is: - \( \mathrm{NO}(g) + \mathrm{CO}_2(g) \rightleftharpoons \mathrm{NO}_2(g) + \mathrm{CO}(g) \).
This equation showcases the law of conservation of mass, where the number of atoms for each element is the same on both sides of the equation. Balancing ensures that the reaction accurately represents the real-life process.
With the equation properly balanced, it's possible to track how the number of molecules changes as the reaction progresses, directly impacting the calculation of the equilibrium state.
Concentration Calculation
Calculating concentrations is a critical step in understanding chemical reactions at equilibrium. Concentration is measured in \( \text{Molarity} \), \( M \), which is moles per liter. In equilibrium calculations, knowing the volume of the reaction mixture is crucial.
In our exercise, we assume the reaction occurs in a 1 L flask, which makes moles equal concentrations. Thus, the equilibrium concentrations of the substances become directly related to their moles:
  • \( [\mathrm{NO}] = 3.13 \)
  • \( [\mathrm{CO}_2] = 0.11 \)
  • \( [\mathrm{NO}_2] = 0.77 \)
  • \( [\mathrm{CO}] = 0.77 \)
These concentrations allow the calculation of the equilibrium constant \( K_c \), providing insights into the reaction's behavior and directionality. Using such calculations sets the foundation for predicting product yields and optimizing conditions for chemical processes.

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

Consider the heterogeneous equilibrium process: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) $$ At \(700^{\circ} \mathrm{C},\) the total pressure of the system is found to be \(4.50 \mathrm{~atm}\). If the equilibrium constant \(K_{P}\) is 1.52 , calculate the equilibrium partial pressures of \(\mathrm{CO}_{2}\) and CO.

The equilibrium constant \(K_{P}\) for the reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}(g) \rightleftarrows 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) $$ the same temperature? (b) The very small value of \(K_{P}\) (and \(K_{\mathrm{c}}\) ) indicates that the reaction overwhelmingly favors the formation of water molecules. Explain why, despite this fact, a mixture of hydrogen and oxygen gases can be kept at room temperature without any change.

When heated at high temperatures, iodine vapor dissociates as follows: $$ \mathrm{I}_{2}(g) \rightleftarrows 2 \mathrm{I}(g) $$ In one experiment, a chemist finds that when 0.054 mole of \(\mathrm{I}_{2}\) was placed in a flask of volume \(0.48 \mathrm{~L}\) at \(587 \mathrm{~K},\) the degree of dissociation (i.e., the fraction of \(\mathrm{I}_{2}\) dissociated) was \(0.0252 .\) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the reaction at this temperature.

A quantity of 0.20 mole of carbon dioxide was heated to a certain temperature with an excess of graphite in a closed container until the following equilibrium was reached: $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g) $$ Under these conditions, the average molar mass of the gases was \(35 \mathrm{~g} / \mathrm{mol}\). (a) Calculate the mole fractions of \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\). (b) What is \(K_{P}\) if the total pressure is 11 atm? (Hint: The average molar mass is the sum of the products of the mole fraction of each gas and its molar mass.)

Eggshells are composed mostly of calcium carbonate \(\left(\mathrm{CaCO}_{3}\right)\) formed by the reaction: $$\mathrm{Ca}^{2+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \rightleftarrows \mathrm{CaCO}_{3}(s)$$ The carbonate ions are supplied by carbon dioxide produced as a result of metabolism. Explain why eggshells are thinner in the summer when the rate of panting by chickens is greater. Suggest a remedy for this situation.

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.