/*! 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 4 The reaction $$ 2 \mathrm{NO... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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?

Short Answer

Expert verified
The system is not at equilibrium (\( Q_c > K_c \)); \( \mathrm{NO}_2 \) concentration will increase.

Step by step solution

01

Determine Initial Concentrations

First, calculate the initial concentrations of \( \mathrm{NO}_{2} \) and \( \mathrm{N}_{2} \mathrm{O}_{4} \) in the flask. The concentration is the amount of substance divided by the volume of the container. So, for \( \mathrm{NO}_{2} \), the concentration is \( \frac{2.0 \times 10^{-3} \text{ mol}}{10 \text{ L}} = 2.0 \times 10^{-4} \text{ M} \). Similarly, for \( \mathrm{N}_{2} \mathrm{O}_{4} \), it is \( \frac{1.5 \times 10^{-3} \text{ mol}}{10 \text{ L}} = 1.5 \times 10^{-4} \text{ M} \).
02

Calculate Reaction Quotient \( Q_c \)

The reaction quotient \( Q_c \) indicates the current state of the reaction with respect to its equilibrium. It is given by \( Q_c = \frac{[\mathrm{N}_2\mathrm{O}_4]}{[\mathrm{NO}_2]^2} \). Plug in the initial concentrations, \[ Q_c = \frac{1.5 \times 10^{-4}}{(2.0 \times 10^{-4})^2} = \frac{1.5 \times 10^{-4}}{4.0 \times 10^{-8}} = 3.75 \times 10^3 \].
03

Compare \( Q_c \) with \( K_c \)

Compare the value of \( Q_c \) with the equilibrium constant \( K_c \). We find \( Q_c = 3.75 \times 10^3 \) and \( K_c = 170 \). Since \( Q_c > K_c \), the reaction quotient is greater than the equilibrium constant, indicating that the system is not at equilibrium.
04

Determine the Direction of Reaction Shift

Because \( Q_c > K_c \), the reaction will shift to the left to reach equilibrium. This means the concentration of \( \mathrm{NO}_{2} \) will increase as the system proceeds to equilibrium.

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.

Reaction Quotient
The Reaction Quotient, often represented as \( Q_c \), is an important concept in understanding chemical equilibrium. It lets us know if a reaction is at equilibrium and which direction it needs to go to achieve equilibrium. For a given chemical reaction, \( Q_c \) is calculated using the same expression as the equilibrium constant \( K_c \), but with the current or initial concentrations of the reactants and products.Here's how you can think of it:- If \( Q_c = K_c \), the reaction is at equilibrium. No shifts needed.- If \( Q_c < K_c \), there are more reactants than needed, so the reaction will shift to the right, towards the formation of products.- If \( Q_c > K_c \), like in our exercise, too many products are present. Hence, the reaction will shift left, towards the formation of more reactants.This tool helps predict how the reaction needs to adjust to attain balance!
Equilibrium Constant
The Equilibrium Constant \( K_c \) is a critical factor that determines the position of equilibrium in a chemical reaction at a given temperature. It provides insight into how far a reaction will proceed at equilibrium.For our reaction, \( 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \), the equilibrium constant expression is:\[ K_c = \frac{[\mathrm{N}_{2}\mathrm{O}_{4}]}{[\mathrm{NO}_{2}]^2} \]A high value of \( K_c \), such as 170, indicates that at equilibrium, the concentration of products \( \mathrm{N}_{2} \mathrm{O}_{4} \) is favored, meaning more product is present than reactants. Conversely, a low \( K_c \) would mean reactants are favored.Knowing \( K_c \) is essential for predicting the composition of the reaction system at equilibrium. It remains constant at a given temperature, allowing for consistency in calculations.
Le Chatelier's Principle
Le Chatelier's Principle is a guiding rule to predict how a change in conditions affects a system's equilibrium. It essentially states that a system at equilibrium will adjust to counteract any imposed change.Key factors that can affect equilibrium include:- **Concentration:** Adding or removing reactants or products shifts the equilibrium to oppose the change.- **Temperature:** Changing the temperature can favor either the endothermic or exothermic direction of the reaction.- **Pressure:** In reactions involving gases, a change in pressure by changing volume can shift the equilibrium when there are different numbers of moles of gases on each side.In our situation, as \( Q_c > K_c \), the principle suggests that the system will respond by shifting to the left, increasing the concentration of \( \mathrm{NO}_{2} \), helping to restore equilibrium. It's all about balance; the system inherently wants to return to a state of equilibrium.

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 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?

Write equilibrium constant expressions for the following reactions. For gases, use either pressures or concentrations. (a) \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g})\) (b) \(\mathrm{Fe}(\mathrm{s})+5 \mathrm{CO}(\mathrm{g}) \rightleftharpoons \mathrm{Fe}(\mathrm{CO})_{5}(\mathrm{g})\) $$\begin{aligned} &\text { (c) }\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(\mathrm{s}) \rightleftharpoons\\\ &&2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \end{aligned}$$$$\text { (d) } \mathrm{Ag}_{2} \mathrm{SO}_{4}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{SO}_{4}^{2-}(\mathrm{aq})$$

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).]

Suppose 0.086 mol of Bra is placed in a 1.26-L. flask and heated to \(1756 \mathrm{K}\), a temperature at which the halogen dissociates to atoms. $$ \mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{Br}(\mathrm{g}) $$ If \(\mathrm{Br}_{2}\) is \(3.7 \%\) dissociated at this temperature, calculate \(K_{c}\)

A mixture of \(\mathrm{CO}\) and \(\mathrm{Cl}_{2}\) is placed in a reaction flask: \(|\mathrm{CO}|=0.0102 \mathrm{mol} / \mathrm{L}\) and \(\left|\mathrm{Cl}_{2}\right|=\) \(0.00609 \mathrm{mol} / \mathrm{L} .\) When the reaction $$ \mathrm{CO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{COCl}_{2}(\mathrm{g}) $$ has come to equilibrium at \(600 \mathrm{R},\left[\mathrm{Cl}_{2}\right]=\) \(0.00301 \mathrm{mol} / \mathrm{L}\) (a) Calculate the concentrations of \(\mathrm{CO}\) and \(\mathrm{COCl}_{2}\) at equilibrium. (b) Calculate \(K_{\mathrm{r}}\)

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.