/*! 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 40 Assume you place \(0.010 \mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

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Assume you place \(0.010 \mathrm{~mol} \mathrm{~N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) in a sealed \(2.0-\mathrm{L}\) flask at \(50 .{ }^{\circ} \mathrm{C}\). After the system reaches equilibrium, \(\left[\mathrm{N}_{2} \mathrm{O}_{4}\right]=0.00090 \mathrm{M} .\) Calculate the value of \(K_{\mathrm{c}}\) for this reaction. $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$

Short Answer

Expert verified
The value of \(K_c\) is approximately 0.075.

Step by step solution

01

Write the Balanced Equation

The given reaction is already balanced: \( \text{N}_{2} \text{O}_{4}(\text{g}) \rightleftharpoons 2\text{NO}_{2}(\text{g}) \).
02

Determine Initial Concentration of N2O4

We start with \(0.010 \text{ mol} \) of \( \text{N}_{2} \text{O}_{4} \) in a \(2.0 \text{ L} \) flask. The initial concentration \([\text{N}_{2} \text{O}_{4}]_0\) is calculated as \( \frac{0.010 \text{ mol}}{2.0 \text{ L}} = 0.0050 \text{ M} \).
03

Calculate Change in Concentration of N2O4

At equilibrium, \([\text{N}_{2} \text{O}_{4}] = 0.00090\text{ M}\). The change in concentration of \( \text{N}_{2} \text{O}_{4} \) is \(0.0050 - 0.00090 = 0.0041 \text{ M}\).
04

Calculate Change in Concentration of NO2

Since \( \text{N}_{2} \text{O}_{4} \) dissociates into 2 \( \text{NO}_{2} \), the increase in concentration of \( \text{NO}_{2} \) is \( 2 \times 0.0041 = 0.0082 \text{ M}\). At equilibrium, \([\text{NO}_{2}] = 0.0082 \text{ M}\).
05

Write the Equilibrium Expression

For the reaction \(\text{N}_{2} \text{O}_{4}(\text{g}) \rightleftharpoons 2\text{NO}_{2}(\text{g})\), \(K_{c} = \frac{[\text{NO}_{2}]^2}{[\text{N}_{2} \text{O}_{4}]}\).
06

Plug in Equilibrium Concentrations

Substitute the equilibrium concentrations into the expression: \(K_{c} = \frac{(0.0082)^2}{0.00090}\) \(= \frac{0.00006724}{0.00090} \approx 0.075\).

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Key Concepts

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

Equilibrium Constant
Chemical reactions often reach a state known as equilibrium, where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time. The equilibrium constant, denoted as \( K_c \), is a numerical value that describes the ratio of the concentrations of products to reactants at equilibrium. It is specific to a particular reaction at a given temperature. For the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \), the equilibrium expression is given by:
  • \( K_{c} = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \)
This equation highlights how the equilibrium constant reflects the balance of concentrations; a large \( K_c \) indicates a higher concentration of products at equilibrium, while a small \( K_c \) suggests more reactants. Understanding \( K_c \) is crucial for predicting how a reaction behaves under different conditions.
Le Chatelier's Principle
Le Chatelier's Principle is a fundamental concept used to predict how a change in conditions can shift the position of equilibrium in a chemical reaction. According to this principle, if a system at equilibrium experiences a change in concentration, temperature, or pressure, the system adjusts to counteract the effect of the change and re-establish equilibrium.
  • If the concentration of a reactant or product is changed, the equilibrium will shift to oppose that change.
  • An increase in temperature usually favors the endothermic direction of a reaction.
  • Changes in pressure mainly affect reactions involving gases with differing numbers of moles on each side.
For example, increasing the amount of \( \text{N}_2\text{O}_4 \) in the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \) will shift the equilibrium position to the right, favoring the production of more \( \text{NO}_2 \).Le Chatelier's Principle helps chemists control reactions to obtain the desired yield.
Equilibrium Concentrations
Equilibrium concentrations are the amounts of reactants and products present in a reaction mixture when the reaction has reached equilibrium. These concentrations are key to calculating the equilibrium constant \( K_c \).In the exercise, we start by calculating the initial concentration of \( \text{N}_2\text{O}_4 \):
  • Initial concentration \([\text{N}_2\text{O}_4]_0 = \frac{0.010 \text{ mol}}{2.0 \text{ L}} = 0.0050 \text{ M} \)
As the reaction proceeds towards equilibrium, the concentration of \( \text{N}_2\text{O}_4 \) decreases, and that of \( \text{NO}_2 \) increases. At equilibrium, the concentrations are:
  • \([\text{N}_2\text{O}_4] = 0.00090\text{ M}\)
  • \([\text{NO}_2] = 0.0082\text{ M}\)
These values are then used in the equilibrium expression to calculate \( K_c \). Understanding how to determine equilibrium concentrations is crucial for analyzing the extent and direction of chemical reactions.
Reaction Quotient
The reaction quotient, denoted as \( Q \), is a measure of the relative amounts of products and reactants present in a reaction at any point in time. It is similar to the equilibrium constant, but \( Q \) is not necessarily at equilibrium. The expression for \( Q \) is identical to that for \( K_c \). For the reaction \( \text{N}_2\text{O}_4(\text{g}) \rightleftharpoons 2\text{NO}_2(\text{g}) \), it is written as:
  • \( Q = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]} \)
Comparison between \( Q \) and \( K_c \) determines the direction in which a reaction will proceed:
  • If \( Q = K_c \), the system is at equilibrium.
  • If \( Q < K_c \), the reaction will shift to the right, forming more products.
  • If \( Q > K_c \), the reaction will shift to the left, forming more reactants.
Understanding the reaction quotient is essential for predicting changes in concentration and managing chemical processes effectively.

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Most popular questions from this chapter

Considering both the enthalpy effect and the entropy effect for the Haber- Bosch process, explain why choosing the temperature at which to run this reaction is very important.

At \(503 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}\) for the dissociation of \(\mathrm{N}_{2} \mathrm{O}_{4}\) $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) $$ has the value 40.0 . (a) Calculate the fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) left undissociated when \(1.00 \mathrm{~mol}\) of this gas is heated to \(503 \mathrm{~K}\) in a \(10.0-\mathrm{L}\) sealed container. (b) If the volume is now reduced to \(2.0 \mathrm{~L},\) calculate the new fraction of \(\mathrm{N}_{2} \mathrm{O}_{4}\) that is undissociated. (c) Calculate all three equilibrium concentrations.

Heating a metal carbonate leads to decomposition. $$ \mathrm{BaCO}_{3}(\mathrm{~s}) \rightleftharpoons \mathrm{BaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) $$ Predict the effect on the equilibrium of each change listed below. Answer by choosing (i) no change, (ii) shifts left, or (iii) shifts right. (Except for part (e), assume that the volume is constant.) (a) Add \(\mathrm{BaCO}_{3}\) (b) Add \(\mathrm{CO}_{2}\) (c) Add \(\mathrm{BaO}\) (d) Raise the temperature (e) Increase the volume of the reaction flask

A small sample of \(c i s\) -dichloroethene in which one carbon atom is the radioactive isotope \({ }^{14} \mathrm{C}\) is added to an equilibrium mixture of the cis and trans isomers at a certain temperature. Eventually, \(40 \%\) of the radioactive molecules are found to be in the trans configuration at any given time. (a) Determine the value of \(K_{\mathrm{c}}\) for the cis \(\rightleftharpoons\) trans equilibrium. (b) What would have happened if a small sample of radioactive trans isomer had been added instead of the cis isomer?

Consider the equilibrium $$ \mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g}) $$ At \(745 \mathrm{~K}\) the equilibrium constant \(K_{\mathrm{c}}=50.0 .\) Suppose that \(0.75 \mathrm{~mol} \mathrm{HI}(\mathrm{g}), 0.025 \mathrm{~mol} \mathrm{H}_{2}(\mathrm{~g})\), and \(0.025 \mathrm{~mol}\) \(\mathrm{I}_{2}(\mathrm{~g})\) are placed into a sealed 20.0 - \(\mathrm{L}\) flask and heated to \(745 \mathrm{~K}\) (a) Is the system at equilibrium? (b) If not, in which direction must the reaction proceed to reach equilibrium? (c) Calculate the equilibrium concentrations of all three substances.

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