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An equilibrium mixture of dinitrogen tetroxide, \(\mathrm{N}_{2} \mathrm{O}_{4}\), and nitrogen dioxide, \(\mathrm{NO}_{2}\), is \(65.8 \% \mathrm{NO}_{2}\) by mass at \(1.00 \mathrm{~atm}\) pressure and \(25^{\circ} \mathrm{C}\). Calculate \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) $$

Short Answer

Expert verified
The equilibrium constant \( K_c \) is 0.87.

Step by step solution

01

Write the equilibrium expression for the reaction

For the reaction \( \mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g) \), the equilibrium constant expression in terms of concentrations is given by:\[K_c = \frac{[\mathrm{NO}_2]^2}{[\mathrm{N}_2\mathrm{O}_4]}\]where \([\mathrm{NO}_2]\) and \([\mathrm{N}_2\mathrm{O}_4]\) are the molar concentrations of nitrogen dioxide and dinitrogen tetroxide respectively at equilibrium.
02

Calculate total mass of the mixture

Assume a 100 g sample of the mixture. Since it is 65.8% NO2 by mass, the mass of NO2 is 65.8 g and the mass of N2O4 is 34.2 g. Use these masses to find the moles of each gas.
03

Determine moles of each component

The molar mass of \( \mathrm{NO}_2 \) is 46 g/mol and that of \( \mathrm{N}_2\mathrm{O}_4 \) is 92 g/mol. Calculate the moles of each as follows:- Moles of \( \mathrm{NO}_2 = \frac{65.8 \text{ g}}{46 \text{ g/mol}} = 1.43 \text{ moles}\)- Moles of \( \mathrm{N}_2\mathrm{O}_4 = \frac{34.2 \text{ g}}{92 \text{ g/mol}} = 0.37 \text{ moles}\)

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Key Concepts

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

Equilibrium Constant
In chemistry, the Equilibrium Constant, denoted as \( K_c \), is a crucial concept when dealing with reversible chemical reactions. It gives us a numerical value that expresses the ratio of concentrations of products to reactants at equilibrium.
For the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), the equilibrium constant expression is:
  • \[K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}\]
This expression helps us understand the balance of products and reactants at equilibrium.
The power to which each concentration is raised is determined by the stoichiometry of the balanced equation.
This means for every 1 mole of \( \text{N}_2\text{O}_4 \), 2 moles of \( \text{NO}_2 \) are produced, hence \([\text{NO}_2]^2\).
Calculating \( K_c \) allows chemists to predict how far a reaction will proceed or to determine the conditions required to achieve the desired concentrations of substances.
It's important to remember that \( K_c \) is specific to a particular temperature, as changes in temperature can shift the position of equilibrium and alter the constant.
Le Chatelier's Principle
Le Chatelier’s Principle is a fundamental concept when studying chemical equilibrium.
It states that if an external change is applied to a system at equilibrium, the system adjusts itself to minimize that change and re-achieve equilibrium.
This can include changes in concentration, temperature, or pressure.For example, if the concentration of \( \text{NO}_2 \) is increased in the reaction \( \text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g) \), according to Le Chatelier’s Principle, the system will shift towards forming more \( \text{N}_2\text{O}_4 \) to reduce the increased \( \text{NO}_2 \) concentration.
This shift continues until equilibrium is restored.Le Chatelier's Principle is extremely useful for predicting the effects of a change in conditions on a chemical system.
It allows chemists to optimize reactions to maximize yield or control the production of specific products by appropriately adjusting the environment in which the reaction occurs.
Reaction Quotient
The Reaction Quotient, represented as \( Q_c \), is similar to the equilibrium constant but applies to non-equilibrium conditions.
It helps predict the direction in which a reaction will proceed to reach equilibrium.This is calculated using the same expression as \( K_c \):
  • \[Q_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}\]
When you calculate \( Q_c \) and compare it to \( K_c \):
  • If \( Q_c = K_c \), the system is at equilibrium.
  • If \( Q_c > K_c \), the reaction will proceed in the reverse direction, forming reactants.
  • If \( Q_c < K_c \), the reaction will proceed forward, forming more products.
By determining \( Q_c \), it is possible to understand current conditions versus equilibrium and predict the necessary changes for equilibrium to be reached.
This makes it an invaluable tool in the manipulation and study of chemical reactions.

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Most popular questions from this chapter

Suppose you start with a mixture of \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(4.00 \mathrm{~mol} \mathrm{H}_{2}\) in a 10.00-L vessel. Find the moles of substances present at equilibrium at \(1200 \mathrm{~K}\) for the reaction $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) ; K_{c}=3.92 $$ You will get an equation of the form $$ f(x)=3.92 $$ where \(f(x)\) is an expression in the unknown \(x\) (the amount of \(\mathrm{CH}_{4}\) ). Solve this equation by guessing values of \(x\), then computing values of \(f(x)\). Find values of \(x\) such that the values of \(f(x)\) bracket \(3.92 .\) Then choose values of \(x\) to get a smaller bracket around 3.92. Obtain \(x\) to two significant figures.

A 4.00-L vessel contained \(0.0148\) mol of phosphorus trichloride, \(0.0126 \mathrm{~mol}\) of phosphorus pentachloride, and \(0.0870 \mathrm{~mol}\) of chlorine at \(230^{\circ} \mathrm{C}\) in an equilibrium mixture. Calculate the value of \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$

Phosgene, \(\mathrm{COCl}_{2}\), is a toxic gas used in the manufacture of urethane plastics. The gas dissociates at high temperature. $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ At \(400^{\circ} \mathrm{C}\), the equilibrium constant \(K_{c}\) is \(8.05 \times 10^{-4}\). Find the percentage of phosgene that dissociates at this temperature when \(1.00\) mol of phosgene is placed in a \(25.0\) - \(\mathrm{L}\) vessel.

Write the expression for the equilibrium constant \(K_{c}\) for each of the following equations: a. \(\mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)\) b. \(\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\) c. \(\mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{CO}_{2}(g)\) d. \(\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-}(a q)\)

Phosphorus(V) chloride, \(\mathrm{PCl}_{5}\), dissociates on heating to give phosphorus(III) chloride, \(\mathrm{PCl}_{3}\), and chlorine. $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ A closed \(2.00-\mathrm{L}\) vessel initially contains \(0.0100 \mathrm{~mol} \mathrm{PCl}_{5}\). What is the total pressure at \(250^{\circ} \mathrm{C}\) when equilibrium is achieved? The value of \(K_{c}\) at \(250^{\circ} \mathrm{C}\) is \(4.15 \times 10^{-2}\).

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