/*! 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 72 Given the following equilibrium ... [FREE SOLUTION] | 91Ó°ÊÓ

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Given the following equilibrium constants at \(427^{\circ} \mathrm{C}\) $$\begin{array}{ll}\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{1}=2 \times 10^{-25} \\\\\mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}(l)+\frac{1}{2} \mathrm{O}_{2}(g) & K_{2}=2 \times 10^{-5} \\\\\mathrm{Na}_{2} \mathrm{O}_{2}(s) \rightleftharpoons 2 \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{3}=5 \times 10^{-29} \\\\\mathrm{NaO}_{2}(s) \rightleftharpoons \mathrm{Na}(l)+\mathrm{O}_{2}(g) & K_{4}=3 \times 10^{-14}\end{array}$$,determine the values for the equilibrium constants for the following reactions: a. \(\mathrm{Na}_{2} \mathrm{O}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) b. \(\mathrm{NaO}(g)+\mathrm{Na}_{2} \mathrm{O}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)+\mathrm{Na}(l)\) c. \(2 \mathrm{NaO}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{O}_{2}(s)\) (Hint: When reaction equations are added, the equilibrium expressions are multiplied.)

Short Answer

Expert verified
The equilibrium constants for the new reactions are: a. \(Ka = 2.5 \times 10^{-4}\) b. \(Kb = 4 \times 10^{-30}\) c. \(Kc = 1.6 \times 10^{-9}\)

Step by step solution

01

Identify the given reactions and their equilibrium constants

We are given the following reactions with their respective equilibrium constants: 1. Na2O(s) <=> 2Na(l) + 0.5O2(g) with K1 = 2 x 10^-25 2. NaO(g) <=> Na(l) + 0.5O2(g) with K2 = 2 x 10^-5 3. Na2O2(s) <=> 2Na(l) + O2(g) with K3 = 5 x 10^-29 4. NaO2(s) <=> Na(l) + O2(g) with K4 = 3 x 10^-14
02

Manipulate the given reactions to obtain the desired reactions

We need to obtain the new reactions: a. Na2O(s) + 0.5O2(g) <=> Na2O2(s) b. NaO(g) + Na2O(s) <=> Na2O2(s) + Na(l) c. 2NaO(g) <=> Na2O2(s) Now, let's manipulate the given reactions: For (a), subtract Reaction 1 from Reaction 3: Na2O2(s) + 2Na(l) + O2(g) - (Na2O(s) + 2Na(l) + 0.5O2(g)) = Na2O(s) + 0.5O2(g) <=> Na2O2(s) For (b) add Reaction 1 and Reaction 2: Na2O(s) + 2Na(l) + 0.5O2(g) + NaO(g) <=> Na(l) + 0.5O2(g) + Na2O2(s) + Na(l) For (c), double Reaction 2 and subtract Reaction 3: (2)(NaO(g) <=> Na(l) + 0.5O2(g)) - (Na2O2(s) <=> 2Na(l) + O2(g)) = 2NaO(g) <=> Na2O2(s)
03

Apply rules for combining equilibrium constants

For (a), when we subtract equilibrium expressions (Reaction 3 - Reaction 1), we divide the equilibrium constants (K3 / K1): Ka = K3 / K1 = (5 x 10^-29) / (2 x 10^-25) = 2.5 x 10^-4 For (b), when we add equilibrium expressions (Reaction 1 + Reaction 2), we multiply the equilibrium constants (K1 * K2): Kb = K1 * K2 = (2 x 10^-25) * (2 x 10^-5) = 4 x 10^-30 For (c), when we double Reaction 2 and subtract Reaction 3, first we square the equilibrium constant of Reaction 2 (K2^2) and then divide it by the equilibrium constant of Reaction 3 (K2^2 / K3): Kc = (K2^2) / K3 = (2 x 10^-5)^2 / (5 x 10^-29) = 1.6 x 10^-9 So, the equilibrium constants for the new reactions are: a. Ka = 2.5 x 10^-4 b. Kb = 4 x 10^-30 c. Kc = 1.6 x 10^-9

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Key Concepts

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

Chemical Equilibrium
Chemical equilibrium is a fundamental concept in chemistry that refers to the state in a chemical reaction when the concentrations of reactants and products remain constant over time. This does not mean that the reactants and products are equal in concentration but that their rates of formation and consumption are balanced. For any reversible reaction, equilibrium can be achieved given enough time.
Reactions reach equilibrium when the forward and reverse reaction rates are equal. At this point, the concentration of reactants and products remains unchanged, leading to a stable reaction environment. It's important to note that reaching equilibrium does not mean the reaction has stopped; it is constantly proceeding in both directions at an equal, steady rate.
Chemical equilibrium can be influenced by several factors, including changes in concentration, pressure, temperature, and the presence of catalysts. Understanding equilibrium helps chemists predict the behavior of a system and manipulate it to achieve desired outcomes in various chemical processes.
Reaction Manipulation
Manipulating reactions involves adjusting components of chemical equations to obtain different reactions with new equilibrium conditions. This is particularly useful when combining known reactions to derive new chemical equations and the associated equilibrium constants. By carefully applying rules of thermodynamics and mathematical operations, chemists can transform the equilibrium expressions and constants to match new reaction setups.
In the context of the given exercise, reaction manipulation is achieved by adding, subtracting, or combining existing reactions to form new desired equations. The manipulation follows laws such as Hess's Law, which states that the total enthalpy change during a complete process is the same regardless of the pathway by which the chemical change occurs.
When reaction equations are manipulated:
  • Addition of reactions results in the multiplication of their equilibrium constants.
  • Subtraction of reactions involves division of their equilibrium constants.
  • Multiplying a reaction by a coefficient raises the equilibrium constant to the power of the coefficient.
This understanding allows chemists to predict the outcome and equilibrium status of reactions that have been experimentally or theoretically derived from simpler known reactions.
Equilibrium Expression
An equilibrium expression is a mathematical representation of the relationship between the concentrations of reactants and products in a chemical equilibrium state. It is crucial to write this expression accurately to understand how changes in conditions affect the position of equilibrium.
For a general reversible reaction:
\[ aA + bB \rightleftharpoons cC + dD \]
The equilibrium expression, also known as the equilibrium constant (\( K \)), is given by:
\[ K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \]
Where brackets denote the concentration of chemical species, and the exponents correspond to the stoichiometric coefficients from the balanced chemical equation.
The value of the equilibrium constant, \( K \), gives insight into the favored direction of the reaction:
  • If \( K \) is much greater than 1, products are favored at equilibrium.
  • If \( K \) is much less than 1, reactants are favored at equilibrium.
By carefully interpreting and using equilibrium expressions, chemists can describe and predict the outcomes of complex reaction systems, thus providing valuable insights in fields like drug development, environmental science, and industrial chemistry.

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Most popular questions from this chapter

Calculate a value for the equilibrium constant for the reaction $$\mathbf{O}_{2}(g)+\mathbf{O}(g) \rightleftharpoons \mathbf{O}_{3}(g)$$.given $$\begin{aligned}& \mathrm{NO}_{2}(g) \stackrel{h v}{\rightleftharpoons} \mathrm{NO}(g)+\mathrm{O}(g) & & K=6.8 \times 10^{-49} \\\\\mathrm{O}_{3}(g)+\mathrm{NO}(g) & \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & & K=5.8 \times 10^{-34}\end{aligned}$$.(Hint: When reactions are added together, the equilibrium expressions are multiplied.) (Hint: When reactions are added together, the equilibrium expressions are multiplied.)

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11}\) If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?

Le Châtelier's principle is stated (Section \(12-7\) ) as follows: "If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change." The system \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) is used as an example in which the addition of nitrogen gas at equilibrium results in a decrease in \(\mathrm{H}_{2}\) concentration and an increase in \(\mathrm{NH}_{3}\) concentration. In the experiment the volume is assumed to be constant. On the other hand, if \(\mathrm{N}_{2}\) is added to the reaction system in a container with a piston so that the pressure can be held constant, the amount of \(\mathrm{NH}_{3}\) actually could decrease, and the concentration of \(\mathrm{H}_{2}\) would increase as equilibrium is reestablished. Explain how this can happen. Also, if you consider this same system at equilibrium, the addition of an inert gas, holding the pressure constant, does affect the equilibrium position. Explain why the addition of an inert gas to this system in a rigid container does not affect the equilibrium position.

The reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ has \(K_{\mathrm{p}}=109\) at \(25^{\circ} \mathrm{C}\). If the equilibrium partial pressure of \(\mathrm{Br}_{2}\) is 0.0159 atm and the equilibrium partial pressure of NOBr is 0.0768 atm, calculate the partial pressure of NO at equilibrium.

Consider the following exothermic reaction at equilibrium: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$. Predict how the following changes affect the number of moles of each component of the system after equilibrium is reestablished by completing the table below. Complete the table with the terms increase, decrease, or no change.

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