/*! 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 37 Consider the following reaction ... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following reaction at a certain temperature:$$4 \mathrm{Fe}(s)+3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s)$$.An equilibrium mixture contains 1.0 mole of \(\mathrm{Fe}, 1.0 \times 10^{-3}\) mole of \(\mathrm{O}_{2},\) and 2.0 moles of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) all in a \(2.0-\mathrm{L}\) container. Calculate the value of \(K\) for this reaction.

Short Answer

Expert verified
The value of the equilibrium constant (K) for the given reaction is \( 8.0 \times 10^{9} \).

Step by step solution

01

Write the balanced chemical equation

For this problem, we are given the balanced chemical equation: \[ 4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(s) \]
02

Calculate the concentrations of Fe, O₂, and Fe₂O₃

We know the amount of each substance and the volume of the container (2.0 L). The concentrations can be calculated as follows: - Fe(s): \(1.0 \,\text{mole} \); \(\frac{1.0 \,\text{mole}}{2.0 \,\text{L}} = 0.5 \,\text{M} \) - O₂(g): \(1.0 \times 10^{-3}\,\text{mole} \); \(\frac{1.0 \times 10^{-3}\,\text{mole}}{2.0 \,\text{L}} = 5.0 \times 10^{-4}\,\text{M} \) - Fe₂O₃(s): \(2.0\,\text{moles}\); \(\frac{2.0\,\text{moles}}{2.0\,\text{L}} = 1.0\,\text{M} \)
03

Apply the equilibrium constant formula

For the reaction equilibrium constant \(K_{c}\), the formula is: \[K_{c} = \frac{[\mathrm{Fe}_{2}\mathrm{O}_{3}]^2}{[\mathrm{Fe}]^4 [\mathrm{O}_{2}]^3}\] However, we don't include solids (Fe and Fe₂O₃) in the equilibrium expression. Hence, the formula becomes: \[K_{c} = \frac{1}{[\mathrm{O}_{2}]^3}\]
04

Substitute the values and calculate K

Now we substitute the concentration of Oâ‚‚ into the formula and calculate K: \[K_{c} = \frac{1}{(5.0 \times 10^{-4})^3} = \frac{1}{1.25 \times 10^{-10}} = 8.0 \times 10^{9}\] Thus, the value of the equilibrium constant (K) for this reaction is \(8.0 \times 10^{9}\).

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Key Concepts

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

Equilibrium Constant
The equilibrium constant, often represented as \( K \), is an important concept in chemical equilibrium. It reflects the ratio of the concentration of products to reactants when a reaction has reached equilibrium. In an equilibrium state, the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of the reactants and products remain constant over time.

To calculate the equilibrium constant for a reaction, use the formula:
  • \( K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b} \)
Here, \([A], [B], [C], [D]\) are the molar concentrations, and \(a, b, c, d\) are the coefficients from the balanced chemical equation. For reactions involving gases, the equilibrium constant can also be expressed in terms of partial pressures, known as \( K_p \). In the example reaction, only equilibrium concentrations of gaseous components should be used in the expression for \( K_c \). Solids and pure liquids are not included.
Balanced Chemical Equation
A balanced chemical equation is a fundamental aspect of understanding chemical reactions. It represents both the reactants and products in a reaction with their respective quantities indicated. Balancing ensures mass conservation by having the same number of each type of atom on both sides of the equation.

For the example reaction:
  • \( 4 \mathrm{Fe}(s) + 3 \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{Fe}_{2}\mathrm{O}_{3}(s) \)
The equation is balanced because both sides have equal numbers of each atom type.

  • 4 iron (Fe) atoms on the left equal 4 iron atoms on the right.
  • 3 oxygen (O) molecules, each having 2 oxygen atoms (total of 6 oxygen atoms), equal the 6 oxygen atoms in the products.
Balancing is crucial for determining the correct stoichiometry for the calculation of the equilibrium constant.
Reaction Concentration
Reaction concentration refers to the amount of a substance in a given volume of a solution or mixture, often expressed in molarity (M), which is moles per liter. To calculate the concentration, divide the number of moles of the substance by the volume in liters.

In the given example:
  • Iron: \( \frac{1.0 \, \text{mole}}{2.0 \, \text{L}} = 0.5 \, \text{M} \)
  • Oxygen: \( \frac{1.0 \times 10^{-3} \, \text{mole}}{2.0 \, \text{L}} = 5.0 \times 10^{-4} \, \text{M} \)
  • Ferric oxide: \( \frac{2.0 \, \text{moles}}{2.0 \, \text{L}} = 1.0 \, \text{M} \)
Understanding reaction concentration is essential for calculating the equilibrium constant and understanding how concentrations change as a reaction approaches equilibrium.
Equilibrium Mixture
An equilibrium mixture contains both reactants and products that exist in concentrations where the rates of the forward and reverse reactions are equal. At equilibrium, there is no net change in the concentration of reactants and products over time.

In the context of the example:
  • The equilibrium mixture includes 0.5 M of Fe, \(5.0 \times 10^{-4} \, \text{M}\) of \( \mathrm{O}_2 \), and 1.0 M of \( \mathrm{Fe}_{2} \mathrm{O}_{3} \) in a 2.0 L container.
The concentrations remain consistent within this mixture as long as conditions such as temperature and pressure remain constant.

Understanding the nature of equilibrium mixtures helps predict how changes in conditions can shift the equilibrium position, altering the concentrations of substances in the mixture.

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Most popular questions from this chapter

Suppose the reaction system $$\mathrm{UO}_{2}(s)+4 \mathrm{HF}(g) \rightleftharpoons \mathrm{UF}_{4}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$$,has already reached equilibrium. Predict the effect that each of the following changes will have on the equilibrium position. Tell whether the equilibrium will shift to the right, will shift to the left, or will not be affected. a. Additional UO \(_{2}(s)\) is added to the system. b. The reaction is performed in a glass reaction vessel; \(\mathrm{HF}(g)\) attacks and reacts with glass. c. Water vapor is removed.

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.

At a particular temperature, \(K_{\mathrm{p}}=0.25\) for the reaction $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$. a. A flask containing only \(\mathrm{N}_{2} \mathrm{O}_{4}\) at an initial pressure of 4.5 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. b. A flask containing only \(\mathrm{NO}_{2}\) at an initial pressure of 9.0 atm is allowed to reach equilibrium. Calculate the equilibrium partial pressures of the gases. c. From your answers to parts a and b, does it matter from which direction an equilibrium position is reached?

The hydrocarbon naphthalene was frequently used in mothballs until recently, when it was discovered that human inhalation of naphthalene vapors can lead to hemolytic anemia. Naphthalene is \(93.71 \%\) carbon by mass, and a 0.256 -mole sample of naphthalene has a mass of 32.8 g. What is the molecular formula of naphthalene? This compound works as a pesticide in mothballs by sublimation of the solid so that it fumigates enclosed spaces with its vapors according to the equation Naphthalene( \(s) \rightleftharpoons\) naphthalene \((g)$$$K=4.29 \times 10^{-6}(\text {at } 298 \mathrm{K})$$.If \)3.00 \mathrm{g}\( solid naphthalene is placed into an enclosed space with a volume of \)5.00 \mathrm{L}\( at \)25^{\circ} \mathrm{C},$ what percentage of the naphthalene will have sublimed once equilibrium has been established?

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

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