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At \(1800 \mathrm{K},\) oxygen dissociates very slightly into its atoms. $$ \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}(\mathrm{g}) \quad K_{\mathrm{p}}=1.2 \times 10^{-10} $$ If you place 1.0 mol of \(\mathrm{O}_{2}\) in a \(10 .\).L. vessel and heat it to \(1800 \mathrm{K}\), how many \(\mathrm{O}\) atoms are present in the flask?

Short Answer

Expert verified
Approximately \(5.43 \times 10^{19}\) \(\mathrm{O}\) atoms are present in the flask.

Step by step solution

01

Set Up Initial Conditions

Start with determining the initial conditions for the gas in the flask. Given that 1.0 mol of \(\mathrm{O}_2\) is initially present in a \(10.0\, L\) vessel, calculate the initial pressure \(P_0\) using the ideal gas law: \(P_0 = \frac{nRT}{V}\). At 1800 K, assuming \(R = 0.0821 \, \text{Latm/molK}\), the initial pressure is \(P_0 = \frac{1.0 \times 0.0821 \times 1800}{10.0} = 14.778 \, \text{atm}\).
02

Express Equilibrium Pressures

Define \(x\) as the change in pressure due to the dissociation of \(\mathrm{O}_2\). At equilibrium, the pressure of \(\mathrm{O}_2\) becomes \((14.778 - x)\) atm and for \(\mathrm{O}\), the pressure is \(2x\) atm.
03

Apply Equilibrium Expression

Use the equilibrium constant \(K_{\mathrm{p}}\) to relate the pressures: \(K_{\mathrm{p}} = \frac{(2x)^2}{14.778 - x}\). Given \(K_{\mathrm{p}} = 1.2 \times 10^{-10}\), set up the equation: \(1.2 \times 10^{-10} = \frac{4x^2}{14.778 - x}\).
04

Solve for x

Assume \(x\) is small because \(K_p\) is very small, simplifying the equation to: \(1.2 \times 10^{-10} = \frac{4x^2}{14.778}\). Solving gives: \(x^2 = \frac{1.2 \times 10^{-10} \times 14.778}{4}\), leading to \(x = \sqrt{\frac{1.7736 \times 10^{-9}}{4}}\). Calculate \(x = 6.657 \times 10^{-5}\).
05

Calculate Number of O Atoms

Find the partial pressure of \(\mathrm{O}\) atoms, which is \(2x = 1.3314 \times 10^{-4}\) atm. Convert this pressure to moles using \(n = \frac{PV}{RT}\), where \(P = 1.3314 \times 10^{-4}\, \text{atm}\), \(V = 10.0 \, L\), \(R = 0.0821 \, \text{Latm/molK}\), and \(T = 1800 \, K\). This calculation gives \(n \approx 9.02 \times 10^{-5}\, \text{mol}\). Since each mole contains \(6.022 \times 10^{23}\) atoms (Avogadro's number), the number of \(\mathrm{O}\) atoms is \(9.02 \times 10^{-5} \times 6.022 \times 10^{23} \approx 5.43 \times 10^{19}\) atoms.

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Key Concepts

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

Equilibrium Constant
In chemical reactions that reach an equilibrium, the reactants and products maintain a constant concentration ratio at a given temperature. This ratio is expressed as the equilibrium constant, denoted as either \( K_c \) for concentrations or \( K_p \) for partial pressures.
For gas-phase reactions, such as the dissociation of oxygen into its atoms, \( K_p \) is typically used. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the expression of \( K_p \) is given by: \[ K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b} \] where \( P \) represents the partial pressures of the gaseous components.
  • A high \( K_p \) value indicates a favorable reaction towards product formation.
  • A low \( K_p \) value, as is in the case of oxygen dissociation, suggests that very few products form at equilibrium.
Understanding \( K_p \) helps predict the extent of a reaction under set conditions.
Dissociation Reaction
Dissociation reactions involve the breaking of complex molecules into simpler entities. In the context of gases, it usually refers to a molecule breaking into its atoms or simpler molecules.
For instance, diatomic oxygen \( \mathrm{O}_2 \) dissociates into individual oxygen atoms: \[ \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{O}(g) \] In the given problem, we observe this dissociation at a high temperature of 1800 K. Factors influencing dissociation include:
  • Temperature: Higher temperatures typically increase the extent of dissociation.
  • Pressure: Affects reaction shifts according to Le Chatelier’s principle.
  • Equilibrium Constants: Indicate dissociation’s extent and favorability.
Comprehending these aspects is key in predicting and controlling dissociative processes.
Ideal Gas Law
The Ideal Gas Law establishes a relationship between the pressure, volume, temperature, and amount of gas present.It is expressed in the well-known formula: \[ PV = nRT \] Here, \( P \) represents pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
In the oxygen dissociation problem, this law is crucial to determine the initial conditions and later convert pressures into moles. Key points regarding ideal gases include:
  • They follow the law precisely only under certain conditions (high temperature, low pressure).
  • Real gases closely adhere to the law under these typical conditions.
  • The equation allows for converting experimental or theoretical data easily from one set of parameters to another.
Mastery of the Ideal Gas Law serves as a foundational pillar for solving problems involving gaseous chemical reactions.
Partial Pressure
Partial pressure refers to the pressure a single gas component in a mixture exerts as if it occupied the entire volume solely. It’s an integral concept in reactions involving gases, helping to evaluate how individual gases influence the overall system. Dalton's Law of Partial Pressures provides that: \[ P_{\text{total}} = P_1 + P_2 + ... + P_n \] For gases ethically behaving as ideal gases, the partial pressure can be calculated by: \[ P_i = \frac{n_iRT}{V} \] In the oxygen dissociation problem, each component's partial pressure defines reaction status and equilibrium calculations.
  • Total pressure is the sum of all partial pressures in a system.
  • Used in calculating equilibrium constants when dealing with reactions involving multiple gaseous species.
  • Essential for defining individual gas roles in mixed-gas conditions.
Knowing how to determine and apply partial pressures is necessary for evaluating gas reactions and system equilibria.

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Most popular questions from this chapter

Which of the following correctly relates the two equilibrium constants for the two reactions shown? \(\mathrm{NOCl}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{Cl}_{2}(\mathrm{g}) \quad K_{1}\) \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{NOCl}(\mathrm{g}) \quad K_{2}\) (a) \(K_{2}=-K_{1}^{2}\) (c) \(K_{2}=1 / K_{1}^{2}\) (b) \(K_{2}=1 /\left(K_{1}\right)^{1 / 2} \quad\) (d) \(K_{2}=2 K_{1}\)

The equilibrium constant, \(K_{p},\) is 0.14 at \(25^{\circ} \mathrm{C}\) for the following reaction: $$ \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g}) $$ If the total pressure of the gas mixture is 2.5 atm at equilibrium, what is the partial pressure of each gas?

The photograph below (a) shows what occurs when a solution of iron(III) nitrate is treated with a few drops of aqueous potassium thiocyanate. The nearly colorless iron(III) ion is converted to a red [Fe(H_O)sSCN] \(^{2+}\) ion. (This is a classic test for the presence of iron(III) ions in solution.) \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}(\mathrm{aq})+\mathrm{SCN}^{-}(\mathrm{aq}) \rightleftharpoons\) $$ \left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{SCN}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ (a) As more \(\mathrm{KSCN}\) is added to the solution, the color becomes even more red. Explain this observation. (b) Silver ions form a white precipitate with SCN " ions. What would you observe on adding a few drops of aqueous silver nitrate to a red solution of \(\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5} \mathrm{SCN}\right]^{+}\) ions? Explain your observation.

The reaction $$ \mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{CO}(\mathrm{g}) $$ occurs at high temperatures. At \(700^{\circ} \mathrm{C},\) a \(2.0-\mathrm{L}\) flask contains 0.10 mol of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2}\) and 0.40 mol of \(\mathrm{C}\) at equilibrium. (a) Calculate \(K_{c}\) for the reaction at \(700^{\circ} \mathrm{C}\) (b) Calculate \(K_{c}\) for the reaction, also at \(700^{\circ} \mathrm{C},\) if the amounts at equilibrium in the 2.0 -I. flask are 0.10 mol of \(\mathrm{CO}, 0.20 \mathrm{mol}\) of \(\mathrm{CO}_{2},\) and \(0.80 \mathrm{mol}\) of C. (c) Compare the results of (a) and (b). Does the quantity of carbon affect the value of \(K_{c} ?\) Explain.

A mixture of \(\mathrm{SO}_{2}, \mathrm{O}_{2},\) and \(\mathrm{SO}_{3}\) at \(1000 \mathrm{K}\) contains the gases at the following concentrations: \(\left[\mathrm{SO}_{2}\right]=\) $$ 5.0 \times 10^{-3} \mathrm{mol} / \mathrm{L},\left[\mathrm{O}_{2}\right]=1.9 \times 10^{-3} \mathrm{mol} / \mathrm{L}, \text { and } $$ \(\left[\mathrm{SO}_{3}\right]=6.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Is the reaction at equilib- rium? If not, which way will the reaction proceed to reach equilibrium? $$ 2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{SO}_{3}(\mathrm{g}) K_{\mathrm{c}}=279 $$

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