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Chapter 13: Problem 57

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$ 2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following original mixtures. a. \(2.0 \mathrm{~mol}\) pure \(\mathrm{NOCl}\) in a \(2.0\) - \(\mathrm{L}\) flask b. \(1.0 \mathrm{~mol} \mathrm{NOCl}\) and \(1.0 \mathrm{~mol} \mathrm{NO}\) in a \(1.0\) -L flask c. \(2.0 \mathrm{~mol} \mathrm{NOCl}\) and \(1.0 \mathrm{~mol} \mathrm{Cl}_{2}\) in a \(1.0\) -L flask

Short Answer

Expert verified
The equilibrium concentrations for each original mixture are: a. [NOCl] ≈ 0.987 M, [NO] ≈ 0.0126 M, and [Cl₂] ≈ 0.0063 M. b. [NOCl] ≈ 0.992 M, [NO] ≈ 1.008 M, and [Cl₂] ≈ 0.00397 M. c. [NOCl] ≈ 1.994 M, [NO] ≈ 0.0064 M, and [Cl₂] ≈ 1.0032 M.

Step by step solution

01

Set up an ICE table

Create an ICE table for the initial, change, and equilibrium concentrations of NOCl, NO, and Cl₂. Since we have 2.0 mol of NOCl in a 2.0 L flask, the initial concentration of NOCl would be \(\frac{2.0\ \text{mol}}{2.0\ \text{L}} = 1.0\ \text{M}\). Initially, we do not have any NO and Cl₂. |\( [NOCl] \)|\( [NO] \)|\( [Cl_2] \)| |-|-|-| |Initial|1.0 M|0 M|0 M| |Change|-2x|+2x|+x| |Equilibrium|1.0 M-2x|2x|x|
02

Write the expression for K

Write the expression for K and use the equilibrium concentrations. \( K = \frac{[NO]^2[Cl_2]}{[NOCl]^2} = \frac{(2x)^2(x)}{(1 - 2x)^2} \) We know K = 1.6 × 10⁻⁵.
03

Solve for x

Solve the equation for x. \( 1.6 \times 10^{-5} = \frac{(2x)^2(x)}{(1 - 2x)^2} \) After solving for x, we get \(x \approx 6.3 \times 10^{-3}\).
04

Calculate equilibrium concentrations

Use the calculated x value to find the equilibrium concentrations. \[ [NOCl] = 1.0 - 2x \approx 0.987\ \text{M} \] \[ [NO] = 2x \approx 0.0126\ \text{M} \] \[ [Cl_2] = x \approx 0.0063\ \text{M} \] So, at equilibrium, the concentrations are approximately [NOCl] = 0.987 M, [NO] = 0.0126 M, and [Cl₂] = 0.0063 M. b. 1.0 mol NOCl and 1.0 mol NO in a 1.0 L flask Follow the same process as described above for this part: Initial concentrations: [NOCl] = 1.0 M, [NO] = 1.0 M, [Cl₂] = 0 M ICE table: |\( [NOCl] \)|\( [NO] \)|\( [Cl_2] \)| |-|-|-| |Initial|1.0 M|1.0 M|0 M| |Change|-2x|+2x|+x| |Equilibrium|1.0 M-2x|1.0 M+2x|x| Expression for K, equation, and x: \( 1.6 \times 10^{-5} = \frac{(1 + 2x)^2(x)}{(1 - 2x)^2} \) Solving this equation, we get \(x \approx 3.97 \times 10^{-3}\). Equilibrium concentrations: [NOCl] = 0.992 M, [NO] = 1.008 M and [Cl₂] = 0.00397 M. c. 2.0 mol NOCl and 1.0 mol Cl₂ in a 1.0 L flask Follow the same process: Initial concentrations: [NOCl] = 2.0 M, [NO] = 0 M, [Cl₂] = 1.0 M ICE table: |\( [NOCl] \)|\( [NO] \)|\( [Cl_2] \)| |-|-|-| |Initial|2.0 M|0 M|1.0 M| |Change|-2x|+2x|+x| |Equilibrium|2.0 M-2x|2x|1.0 M+x| Expression for K, equation, and x: \( 1.6 \times 10^{-5} = \frac{(2x)^2(1 + x)}{(2 - 2x)^2} \) Solving this equation, we get \(x \approx 3.2 \times 10^{-3}\). Equilibrium concentrations: [NOCl] = 1.994 M, [NO] = 0.0064 M and [Cl₂] = 1.0032 M.

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Key Concepts

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

ICE table
The ICE table, an acronym for Initial, Change, and Equilibrium, is an essential tool for solving chemical equilibrium problems. It helps organize data and visualize how concentrations of reactants and products shift from their initial values to their equilibrium state. To start, you list the initial concentrations of all species involved in the reaction. Then, you denote the changes that occur as the system moves towards equilibrium using variables, typically represented by 'x'. Lastly, you add these changes to the initial values to find the equilibrium concentrations.

For example, if the initial concentration of a reactant is 1.0 M and it decreases by '2x' to reach equilibrium, the equilibrium concentration is expressed as '1.0 M - 2x'. The '2' in '2x' correlates with the stoichiometry of the reaction, indicating that two moles of this reactant are involved. The ICE table is crucial for solving equilibrium problems as it systematically provides a clear view of the changes during the reaction.
Equilibrium constant (K)
The equilibrium constant, designated as 'K', quantifies the ratio of concentrations of products to reactants at chemical equilibrium, raised to the power of their respective coefficients in the balanced equation. Its value is determined by the reaction at a specific temperature and reflects the extent to which a reaction proceeds before reaching equilibrium.

For the reaction where reactants A and B form products C and D in the form \(aA + bB \rightleftharpoons cC + dD\), the equilibrium constant expression is \(K = \frac{[C]^c[D]^d}{[A]^a[B]^b}\). The square brackets denote the molar concentrations of each species. It is crucial to understand that 'K' is a constant at a given temperature and only changes with temperature variations. If the value of 'K' is much less than 1, it indicates that the reactants are favored at equilibrium, whereas a 'K' much larger than 1 signifies that the products are favored.
Equilibrium concentrations
Equilibrium concentrations refer to the molar amounts of reactants and products in a reaction mixture when the rate of the forward reaction equals the rate of the reverse reaction, and no further changes in concentrations occur. These concentrations can be calculated using an ICE table and the K expression once the value of 'x' has been determined.

Returning to the initially set up ICE table, after solving for 'x' through the equilibrium constant expression, you substitute 'x' back into the expressions for equilibrium concentrations in the ICE table. This provides the final concentrations of all species present in the mixture at equilibrium. The result reflects the dynamic balance between the forward and reverse reactions, where the chemical system has reached a state of minimum free energy.
Stoichiometry
Stoichiometry is the study of the quantitative relationships, or ratios, between reactants and products in chemical reactions based on the balanced chemical equations. In an equilibrium calculation, stoichiometry provides the coefficients that indicate the proportional changes in concentration for each reactant and product as the system moves towards equilibrium.

For instance, in the reaction \(2NOCl(g) \rightleftharpoons 2NO(g) + Cl_2(g)\), the stoichiometry tells us that two moles of NOCl decompose to form two moles of NO and one mole of Cl2 at equilibrium. This stoichiometric ratio is critical in setting up the 'Change' row of the ICE table, where for every '2x' moles of NOCl that react, '2x' moles of NO and 'x' moles of Cl2 are produced. Stoichiometry, therefore, directly influences the ICE table and the subsequent calculations for equilibrium concentrations.

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Most popular questions from this chapter

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.

For the following reaction at a certain temperature $$ \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{HF}(g) $$ it is found that the equilibrium concentrations in a 5.00-L rigid container are \(\left[\mathrm{H}_{2}\right]=0.0500 M,\left[\mathrm{~F}_{2}\right]=0.0100 M\), and \([\mathrm{HF}]=\) \(0.400 \mathrm{M}\). If \(0.200 \mathrm{~mol} \mathrm{~F}_{2}\) is added to this equilibrium mixture, calculate the concentrations of all gases once equilibrium is reestablished.

Given \(K=3.50\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g) $$ and \(K=7.10\) at \(45^{\circ} \mathrm{C}\) for the reaction $$ 2 \mathrm{~A}(g)+\mathrm{D}(g) \rightleftharpoons \mathrm{C}(g) $$ what is the value of \(K\) at the same temperature for the reaction $$ \mathrm{C}(g)+\mathrm{D}(g) \rightleftharpoons 2 \mathrm{~B}(g) $$ What is the value of \(K_{\mathrm{p}}\) at \(45^{\circ} \mathrm{C}\) for the reaction? Starting with \(1.50\) atm partial pressures of both \(\mathrm{C}\) and \(\mathrm{D}\), what is the mole fraction of \(\mathrm{B}\) once equilibrium is reached?

The compound \(\mathrm{VCl}_{4}\) undergoes dimerization in solution: $$ 2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8} $$ When \(6.6834 \mathrm{~g} \mathrm{VCl}_{4}\) is dissolved in \(100.0 \mathrm{~g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\). Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(1.696\) \(\mathrm{g} / \mathrm{cm}^{3}\), and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} .\).

A sample of \(\mathrm{S}_{8}(g)\) is placed in an otherwise empty rigid container at \(1325 \mathrm{~K}\) at an initial pressure of \(1.00 \mathrm{~atm}\), where it decomposes to \(\mathrm{S}_{2}(g)\) by the reaction $$ \mathrm{S}_{8}(g) \rightleftharpoons 4 \mathrm{~S}_{2}(g) $$ At equilibrium, the partial pressure of \(S_{8}\) is \(0.25\) atm. Calculate \(K_{\mathrm{p}}\) for this reaction at \(1325 \mathrm{~K}\).
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