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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 moles of pure NOCl in a 2.0 -L flask b. 1.0 mole of NOCI and 1.0 mole of \(\mathrm{NO}\) in a 1.0 -L flask c. 2.0 moles of NOCl and 1.0 mole of \(\mathrm{Cl}_{2}\) in a 1.0 -L flask

Short Answer

Expert verified
a. [NOCl] ≈ 0.998 M, [NO] ≈ 0.002 M, [Cl₂] ≈ 0.001 M b. [NOCl] ≈ 0.0078 M, [NO] ≈ 0.992 M, [Cl₂] ≈ 0.9939 M c. [NOCl] ≈ 1.0025 M, [NO] ≈ 0.9976 M, [Cl₂] ≈ 0.0013 M

Step by step solution

01

1. Find initial concentrations

: For this scenario, the initial concentrations are: [NOCl]=2.0 mol / 2.0 L = 1.0 M, [NO]=[Cl₂]=0.
02

2. Set up the ICE table

: An ICE table helps us keep track of the changes in concentrations as the reaction proceeds towards equilibrium. In this case, we let "x" be the change in concentration of NOCl. NOCl → NO + 1/2 Cl₂ Initial 1.0 0 0 Change -2x +2x +x Equilibrium 1.0-2x 2x x
03

3. Write the expression for the equilibrium constant

: The equilibrium constant for this reaction is given as K = \(\frac{[NO]^2[Cl2]}{[NOCl]^2}\).
04

4. Substitute the equilibrium concentrations into the equilibrium expression

: Substitute the equilibrium concentrations from the ICE table into the equilibrium expression: K = \(\frac{(2x)^2(x)}{(1-2x)^2}\)
05

5. Input the given value of K and solve for x

: The given value of K is 1.6 x 10⁻⁵. We substitute this value into the expression and solve for x: 1.6 x 10⁻⁵ = \(\frac{4x^3}{(1-2x)^2}\) Solving this equation, we find that x ≈ 0.001.
06

6. Calculate the equilibrium concentrations

: Now that we have the value of x, we can determine the equilibrium concentrations: [NOCl] = 1.0 - 2x ≈ 0.998 M [NO] = 2x ≈ 0.002 M [Cl₂] = x ≈ 0.001 M #Scenario b# a. 1.0 mole of NOCI and 1.0 mole of NO in a 1.0 -L flask Repeat the steps 1-6 using the initial conditions given for the second scenario to obtain the equilibrium concentrations. #Scenario c# a. 2.0 moles of NOCl and 1.0 mole of Cl₂ in a 1.0 -L flask Repeat the steps 1-6 using the initial conditions given for the third scenario to obtain the equilibrium concentrations.

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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, denoted as K, is a number reflecting the ratio of the concentrations of products to reactants in a reversible chemical reaction at equilibrium. Each substance's concentration is raised to the power of its coefficient in the balanced chemical equation. For the reaction given in the exercise, where dinitrogen monoxide chloride (NOCl) decomposes into nitrogen monoxide (NO) and chlorine gas (Cl₂), the equilibrium constant can be expressed as:

\[\begin{equation} K = \frac{[NO]^2[Cl_2]}{[NOCl]^2} \end{equation}\] where [NO], [Cl₂], and [NOCl] indicate the molar concentrations of these gases at equilibrium.

In simple terms, if K is greater than 1, the reaction favors the formation of products, while a K less than 1 suggests that the reactants are favored at equilibrium. When K is approximately 1, neither the reactants nor the products are favored, indicating a balance between the two. Understanding the magnitude of the equilibrium constant is key to predicting the position of equilibrium and thus the concentrations of reactants and products at equilibrium.
ICE Table Method
The ICE table method stands for Initial concentration, Change in concentration, and Equilibrium concentration. It's a systematic approach to solve chemical equilibrium problems. This method is extremely useful for tracking how the concentrations of reactants and products change over the course of a reaction and for calculating the final concentrations at equilibrium. The exercise provided shows a perfect application of the ICE method, wherein:
  • 'Initial' refers to the starting concentrations of reactants and products, before the reaction has reached equilibrium.
  • 'Change' represents the amount by which these concentrations change as the reactants convert into products. A negative value indicates a decrease in reactant concentration while a positive value shows an increase in product concentration.
  • 'Equilibrium' indicates the concentrations after the reaction has reached a state of balance and no further change is observed.
It's important to pay careful attention to stoichiometry when determining the 'Change' row in the ICE table. As reactants are converted to products, their concentrations decrease by a multiple of the stoichiometric coefficients, whereas the products increase by their respective coefficients. For the given reaction, where NOCl decomposes to 2 NO and Cl₂, the 'Change' for NOCl is -2x because it decreases by two moles for every mole of Cl₂ produced.
Reaction Quotient
The reaction quotient, Q, plays a vital role in predicting the direction in which a reaction mixture will proceed to reach equilibrium. It is calculated using the same formula as the equilibrium constant, K, but with the initial concentrations of the reactants and products before the reaction has reached equilibrium:\[\begin{equation} Q = \frac{[NO]_{initial}^2[Cl_2]_{initial}}{[NOCl]_{initial}^2} \end{equation}\]Comparing Q to K can inform whether the reaction will proceed forwards or backwards to reach equilibrium:
  • If Q < K, the forward reaction is favored, and the concentrations of products will increase while those of reactants will decrease until equilibrium is reached.
  • If Q = K, the reaction is at equilibrium, and no net change in concentrations will be observed.
  • If Q > K, the reverse reaction is favored, and the concentrations of reactants will increase at the expense of the products until equilibrium is reached.
In the exercise, once the initial concentrations are known, students can check Q to predict how the system will reach equilibrium. However, because the exercise starts with pure reactants or products, Q is initially either 0 or infinity, indicating, as expected, that the reaction must proceed to achieve equilibrium.

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Most popular questions from this chapter

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 reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)+\mathrm{D}(g) .\) A friend asks the following: "I know we have been told that if a mixture of \(\mathrm{A}, \mathrm{B}, \mathrm{C},\) and \(\mathrm{D}\) is at equilibrium and more of \(\mathrm{A}\) is added, more \(\mathrm{C}\) and \(\mathrm{D}\) will form. But how can more \(\mathrm{C}\) and \(\mathrm{D}\) form if we do not add more \(\mathrm{B} ?^{\prime \prime}\) What do you tell your friend?

Lexan is a plastic used to make compact discs, eyeglass lenses, and bulletproof glass. One of the compounds used to make Lexan is phosgene \(\left(\mathrm{COCl}_{2}\right),\) an extremely poisonous gas. Phosgene decomposes by the reaction,$$\operatorname{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g)$$,for which \(K_{\mathrm{p}}=6.8 \times 10^{-9}\) at \(100^{\circ} \mathrm{C}\). If pure phosgene at an initial pressure of 1.0 atm decomposes, calculate the equilibrium pressures of all species.

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

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?

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