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When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(2.00\) -L flask at \(303 \mathrm{~K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Calculate \(K_{c}\) for this reaction at this temperature.

Short Answer

Expert verified
The equilibrium constant \(K_c\) for this reaction at 303 K is 0.716 M.

Step by step solution

01

Calculate Initial concentrations

First, let's determine the initial concentration of SOâ‚‚Clâ‚‚. We can do this using the given moles and volume of the container. Initial concentration of SOâ‚‚Clâ‚‚ = moles/volume = 2.00 mol / 2.00 L = 1.00 M Since initially, there are no SOâ‚‚ and Clâ‚‚ present, their initial concentrations will be 0 M.
02

Calculate Equilibrium concentrations

Next, we need to determine the equilibrium concentrations of SOâ‚‚Clâ‚‚, SOâ‚‚, and Clâ‚‚. Given that 56% of SOâ‚‚Clâ‚‚ decomposes, we can calculate the equilibrium concentration of it by using the initial concentration. Concentration decrease in SOâ‚‚Clâ‚‚ = 1.00 M * 56% = 0.56 M Now, the equilibrium concentration of SOâ‚‚Clâ‚‚ will be: Equilibrium concentration of SOâ‚‚Clâ‚‚ = Initial concentration - Concentration decrease = 1.00 M - 0.56 M = 0.44 M For each molecule of SOâ‚‚Clâ‚‚ that decomposes, one molecule of SOâ‚‚ and one molecule of Clâ‚‚ are formed. Thus, the concentration increase in SOâ‚‚ and Clâ‚‚ will be equal to the concentration decrease in SOâ‚‚Clâ‚‚. Equilibrium concentration of SOâ‚‚ = 0 + 0.56 M = 0.56 M Equilibrium concentration of Clâ‚‚ = 0 + 0.56 M = 0.56 M
03

Calculate Kc using the equilibrium concentrations

Now, we can write the equilibrium expression for Kc. Kc = [SOâ‚‚][Clâ‚‚] / [SOâ‚‚Clâ‚‚] Now substitute the equilibrium concentrations obtained in Step 2 to calculate Kc: Kc = (0.56 M)(0.56 M) / (0.44 M) = 0.716 M The equilibrium constant Kc for this reaction at 303 K is 0.716 M.

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Key Concepts

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

Equilibrium Concentrations
Understanding equilibrium concentrations is crucial in the study of chemical reactions. When a system is at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction, leading to a constant ratio of product and reactant concentrations. In the provided exercise, SO_2Cl_2 decomposes to form SO_2 and Cl_2, reaching a state where both the forward and reverse reactions occur at the same rate.

To clarify, when we refer to equilibrium concentrations, we're talking about the amounts of reactants and products present when the reaction has reached this state. In the example, it's stated that 56% of SO_2Cl_2 decomposes. This is crucial for calculating the equilibrium concentrations, as it determines how much product is formed and how much reactant remains. It’s a common mistake to overlook the stoichiometry of the reaction when calculating these changes; always remember that the stoichiometric coefficients affect the concentration changes of all species involved.
Kc Calculation
The equilibrium constant, denoted as Kc, is a dimensionless number that gives insight into the extent of a reaction; it is calculated using the concentrations of the products and reactants at equilibrium. The higher the value of Kc, the further the reaction proceeds to form products.

During a Kc calculation, it's vital to ensure that you're using equilibrium concentrations, keeping in mind reaction stoichiometry. For the given exercise where Kc = [SO_2][Cl_2] / [SO_2Cl_2], you must substitute the equilibrium concentrations into this expression to solve for Kc. Another common mistake here is not paying attention to the physical states of the reactants and products. Only species in the aqueous or gaseous state are included in the Kc expression; solids and liquids do not appear. The correct handling of the reaction's coefficients as exponent values when calculating concentrations is also an essential step for the accurate computation of Kc.
Chemical Kinetics
While the exercise focuses on equilibrium, it's supported by principles of chemical kinetics—the study of the rates of chemical processes. Chemical kinetics gives insight into the 'how fast' of a reaction, which directly affects how quickly equilibrium can be reached. It involves understanding factors such as temperature, concentration, surface area, and catalysts that can influence the speed of the reaction.

In the context of our problem, although kinetics is not directly addressed, the underlying assumption is that equilibrium has been established at a certain temperature (303 K). It's useful to remember that if the temperature were changed, the kinetic energy of the molecules would be affected, potentially altering the rate at which equilibrium is achieved and possibly changing the value of the equilibrium constant Kc. When interpreting the results of an equilibrium constant calculation, always consider the potential kinetic pathways and rate-determining steps that could impact the reaction in question.

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Most popular questions from this chapter

(a) At \(800 \mathrm{~K}\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0-L vessel contains \(2.67 \times 10^{-2} \mathrm{~g}\) of \(\mathrm{I}(\mathrm{g})\), how many grams of \(I_{2}\) are in the mixture? (b) For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), \quad K_{p}=3.0 \times 10^{4} \mathrm{at}\) \(700 \mathrm{~K} .\) In a 2.00-L vessel the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g) ; K_{p}=5.0 \times 10^{12}\) (b) \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) ; K_{c}=5.8 \times 10^{-18}\)

A \(0.831-g\) sample of \(\mathrm{SO}_{3}\) is placed in a 1.00-Lcontainer and heated to \(1100 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\). $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the total pressure in the container is \(1.300 \mathrm{~atm}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

Write the expression for \(K_{c}\) for the following reactions. In each case indicate whether the reaction is homogeneous or heterogeneous. (a) \(3 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g)\) (b) \(\mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g) \rightleftharpoons \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g)\) (c) \(\mathrm{Ni}(\mathrm{CO})_{4}(g) \rightleftharpoons \mathrm{Ni}(s)+4 \mathrm{CO}(g)\) (d) \(\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)\) (e) \(2 \mathrm{Ag}(s)+\mathrm{Zn}^{2+}(a q) \rightleftharpoons 2 \mathrm{Ag}^{+}(a q)+\mathrm{Zn}(s)\)

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by increasing the temperature and increasing the volume of the reaction vessel. (a) What can you conclude about the reaction from the influence of temperature on the equilibrium? (b) What can you conclude from the influence of increasing the volume?

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