For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{4}\) at 700 \(\mathrm{K} .\) In a \(2.00-\mathrm{L}\) 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?

Short Answer

Expert verified
There are 28.5 grams of SO2 in the vessel at equilibrium.

Step by step solution

01

Write down the given data

We are given the following data: - The balanced chemical reaction: \(2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) - The Kp of the reaction: \(K_{p}=3.0 \times 10^{4}\) - Volume of the vessel: \(V = 2.00 \mathrm{L}\) - Mass of SO3 at equilibrium: \(m_{SO_3} = 1.17 \mathrm{g}\) - Mass of O2 at equilibrium: \(m_{O_2} = 0.105 \mathrm{g}\)
02

Calculate the moles of SO3 and O2 at equilibrium

To calculate the moles of any substance, we use the formula: Moles = mass / molar mass For SO3 (molar mass: 80.07 g/mol) and O2 (molar mass: 32 g/mol), we calculate the moles as follows: \(n_{SO_3} = \frac{1.17 g}{80.07 g/mol} = 0.0146 mol\) \(n_{O_2} = \frac{0.105 g}{32 g/mol} = 0.00328 mol\)
03

Write down the expression for Kp

The expression for Kp for the given balanced chemical equation is: \(K_p = \frac{[SO3]^2}{[SO2]^2[O2]}\) where [SO3], [SO2], and [O2] are the equilibrium molar concentrations of the respective compounds.
04

Calculate the equilibrium concentrations

To find the equilibrium concentrations of the compounds, we divide their equilibrium moles by the volume of the vessel (2 L): \([SO3] = \frac{0.0146 mol}{2 L} = 0.00730 M\) \([O2] = \frac{0.00328 mol}{2 L} = 0.00164 M\) Now, we can relate the initial and equilibrium concentrations of SO2 and find its concentration at equilibrium. Let x be the initial concentration of SO2 before the reaction reached equilibrium: \([SO2]_{initial} = x M\) According to the balanced equation, every two moles of SO2 that react produces two moles of SO3. Hence, the change in the concentrations of SO2 and SO3 are equal. \([SO2] = x - 0.0073 M\)
05

Substitute the equilibrium concentrations into the Kp expression and solve for SO2 concentration

Substitute the equilibrium concentrations of SO3 and O2 into the Kp expression: \(3.0 \times 10^{4} = \frac{(0.00730)^2}{(x - 0.0073)^2 (0.00164)}\) Solve for x: \(x - 0.0073 = 0.215\) \(x = 0.222 M\)
06

Calculate the moles and mass of SO2

Now that we have the equilibrium concentration of SO2, we can calculate its moles and mass: Moles of SO2: \(n_{SO2} = [SO2]V = 0.222 M \times 2 L = 0.444 mol\) Mass of SO2: \(m_{SO2} = n_{SO2} \times M_{SO2} = 0.444 mol \times 64.07 g/mol = 28.5 g\) Hence, there are 28.5 grams of SO2 in the vessel at equilibrium.

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Key Concepts

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

Equilibrium Constant (Kp)
The equilibrium constant, expressed as Kp when dealing with gases, is a crucial concept in chemical equilibrium that quantifies the relationship between the concentrations of reactants and products in a gaseous reaction at equilibrium. Given the equilibrium state of a chemical reaction, Kp is calculated using the partial pressures of the gases involved.

Kp has a fixed value at a constant temperature, which implies that at equilibrium, no matter the initial concentrations of reactants and products, the ratio of their concentrations raised to the power of their stoichiometric coefficients will always be the same. In simple terms, higher Kp values indicate a greater quantity of products relative to reactants at equilibrium, suggesting the reaction favors the product side. Conversely, a lower Kp value suggests the reaction favors the reactants. Understanding Kp is essential in predicting the direction of the reaction's shift when external conditions change, such as pressure or temperature.
Molar Concentration
Molar concentration is a way to express the concentration of a solute in a solution, defined as the number of moles of solute per liter of solution, with units typically in moles per liter (M). For gas-phase reactions, as in our chemical equilibrium example, molar concentration can also be used to describe the concentration of a gaseous compound in a mixture of gases.

It is calculated by dividing the number of moles of the compound by the volume of the gas or solution in which it is contained. Knowing the molar concentration of the reactants and products at equilibrium is essential for calculating the equilibrium constant, as these values are directly plugged into the expression for Kp or Kc (the equilibrium constant in terms of concentration). Correctly finding these concentrations is pivotal in understanding the system's balance and in predicting how it will react to external changes.
Molar Mass
Molar mass is the mass of one mole of a substance, and it is a bridge between the macroscopic and molecular worlds, allowing chemists to count molecules by weighing. Measured in grams per mole (g/mol), it is determined by summing the atomic masses of all atoms in a molecule of the substance.

Molar mass can help us convert between the mass of a substance and the number of moles, a fundamental step in stoichiometry. For example, knowing the molar mass of SO3 and O2 allowed us to calculate their moles from the mass in the original exercise, a necessary step before finding their molar concentrations. Understanding molar mass is not only vital for reaction stoichiometry but also for interpreting various physical, chemical, and biological processes.
Stoichiometry
Stoichiometry is the field of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It is the roadmap that allows chemists to predict the amounts of substances consumed and produced in a given reaction.

Through stoichiometry, we can deduce how many moles of each reactant are needed to form a certain number of moles of a product and vice versa. The coefficients in a balanced chemical equation provide the necessary ratios to perform these calculations. In the context of our exercise, the stoichiometric coefficients allowed us to relate the changes in molar concentrations of SO2, O2, and SO3 during the reaction. A firm grasp of stoichiometry is essential for making accurate predictions in chemistry, ranging from the laboratory to industrial processes.

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Most popular questions from this chapter

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 \(\mathrm{Cl}_{2} :\) $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at 303 \(\mathrm{K}\) . (c) According to Le Chatelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00-\mathrm{L}\) . vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00-\mathrm{L}\) vessel at 303 \(\mathrm{K}\) .

For the equilibrium $$2 \operatorname{IBr}(g) \Longrightarrow \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C} .\) If 0.025 atm of IBr is placed in a 2.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

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 I(g), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

Which of the following statements are true and which are false? (a) For the reaction \(2 \mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{A}_{2} \mathrm{B}(g) K_{c}\) and \(K_{p}\) are numerically the same. (b) It is possible to distinguish \(K_{c}\) from \(K_{p}\) by comparing the units used to express the equilibrium constant. (c) For the equilibrium in (a), the value of \(K_{c}\) increases with increasing pressure.

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