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For the reaction $$\mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g)$$ \(K_{\mathrm{p}}=2.00\) at some temperature. If this reaction at equilibrium has a total pressure of 6.00 \(\mathrm{atm}\) , determine the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the reaction container.

Short Answer

Expert verified
The partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) at equilibrium are 1 atm and 4 atm, respectively.

Step by step solution

01

Set up the equilibrium expression

According to the given equilibrium constant, we can write the following equilibrium expression for this reaction: \[K_{\mathrm{p}} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_{2}}}\] where \(P_{\mathrm{CO_{2}}}\) and \(P_{\mathrm{CO}}\) represent the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\).
02

Write expressions for partial pressures in terms of a single variable

Let's define the variable x as the partial pressure of \(\mathrm{CO}_{2}\) at equilibrium: \(x=P_{\mathrm{CO_{2}}}\) Since we know that the total pressure at equilibrium is 6.00 atm, and there are 2 moles of \(\mathrm{CO}\) formed for every mole of \(\mathrm{CO}_{2}\) consumed, we can express the partial pressure of \(\mathrm{CO}\) as: \(P_{\mathrm{CO}}=6.00 - 2x\)
03

Substitute the expressions into the equilibrium equation

Now, we substitute the expressions for the partial pressures into the equilibrium expression: \[2.00 = K_{\mathrm{p}} = \frac{(6.00 - 2x)^2}{x}\]
04

Solve for the variable and determine the partial pressures

To solve for x, we first multiply both sides of the equation by x: \[2.00x= (6.00-2x)^2\] Now, we can solve for x using either the quadratic formula or factoring (we'll use factoring): \[4x(3-x) = 2x(6-4x) = (6-2x)^2\] This equation simplifies to: \[6-4x = 2x\] Solving for x: \[6 = 6x\] \[x = 1\] Now, using the expressions from Step 2, we can determine the partial pressures at equilibrium: \(P_{\mathrm{CO}_{2}} = x = 1\,\mathrm{atm}\) \(P_{\mathrm{CO}}= 6 - 2x = 6 - 2(1) = 4\,\mathrm{atm}\) Therefore, the partial pressures of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) at equilibrium are 1 atm and 4 atm, respectively.

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Key Concepts

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

Equilibrium Constant
In chemical equilibrium, the equilibrium constant is a key concept that helps in quantifying how a reaction mixture reaches a state of balance. The equilibrium constant, denoted as \( K_{p} \) for reactions involving gases, is the ratio of the concentrations (or partial pressures) of the products to the reactants, each raised to the power of their stoichiometric coefficients. For instance, in the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2 \mathrm{CO}(g) \), the equilibrium constant expression is \( K_{p} = \frac{P_{\mathrm{CO}}^2}{P_{\mathrm{CO}_{2}}} \).
Comparing this calculated ratio with the given \( K_{p} \) allows determination of reaction direction or extent, indicating how far a reaction has "gone forward" at a given temperature and pressure.
It's important to note that \( K_{p} \) is influenced only by temperature, not by changes in pressure or concentration. Thus, understanding the equilibrium constant helps in predicting the amounts of reactants and products during a chemical reaction at equilibrium.
Partial Pressure
Partial pressure describes the pressure exerted by an individual gas in a mixture of gases, each gas contributing to the total pressure independently. In a given reaction at equilibrium, the total pressure is the sum of the partial pressures of all gas components involved.
Let's consider our exercise where the reaction \( \mathrm{C}(s)+\mathrm{CO}_{2}(g) \leftrightharpoons 2\mathrm{CO}(g) \) reaches equilibrium with a total pressure of 6 atm. If \( P_{\mathrm{CO}_{2}} \) is the partial pressure of carbon dioxide and \( P_{\mathrm{CO}} \) is that of carbon monoxide, these can be expressed as \( x \) and \( 6-2x \) respectively, relying on stoichiometry where 2 moles of \( \mathrm{CO} \) are formed per mole of \( \mathrm{CO}_{2} \) consumed.
By solving the equilibrium expression with these variables, one can deduce the individual partial pressures, providing insights into the composition of the gas mixture at equilibrium.
Le Chatelier's Principle
Le Chatelier's Principle is an essential concept for predicting how a change in conditions can affect a system at equilibrium. It posits that if a stress is applied to a system in equilibrium, the system will adjust in a way that counteracts the stress in order to restore equilibrium.
This can involve changes like variations in concentration, pressure, volume, or temperature which affect the balance between reactants and products. For instance, increasing the pressure by decreasing volume in a gas reaction would favor the direction with fewer gas molecules.
While our specific example doesn't involve applying stress directly, understanding this principle helps determine how establishing or restoring equilibrium proceeds when conditions change. In practice, by comprehensively understanding Le Chatelier's Principle, chemists can manipulate conditions to favor the formation of desired products or minimize unwanted ones.

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Most popular questions from this chapter

Hydrogen for use in ammonia production is produced by the reaction $$\mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \frac{\text { Nicatalyst }}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g)$$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

The equilibrium constant is 0.0900 at \(25^{\circ} \mathrm{C}\) for the reaction $$\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those that are not at equilibrium, in which direction will the system shift? a. A 1.0 -L flask contains 1.0 mole of HOCl, 0.10 mole of \(\mathrm{Cl}_{2} \mathrm{O}\) , and 0.10 mole of \(\mathrm{H}_{2} \mathrm{O}\) . b. A 2.0 -L flask contains 0.084 mole of HOCl, 0.080 mole of \(\mathrm{Cl}_{2} \mathrm{O}\) , and 0.98 mole of \(\mathrm{H}_{2} \mathrm{O}\) . c. A 3.0 - flask contains 0.25 mole of HOCl, 0.0010 mole of \(\mathrm{Cl}_{2} \mathrm{O},\) and 0.56 mole of \(\mathrm{H}_{2} \mathrm{O}\) .

Consider the reaction $$\mathrm{P}_{4}(g) \rightleftharpoons 2 \mathrm{P}_{2}(g)$$ where \(K_{\mathrm{p}}=1.00 \times 10^{-1}\) at 1325 \(\mathrm{K}\) . In an experiment where \(\mathrm{P}_{4}(g)\) is placed into a container at 1325 \(\mathrm{K}\) , the equilibrium mixture of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g)\) has a total pressure of 1.00 atm. Calculate the equilibrium pressures of \(\mathrm{P}_{4}(g)\) and \(\mathrm{P}_{2}(g) .\) Calculate the fraction (by moles) of \(\mathrm{P}_{4}(g)\) that has dissociated to reach equilibrium.

Write expressions for \(K\) and \(K_{\mathrm{p}}\) for the following reactions. a. \(2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{CH}_{4} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) b. \(2 \mathrm{NBr}_{3}(s) \Longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{Br}_{2}(g)\) c. \(2 \mathrm{KClO}_{3}(s) \Longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g)\) d. \(\mathrm{CuO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Cu}(l)+\mathrm{H}_{2} \mathrm{O}(g)\)

A sample of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is placed in an empty cylinder at \(25^{\circ} \mathrm{C}\) . After equilibrium is reached the total pressure is 1.5 atm and 16\(\%\) (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) has dissociated to \(\mathrm{NO}_{2}(g) .\) a. Calculate the value of \(K_{\mathrm{p}}\) for this dissociation reaction at \(25^{\circ} \mathrm{C} .\) b. If the volume of the cylinder is increased until the total pressure is 1.0 atm (the temperature of the system remains constant), calculate the equilibrium pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and \(\mathrm{NO}_{2}(g) .\) c. What percentage (by moles) of the original \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) is dissociated at the new equilibrium position (total pressure \(=1.00 \mathrm{atm} ) ?\)

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