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Chapter 19: Problem 76

The equilibrium constant at \(823 \mathrm{~K}\) for the chemical equation $$ \mathrm{MgCl}_{2}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{Mg} \mathrm{O}(s)+\mathrm{Cl}_{2}(g) $$ is \(K_{\mathrm{p}}=1.75 \mathrm{~atm}^{1 / 2}\). Suppose that \(50.0\) grams of \(\mathrm{MgCl}_{2}(s)\) is placed in a reaction vessel with \(2.00\) liters of oxygen at \(25^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) and that the reaction vessel is sealed and heated to \(823 \mathrm{~K}\) until equilibrium is attained. Calculate \(P_{\mathrm{Cl}_{2}}\) and \(P_{\mathrm{O}_{2}}\) at equilibrium in atm.

Short Answer

Expert verified
At equilibrium, \( P_{\mathrm{Cl}_2} = 0.875 \, \text{atm} \) and \( P_{\mathrm{O}_2} = 0.5625 \, \text{atm} \).

Step by step solution

01

Write the Reaction and Assign Initial Partial Pressures

The given reaction is \( \mathrm{MgCl}_{2}(s) + \frac{1}{2} \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{MgO}(s) + \mathrm{Cl}_{2}(g) \). Since \( \mathrm{MgCl}_{2}(s) \) and \( \mathrm{MgO}(s) \) are solids, their concentrations do not appear in the equilibrium expression. Initially, \( P_{\mathrm{O}_2}^{\text{initial}} = 1.00 \; \text{atm} \) and \( P_{\mathrm{Cl}_2}^{\text{initial}} = 0 \, \text{atm} \).
02

Set Up the Expression for Equilibrium Constant

The equilibrium expression for the reaction in terms of partial pressures is:\[ K_{\mathrm{p}} = \frac{P_{\mathrm{Cl}_2}}{(P_{\mathrm{O}_2})^{1/2}} \]Given, \( K_{\mathrm{p}} = 1.75 \; \text{atm}^{1/2} \).
03

Define Changes in Partial Pressures at Equilibrium

Assume that \( x \; \text{atm} \) of \( \mathrm{O}_2 \) is consumed in reaching equilibrium. Therefore, the change in partial pressure of \( \mathrm{Cl}_2 \) is \( 2x \) because for every \( \frac{1}{2} \, \mathrm{O}_2 \) consumed, \( 1 \, \mathrm{Cl}_2 \) is produced. At equilibrium, \( P_{\mathrm{O}_2}^{\text{eq}} = 1.00 - x \; \text{atm} \) and \( P_{\mathrm{Cl}_2}^{\text{eq}} = 2x \; \text{atm} \).
04

Substitute into the Equilibrium Expression and Solve for x

Substitute \( P_{\mathrm{Cl}_2}^{\text{eq}} = 2x \) and \( P_{\mathrm{O}_2}^{\text{eq}} = 1.00 - x \) into the equilibrium expression:\[K_{\mathrm{p}} = \frac{2x}{(1.00 - x)^{1/2}} = 1.75\]Squaring both sides:\[(\frac{2x}{(1.00 - x)^{1/2}})^2 = 1.75^2 = 3.0625\]Solve for \( x \):\[4x^2 = 3.0625(1.00 - x)\]\[4x^2 + 3.0625x - 3.0625 = 0\]Use the quadratic formula to find \( x \).
05

Calculate Partial Pressures at Equilibrium

From the quadratic solution, select the physically meaningful (positive) value of \( x \). Calculate:- \( P_{\mathrm{O}_2}^{\text{eq}} = 1.00 - x \, \text{atm} \)- \( P_{\mathrm{Cl}_2}^{\text{eq}} = 2x \, \text{atm} \)

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Key Concepts

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

Partial Pressure in Chemical Reactions
Partial pressure is an important concept when dealing with gases in chemical reactions. It refers to the pressure that each gas in a mixture would exert if it were alone in the container. The total pressure of a gas mixture is the sum of the partial pressures of each component gas.
  • In the given exercise, the partial pressure of oxygen initially is 1.00 atm since it is the only gas present at the start.
  • Partial pressures change as the reaction progresses and reaches equilibrium.
It's crucial to accurately track these changes to find the pressures at equilibrium for the reaction involved. Understanding partial pressures allows us to quantify how much of each reactant and product gas is present in the system.
Understanding Chemical Equilibrium
Chemical equilibrium occurs when a reversible chemical reaction reaches a state where the rates of the forward and reverse reactions are equal. At this point, the concentrations (or partial pressures for gases) of the reactants and products remain constant over time.
For our exercise, the equilibrium constant, denoted as \( K_p \), relates the partial pressures of gases at equilibrium. It is calculated using the expression: \[ K_p = \frac{P_{\text{Cl}_2}}{(P_{\text{O}_2})^{1/2}} \]
  • This formula shows how the equilibrium constant depends on the partial pressures of chlorine and oxygen.
  • The given equilibrium constant value, 1.75 atm\(^{1/2}\), helps us find the unknown partial pressures at equilibrium.
This represents the balance point of the reaction where the chemical system is stable and no net change is occurring.
Solving with the Quadratic Formula
The quadratic formula is a mathematical tool used to solve quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). In chemical equilibrium problems, changes in concentrations or pressures at equilibrium often lead to quadratic equations.
In our example, after substituting values into the equilibrium expression, the equation becomes: \[ 4x^2 + 3.0625x - 3.0625 = 0 \]
To solve this, we apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where, in our equation, \( a = 4 \), \( b = 3.0625 \), \( c = -3.0625 \).
  • Solving this equation gives us the value(s) of x, which represents the change in partial pressures.
  • Only the positive value of x is physically meaningful, as pressures cannot be negative.
Mastery of using the quadratic formula is vital for finding equilibrium concentrations and pressures in complex reactions.

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Most popular questions from this chapter

Prior to learning about equilibrium states, we solved stoichiometric problems using the concept of "limiting reactants." Under what conditions does the method of limiting reactants apply?

Sodium hydrogen carbonate, commonly called sodium bicarbonate, is used in baking soda and in fire extinguishers as a source of \(\mathrm{CO}_{2}(g) .\) It decomposes according to the equation $$ 2 \mathrm{NaHCO}_{3}(s) \leftrightharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Given that \(K_{\mathrm{p}}=0.26 \mathrm{bar}^{2}\) at \(125^{\circ} \mathrm{C}\), calculate the partial pressures of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) in units of bars at equilibrium when \(\mathrm{NaHCO}_{3}(s)\) is heated to \(125^{\circ} \mathrm{C}\) in a closed vessel.

Write the equilibrium-constant expression \(\left(K_{c}\right)\) for each of the following equations: (a) \(\mathrm{NH}_{2} \mathrm{COONH}_{4}(s) \leftrightharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)\) (b) \(2 \mathrm{HgO}(s) \leftrightharpoons 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+2 \mathrm{O}_{2}(g) \leftrightharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)\) What are the units of \(K_{c}\) in each case?

Consider the chemical equilibrium described by the equation $$ \mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \leftrightharpoons \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) $$ Use Le Ch芒telier's principle to predict the effect on the equilibrium pressure of \(\mathrm{CO}_{2}(g)\) and of \(\mathrm{CO}(g)\) resulting from (a) an increase in the pressure of \(\mathrm{H}_{2} \mathrm{O}(g)\) (b) an increase in the reaction volume

An equilibrium mixture of \(\mathrm{CO}(g), \mathrm{Cl}_{2}(g)\), and \(\mathrm{COCl}_{2}(g)\) has partial pressures \(P_{\mathrm{co}}=P_{\mathrm{Cl}_{2}}=1.09 \mathrm{bar}\) and \(P_{\mathrm{Cocl}_{2}}=0.144\) bar. A quantity of \(\mathrm{CO}(g)\) is suddenly injected into the reaction vessel and the total pressure jumps to \(3.31\) bar. Calculate the total pressure after equilibrium is reestablished. The relevant chemical equation is $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \leftrightharpoons \operatorname{COCl}_{2}(g) $$
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