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Chapter 15: Problem 52

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g)$$ at \(400 \mathrm{K}, K_{c}=7.0 .\) If 0.25 mol of \(\mathrm{Br}_{2}\) and 0.55 \(\mathrm{mol}\) of \(\mathrm{Cl}_{2}\) are introduced into a 3.0 - container at \(400 \mathrm{K},\) what will be the equilibrium concentrations of \(\mathrm{Br}_{2}, \mathrm{Cl}_{2},\) and BrCl?

Short Answer

Expert verified
The equilibrium concentrations of Br鈧�, Cl鈧�, and BrCl are 0.068 M, 0.168 M, and 0.031 M, respectively.

Step by step solution

01

Write the balanced chemical equation

The balanced chemical equation is given in the exercise as: \[\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2\mathrm{BrCl}(g)\]
02

Set up the ICE table

An ICE table will help us organize the data and determine the change in the concentration of the reactants and products. The table is set up as follows: | | Br鈧� | Cl鈧� | BrCl | |-----------|-----------|-----------|-----------| | Initial | 0.25 mol | 0.55 mol | 0 mol | | Change | -x mol | -x mol | +2x mol | | Equilibrium| 0.25-x mol | 0.55-x mol | 2x mol | Now we need to find the equilibrium concentrations in terms of x by dividing the moles by the volume of the container (3.0 L). | | Br鈧� | Cl鈧� | BrCl | |-----------|-----------|-----------|-----------| | Equilibrium| \(\frac{(0.25-x)}{3}\) M | \(\frac{(0.55-x)}{3}\) M | \(\frac{2x}{3}\) M |
03

Write the Kc expression

The equilibrium constant expression, Kc, can now be written based on the balanced chemical equation and the equilibrium concentrations: \[K_{c} = \frac{[\mathrm{BrCl}]^{2}}{[\mathrm{Br}_{2}] \times [\mathrm{Cl}_{2}]} = \frac{\left(\frac{2x}{3}\right)^{2}}{\frac{(0.25-x)}{3} \times \frac{(0.55-x)}{3}}\] We are given that at 400 K, Kc = 7.0.
04

Solve for x

Substitute the known value of Kc and solve for x: \[7.0 = \frac{(2x)^{2}}{(0.25-x)(0.55-x)}\] Now, solve the equation to find the value of x. Note that there might be multiple solutions, but we need to choose the one that gives a valid value for the equilibrium concentrations. After solving the equation, we obtain: \(x = 0.046\)
05

Find the equilibrium concentrations

Now that we have calculated the value of x, we can substitute it back into the equilibrium concentrations of the reactants and products. For Br鈧�: \[[\mathrm{Br}_{2}] = \frac{0.25-x}{3} = \frac{0.25-0.046}{3} = 0.068\: M\] For Cl鈧�: \[[\mathrm{Cl}_{2}] = \frac{0.55-x}{3} = \frac{0.55-0.046}{3} = 0.168\: M\] For BrCl: \[[\mathrm{BrCl}] = \frac{2x}{3} = \frac{2 \times 0.046}{3} = 0.031\: M\] The equilibrium concentrations of Br鈧�, Cl鈧�, and BrCl are 0.068 M, 0.168 M, and 0.031 M, respectively.

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Key Concepts

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

ICE table
To understand how an ICE table works, let's break it down step-by-step. ICE stands for Initial, Change, and Equilibrium. These tables help us track the concentrations of reactants and products over time.

In this exercise, we start with the initial concentrations of Br鈧� and Cl鈧�. At the start, there is no BrCl. The changes in concentration are represented by variables (like \(-x\) and \(+2x\)) to show how the equation shifts toward equilibrium.

By applying the given volumes, we convert the moles to molarity (mol/L) for the equilibrium state.

An ICE table simplifies the process of tracking how reactants and products evolve, making it easier to calculate their balanced states.
Equilibrium Concentrations
Equilibrium concentrations show us the amounts of each substance once the reaction reaches equilibrium. They tell us what the balanced state looks like.

In this exercise, after using the ICE table and finding \(x\), we substitute \(x\) back into our expressions for Br鈧�, Cl鈧�, and BrCl. This gives the equilibrium values.

For Br鈧�, the equilibrium concentration is calculated as \(\frac{0.25-x}{3}\) M. Similarly, Cl鈧� and BrCl are calculated using their respective formulas.

Obtaining these equilibrium values helps us understand the reaction balance and predict the behavior of the chemical system.
Equilibrium Constant (Kc)
The equilibrium constant, \(K_c\), is a crucial concept in chemical equilibrium. It provides a ratio of the concentrations of products to reactants at equilibrium, raised to the power of their coefficients.

Here, we use the expression:
  • \(K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_{2}] \times [\mathrm{Cl}_{2}]}\)
Plugging the equilibrium concentrations into this formula helps confirm the reaction's balance.

Knowing \(K_c\), which is given as 7.0 in the exercise, helps solve for unknowns in the equation. This constant value is specific to a particular reaction at a given temperature, and it guides us to the equilibrium point for the system being considered.

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Most popular questions from this chapter

An equilibrium mixture of \(\mathrm{H}_{2}, \mathrm{I}_{2},\) and \(\mathrm{HI}\) at \(458^{\circ} \mathrm{C}\) contains \(0.112 \mathrm{mol} \mathrm{H}_{2}, 0.112 \mathrm{mol} \mathrm{I}_{2},\) and 0.775 \(\mathrm{mol}\) HI in a 5.00 -L. vessel. What are the equilibrium partial pressures when equilibrium is reestablished following the addition of 0.200 \(\mathrm{mol}\) of \(\mathrm{HI}\) ?

Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons\) \(\mathrm{PCl}_{5}(g) . \mathrm{A} 7.5-\mathrm{L}\) gas vessel is charged with a mixture of \(\mathrm{PCl}_{3}(g)\) and \(\mathrm{Cl}_{2}(g),\) which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are \(P_{\mathrm{PCl}_{3}}=0.124 \mathrm{atm}, P_{\mathrm{Cl}_{2}}=0.157 \mathrm{atm},\) and \(P_{\mathrm{PCl}_{\mathrm{s}}}=1.30 \mathrm{atm}\) (a) What is the value of \(K_{p}\) at this temperature? (b) Does the equilibrium favor reactants or products? (c) Calculate \(K_{c}\) for this reaction at 450 \(\mathrm{K}\)

Consider the equilibrium \(\mathrm{Na}_{2} \mathrm{O}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons\) \(\mathrm{Na}_{2} \mathrm{SO}_{3}(s) .(\mathbf{a})\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in this reaction are soluble in water. Rewrite the equilibrium-constant expression in terms of molarities for the aqueous reaction.

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

A \(0.831-\) g sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 -L container and heated to 1100 \(\mathrm{K}\) . The 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 atm. Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at 1100 \(\mathrm{K}\) .
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