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Chapter 15: Problem 34

The equilibrium constant, \(K_{c}\) for the following reaction is 1.05 at \(350 \mathrm{K}\). $$ 2 \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CCl}_{4}(\mathrm{g}) $$ If an equilibrium mixture of the three gases at \(350 \mathrm{K}\) contains \(0.0206 \mathrm{M} \mathrm{CH}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) and \(0.0163 \mathrm{M}\) \(\mathrm{CH}_{4},\) what is the equilibrium concentration of \(\mathrm{CCl}_{4} ?\)

Short Answer

Expert verified
The equilibrium concentration of \(\text{CCl}_4\) is approximately \(0.0273\,\text{M}\).

Step by step solution

01

Write the Equilibrium Expression

For the given reaction, the equilibrium expression can be written in terms of concentrations as \( K_c = \frac{[\text{CH}_4] [\text{CCl}_4]}{[\text{CH}_2\text{Cl}_2]^2} \). We know \( K_c = 1.05 \).
02

Plug Known Values into the Expression

Substitute the known equilibrium concentrations into the equilibrium expression. \(0.0206\,\text{M}\) for \([\text{CH}_2\text{Cl}_2]\) and \(0.0163\,\text{M}\) for \([\text{CH}_4]\). The expression becomes \[ 1.05 = \frac{0.0163 \times [\text{CCl}_4]}{(0.0206)^2} \].
03

Solve for [CCl4]

Rearrange the equation to solve for \([\text{CCl}_4]\). This gives us \([\text{CCl}_4] = \frac{1.05 \times (0.0206)^2}{0.0163}\).
04

Calculate the Equilibrium Concentration

Substitute the values into the equation obtained from Step 3 to calculate \([\text{CCl}_4]\). Perform the math: \([\text{CCl}_4] = \frac{1.05 \times 0.00042436}{0.0163} \approx 0.0273\,\text{M}\).

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Key Concepts

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

Equilibrium Constant
Every reversible chemical reaction has an equilibrium state, where the rates of the forward and reverse reactions are equal, resulting in no net change in the concentrations of the reactants and products. The equilibrium constant, denoted as \( K_c \), is a value that expresses the ratio of the concentrations of products to reactants at this state. It is specific for a given reaction at a particular temperature. For the example reaction, the equilibrium constant \( K_c \) is given as 1.05 at 350 K. This number tells us how much of the reactants and products are present at equilibrium. It's calculated using the concentrations of gases involved in the reaction.
Concentration Calculation
To find unknown equilibrium concentrations, we solve equations based on the equilibrium constant expression. In the given example, we needed to find the concentration of \( \text{CCl}_4 \) at equilibrium. By substituting the known values of \( [\text{CH}_2\text{Cl}_2] = 0.0206 \text{ M} \) and \( [\text{CH}_4] = 0.0163 \text{ M} \) into the expression for \( K_c \), we form an equation which helps us find \( [\text{CCl}_4] \). The calculation is straightforward: rearrange the equation to isolate \( [\text{CCl}_4] \), then perform the arithmetic to get the value in Molarity (M), which is the concentration of \( \text{CCl}_4 \) at equilibrium.
Equilibrium Expression
The equilibrium expression for a chemical reaction is a mathematical relationship that shows how the equilibrium constant relates to the concentrations of reactants and products. For our gaseous reaction,
  • \( K_c = \frac{[\text{CH}_4] [\text{CCl}_4]}{[\text{CH}_2\text{Cl}_2]^2} \)
This expression is derived from the balanced equation. The concentration of products \( [\text{CH}_4] \) and \( [\text{CCl}_4] \) are in the numerator, and the concentration of the reactant \( [\text{CH}_2\text{Cl}_2] \) raised to the power of its stoichiometric coefficient appears in the denominator. Such expressions help us quantify states of equilibrium for the reaction by showing how concentrations are interrelated.
Gaseous Reactions
Gaseous reactions involve reactants and products in the gas phase and are often influenced by changes in pressure and temperature. In the context of equilibrium, these reactions adjust to maintain a balance, governed by the equilibrium constant. For example, in our reaction, all species \( \text{CH}_2\text{Cl}_2 \), \( \text{CH}_4 \), and \( \text{CCl}_4 \) are gases.
  • Gases expand to fill their container, making concentration adjustments critical for maintaining equilibrium.
  • Pressure changes can affect gaseous equilibria, though it's assumed constant here.
  • Temperature changes can shift the equilibrium constant, which is why \( K_c \) is specified at a certain temperature, like 350 K for this reaction.
The understanding of gaseous reactions is vital, as they are prevalent in industrial and laboratory processes.

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Most popular questions from this chapter

Dinitrogen trioxide decomposes to \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) in an endothermic process \(\left(\Delta_{r} H^{\circ}=40.5 \mathrm{kJ} / \mathrm{mol}-\mathrm{rxn}\right)\) $$ \mathrm{N}, \mathrm{O}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more \(\mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g})\) (b) adding more \(\mathrm{NO}_{2}(\mathrm{g})\) (c) increasing the volume of the reaction flask (d) lowering the temperature

\(K_{p}\) for the following reaction is 0.16 at \(25^{\circ} \mathrm{C}\) $$ 2 \mathrm{NOBr}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ The enthalpy change for the reaction at standard conditions is \(+16.3 \mathrm{kJ} / \mathrm{mol}\) -ran. Predict the effect of the following changes on the position of the equilibrium; that is, state which way the equilibrium will shift (left, right, or no change) when each of the following changes is made. (a) adding more \(\mathrm{Br}_{2}(\mathrm{g})\) (b) removing some \(\mathrm{NOBr}(\mathrm{g})\) (c) decreasing the temperature (d) increasing the container volume

The reaction $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) $$ has an equilibrium constant, \(K_{c}\) of 170 at \(25^{\circ} \mathrm{C}\). If \(2.0 \times 10^{-3}\) mol of \(\mathrm{NO}_{2}\) is present in a \(10 .\) -L. flask along with \(1.5 \times 10^{-3}\) mol of \(\mathrm{N}_{2} \mathrm{O}_{4}\), is the system at equilibrium? If it is not at equilibrium, does the concentration of \(\mathrm{NO}_{2}\) increase or decrease as the system proceeds to equilibrium?

The decomposition of \(\mathrm{NH}_{4} \mathrm{HS}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftharpoons \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ is an endothermic process. Using Le Chatelier's principle, explain how increasing the temperature would affect the equilibrium. If more \(\mathrm{NH}_{4} \mathrm{HS}\) is added to a flask in which this equilibrium exists, how is the equilibrium affected? What if some additional \(\mathrm{NH}_{3}\) is placed in the flask? What will happen to the pressure of \(\mathrm{NH}_{3}\) if some \(\mathrm{H}_{2} \mathrm{S}\) is removed from the flask?

Neither \(\mathrm{PbCl}_{2}\) nor \(\mathrm{PbF}_{2}\) is appreciably soluble in water. If solid \(\mathrm{PbCl}_{2}\) and solid \(\mathrm{PbF}_{2}\) are placed in equal amounts of water in separate beakers, in which beaker is the concentration of \(\mathrm{Pb}^{2+}\) greater? Equilibrium constants for these solids dissolving in water are as follows: $$\begin{aligned} \mathrm{PbCl}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq}) & & K_{c}=1.7 \times 10^{-5} \\ \mathrm{PbF}_{2}(\mathrm{s}) & \rightleftharpoons \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{F}^{-}(\mathrm{aq}) & & K_{\mathrm{c}}=3.7 \times 10^{-8} \end{aligned}$$
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