Problem 32 - Chemistry The Molecular Nature of Matter

Question

Use the following equilibria $$ \begin{aligned} 2 \mathrm{CH}_{4}(g) & \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g) & K_{\mathrm{c}} &=9.5 \times 10^{-13} \\ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) & \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) K_{\mathrm{c}} &=2.8 \times 10^{-21} \end{aligned} $$ to calculate \(K_{\mathrm{c}}\) for the reaction $$ 2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$

Step-by-Step Solution

Answer

\(K_c = 1.2 \times 10^{30}\) for the target reaction

Step 1 - Write down given equilibrium reactions

First, identify the given equilibria and their equilibrium constants (K_c):\[{2 CH_{4}(g) \rightleftharpoons C_{2}H_{6}(g)+H_{2}(g), K_{c} = 9.5 \times 10^{-13}}\]\[{CH_{4}(g)+H_{2}O(g) \rightleftharpoons CH_{3}OH(g)+H_{2}(g), K_{c} = 2.8 \times 10^{-21}}\]

Step 2 - Determine the required reaction from known equilibria

Find the reaction that can be obtained by manipulating the given equations to match the target reaction: \[{2 CH_{3}OH(g) + H_{2}(g) \rightleftharpoons C_{2}H_{6}(g) + 2 H_{2}O(g)}\]

Step 3 - Reverse the second reaction

To obtain the required products, the second reaction must be reversed. Reversing the reaction will invert the equilibrium constant.\[{CH_{3}OH(g)+H_{2}(g) \rightleftharpoons CH_{4}(g)+H_{2}O(g)}\]The new K_c for the reversed reaction is: \[{K'_{c} = \frac{1}{K_{c}} = \frac{1}{2.8 \times 10^{-21}}}\]

Step 4 - Multiply the reversed reaction to match the target coefficients

The reversed second reaction needs to be multiplied by 2 to match the coefficients of the target reaction:\[{2 CH_{3}OH(g) + 2 H_{2}(g) \rightleftharpoons 2 CH_{4}(g) + 2 H_{2}O(g)}\]The new K_c will be the square of the reversed equilibrium constant: \[{K''_{c} = (K'_{c})^2 = (\frac{1}{2.8 \times 10^{-21}})^2}\]

Step 5 - Write the combined equilibrium equation

Add the first reaction to the modified second reaction to get the target reaction:\[2 CH_{4}(g) - 2 CH_{4}(g) + 2 CH_{3}OH(g) + H_{2}(g) \rightleftharpoons C_{2}H_{6}(g) + 2 H_{2}O(g) + H_{2}(g)\]After canceling the common elements on both sides, we obtain the target reaction.\[{2 CH_{3}OH(g) + H_{2}(g) \rightleftharpoons C_{2}H_{6}(g) + 2 H_{2}O(g)}\]

Step 6 - Combine equilibrium constants for the overall reaction

The overall equilibrium constant K_c for the target reaction will be the product of both modified K_c values for the two reactions involved. \[{K_c = K_c \text{(first reaction)} \times K''_{c}}\] Substitute the known values and calculate K_c for the target reaction.\[{K_c = (9.5 \times 10^{-13}) \times (\frac{1}{2.8 \times 10^{-21}})^2}\]

Key Concepts

Understanding Equilibrium Reactions

When we talk about equilibrium reactions in chemistry, we're referring to a dynamic state where the rate of the forward reaction equals the rate of the reverse reaction, meaning that the concentration of reactants and products remains constant over time. Although it may seem as if nothing is happening, in reality, molecules are continuously reacting, forming products and reconverting into reactants.

An example of an equilibrium reaction is the synthesis of methane (CH₄) and hydrogen (H₂) to form ethane (C₂H₆), as represented by the equation: \[2\mathrm{{CH}}_4(g) \rightleftharpoons \mathrm{{C}}_2\mathrm{{H}}_6(g)+\mathrm{{H}}_2(g)\].

The symbol \(\rightleftharpoons\) indicates a dynamic equilibrium. In the given exercise, understanding this concept is critical to manipulate the equations correctly and calculate the new equilibrium constant for a different reaction.

Manipulating Chemical Equations

Manipulating chemical equations is a useful skill when looking to find the equilibrium constant for a new reaction derived from one or more given reactions. To do this, we may need to reverse the direction of a reaction or multiply the coefficients of the entire reaction to match the stoichiometry of the target reaction.

For instance, the reaction \[\mathrm{{CH}}_4(g)+\mathrm{{H}}_2\mathrm{{O}}(g) \rightleftharpoons \mathrm{{CH}}_3\mathrm{{OH}}(g)+\mathrm{{H}}_2(g)\] can be reversed, resulting in a new reaction and a new equilibrium constant, which is the reciprocal of the original one. If this reversed reaction is then multiplied to match the stoichiometry, the new equilibrium constant is the square of this reciprocal, because the reaction has been multiplied by two.

Through careful application of these manipulations, students can learn to create new equilibrium reactions from existing ones, a skill that was central to solving our main exercise.

Calculating Equilibrium Constants

The equilibrium constant, denoted as \(K_c\), is a number that expresses the ratio of the concentration of the products to reactants at equilibrium, each raised to the power equivalent to their coefficients in the balanced equation. It's a direct measurement of the extent of a reaction; the larger the value of \(K_c\), the more products are favored at equilibrium.

To calculate \(K_c\) for a new reaction from known constants, as demonstrated in the exercise, we have to multiply the equilibrium constants of the manipulated reactions involved, because their effects on concentration are multiplicative. Therefore, \[K_c = K_c \text{(first reaction)} \times K''_{c}\] is the mathematical translation of chemical reactions' behavior when combined.

Using such calculations can enhance students' comprehension of chemical reaction systems and their ability to apply this knowledge in both theoretical problems and practical experiments.