Methane, \(\mathrm{CH}_{4}\), reacts with \(\mathrm{I}_{2}\) according to the reaction \(\mathrm{CH}_{4}(g)+\mathrm{l}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{l}(g)+\mathrm{HI}(g) .\) At \(630 \mathrm{~K}, K_{p}\) for this reaction is \(2.26 \times 10^{-4}\). A reaction was set up at \(630 \mathrm{~K}\) with initial partial pressures of methane of 105.1 torr and of 7.96 torr for \(\mathrm{I}_{2}\). Calculate the pressures, in torr, of all reactants and products at equilibrium.

Short Answer

Expert verified
The equilibrium partial pressures of methane (CH₄), iodine (I₂), methyl iodide (CH₃I), and hydrogen iodide (HI) are approximately 104.717 torr, 7.577 torr, 0.383 torr, and 0.383 torr, respectively.

Step by step solution

01

1. Write the balanced chemical equation

Write the balanced chemical equation for the given reaction: \(CH_4(g) + I_2(g) \rightleftharpoons CH_3I(g) + HI(g)\)
02

2. Set up an ICE table

Create an ICE table to keep track of the changes in partial pressures as the reaction occurs: | | CH₄ | I₂ | CH₃I | HI | |:------:|:---:|:---:|:----:|:--:| | Initial| 105.1| 7.96| 0 | 0 | | Change | -x | -x | +x | +x | |Equilibrium| 105.1 - x | 7.96 - x | x | x |
03

3. Write the equilibrium expression

Write the equilibrium expression for the reaction using the given Kp value: \(K_p = \frac{[CH_3I][HI]}{[CH_4][I_2]} = 2.26 \times 10^{-4}\)
04

4. Substitute the equilibrium pressures

Substitute the equilibrium pressures from the ICE table into the equilibrium expression: \(2.26 \times 10^{-4} = \frac{x^2}{(105.1-x)(7.96-x)}\)
05

5. Solve the equation

Solve the equation for the value of x. Since Kp is very small, the reaction shifts very little towards the products, so we can assume that x is very small compared to 105.1 and 7.96. Thus, we can write: \(2.26 \times 10^{-4} = \frac{x^2}{(105.1)(7.96)}\) Now, solve for x: \(x^2 = 2.26 \times 10^{-4} \times (105.1)(7.96)\) \(x = \sqrt{2.26 \times 10^{-4} \times (105.1)(7.96)} \approx 0.383\)
06

6. Calculate the equilibrium pressure

Now that we have the value of x, we can substitute it back into the equilibrium pressures in the ICE table to find the equilibrium partial pressures of all reactants and products: CH₄: 105.1 - x = 105.1 - 0.383 ≈ 104.717 torr I₂: 7.96 - x = 7.96 - 0.383 ≈ 7.577 torr CH₃I: x ≈ 0.383 torr HI: x ≈ 0.383 torr So, the equilibrium partial pressures of methane, iodine, methyl iodide, and hydrogen iodide are approximately 104.717 torr, 7.577 torr, 0.383 torr, and 0.383 torr, respectively.

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Key Concepts

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

Understanding ICE Tables
To grasp the complexities of a chemical reaction reaching equilibrium, employing an ICE table (Initial, Change, Equilibrium) is indispensable. This tabular approach is a powerful tool for tracking the concentrations or partial pressures of reactants and products throughout the reaction. Initially, you input the known starting conditions of the reactants and, if available, products. As the reaction moves forward, it undergoes changes depicted in the 'Change' row, typically represented by '+x' or '-x', where 'x' is the amount of reactant consumed or product formed. Ultimately, the 'Equilibrium' row displays the final concentrations or partial pressures when the reaction has stabilized, and no further changes occur. The use of an ICE table is crucial for visualizing the shift in reactants and products and clearly identifies the variables involved in the equilibrium expression, laying the foundation for accurate calculations.

When approaching problems like the one detailed above, creating an ICE table should be your first step. It ensures organization and clarity, setting you up for success as you solve for equilibrium pressures or concentrations.
Determining Equilibrium Constant (Kp)
The equilibrium constant, particularly Kp when dealing with gases, signifies the ratio of partial pressures of the products to the reactants at equilibrium. It is vital because it reveals the direction of the reaction and the extent to which products are formed. For gas-phase reactions, Kp offers insight into how the system responds when conditions such as pressure or temperature vary.

When partial pressures are used, the formula for Kp is expressed as:
\[ K_p = \frac{\text{Products' partial pressures}}{\text{Reactants' partial pressures}} \]
In the context of the provided methane reaction, the Kp expression incorporates the partial pressures at equilibrium of methane, iodine, methyl iodide, and hydrogen iodide. By plugging these values into the formula, you can solve for the unknown variable and find the equilibrium pressures. Remember, as shown in the solution, assumptions that simplify calculations are permissible when justified, such as the smallness of Kp implying minimal changes in initial pressures.
Working with Partial Pressures
Partial pressures play a pivotal role when we dive into the world of chemical reactions involving gases. Dalton's Law of Partial Pressures states that in a mixture of non-reacting gases, each gas exerts pressure as if the other gases were not present, and the total pressure is the sum of these individual pressures.

In the realm of equilibrium calculations, partial pressures represent the pressure contributed by each gas in a reaction mixture. They are directly proportionate to the moles of gas and temperature but inversely proportionate to the volume of the system. When initial partial pressures are given, as in our methane reaction exercise, they serve as the starting point for understanding how the reaction will shift to reach equilibrium. Calculating the changes in these partial pressures helps in determining the direction of the reaction and the conditions at equilibrium, making them an essential part of predicting how chemical reactions behave under varying conditions.
Applying Le Chatelier's Principle
Le Chatelier's Principle is the guiding compass for predicting the behavior of a reaction in response to external changes. This principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and re-establish equilibrium.

In practice, if you increase the pressure on a gas reaction, the system shifts towards the side with fewer moles of gas. If temperature is increased, the reaction will favor the endothermic direction, as it absorbs heat. This principle also holds when concentrations of reactants or products change; the system shifts to oppose the concentration changes. Understanding and applying Le Chatelier's Principle allows for deliberate manipulation of reaction conditions in order to achieve desired outcomes, making it an invaluable concept for chemists and a critical component for solving equilibrium problems like the one we just explored.

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Most popular questions from this chapter

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the reaction of \(\mathrm{CO}\) with \(\mathrm{H}_{2}\) : $$\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ (a) Use thermochemical data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}\) for this reaction. (b) To maximize the equilibrium yield of methanol, would you use a high or low temperature? (c) To maximize the equilibrium yield of methanol, would you use a high or low pressure?

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: \(2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g) .\) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium- constant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each \(\mathrm{Hb}\) can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) -binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygenbinding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of 19 torr, and adult hemoglobin has a P50 value of 26.8 torr. Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction \(4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \longrightarrow\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\) .

A mixture of \(\mathrm{H}_{2}, \mathrm{~S},\) and \(\mathrm{H}_{2} \mathrm{~S}\) is held in a \(1.0-\mathrm{L}\) vessel at \(90{ }^{\circ} \mathrm{C}\) and reacts according to the equation: $$\mathrm{H}_{2}(g)+\mathrm{S}(s) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium the mixture contains \(0.46 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(0.40 \mathrm{~g}\) \(\mathrm{H}_{2}\) (a) Write the equilibrium-constant expression for this reaction. (b) What is the value of \(K_{c}\) for the reaction at this temperature? (c) Why can we ignore the amount of \(\mathrm{S}\) when doing the calculation in part (b)?

(a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

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