Question

At a particular temperature, \(K_{\mathrm{p}}=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ If 2.00 atm of \(\mathrm{H}_{2}(g)\) and 2.00 atm of \(\mathrm{I}_{2}(g)\) are introduced into a \(1.00-\mathrm{L}\) container, calculate the equilibrium partial pressures of all species.

Short answer

The equilibrium partial pressures of the species are as follows: \(H_2: 0.92\, atm\), \(I_2: 0.92\, atm\), and \(HI: 2.16\, atm\).

Step-by-step solution

Write down the initial given information

We are given the following initial information: - Reaction: \(H_2(g) + I_2(g) \leftrightharpoons 2 HI(g)\) - \(K_p = 1.00 \times 10^2\) - Initial partial pressure of \(H_2 = 2.00\, atm\) - Initial partial pressure of \(I_2 =2.00\, atm\) - Volume of the container = \(1.00\,L\)

Set up an ICE table

An ICE table is a useful way to organize the information we have about how concentrations or pressures change during a chemical reaction. In this table, the columns represent the species involved in the reaction, and the rows represent the initial partial pressures (I), the change in partial pressures during the reaction (C), and the equilibrium partial pressures (E). The ICE table looks like this: | | H_2 | I_2 | 2 HI | |---|-----|-----|-----| | I | 2.00 atm | 2.00 atm | 0.00 atm | | C | -x | -x | +2x | | E | 2.00-x | 2.00-x | 2x | Here, "x" represents the decrease in the partial pressures of H_2 and I_2. Since the stoichiometry of the reaction shows that for every one mole of H_2 and one mole of I_2 consumed, two moles of HI are produced, we represent the change in HI with +2x.

Write the K_p expression and substitute the equilibrium values from the ICE table

Recall that the K_p is the ratio of the products to the reactants raised to their stoichiometric coefficients represented by partial pressures. For the given reaction: \(K_p = \frac{[HI]^2}{[H_2][I_2]}\) Substitute the equilibrium values from the ICE table: \(1.00x10^2 = \frac{(2x)^2}{(2-x)(2-x)}\)

Solve for x and find the equilibrium partial pressures

Now we can solve the equation for x: \(100 = \frac{4x^2}{(2-x)^2}\) \(100(2-x)^2 = 4x^2\) \(100(4 - 4x + x^2) = 4x^2\) \(400 - 400x + 100x^2 = 4x^2\) \(96x^2 - 400x + 400 = 0\) This is a quadratic equation, which we can solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) Where a = 96, b = -400, and c = 400. After solving for x, we get two possible values: \(x = 4.08\) and \(x = 1.08\). However, x cannot be greater than the initial partial pressures of H_2 and I_2 (2.00 atm), so we reject the value 4.08. Therefore, \(x = 1.08\). Finally, we can determine the equilibrium partial pressures for all species: - \(H_2\) and \(I_2\): \(2.00 - 1.08 = 0.92\, atm\) - \(HI\): \(2 \times 1.08 = 2.16\, atm\) The equilibrium partial pressures of the species are as follows: \(H_2: 0.92\, atm\) \(I_2: 0.92\, atm\) \(HI: 2.16\, atm\)

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction in which the concentrations or pressures of reactants and products remain constant over time. When a reaction reaches equilibrium, the rate of the forward reaction equals the rate of the reverse reaction.
This means that the amount of each substance does not change, even though the reactions continue to occur.
Equilibrium is dynamic; it involves continuous and opposing processes that occur at equal rates. In the context of reactions involving gases, equilibrium can be expressed using the equilibrium constant, known as \(K_p\) for pressures.

• The value of \(K_p\) helps determine the position of equilibrium. A large \(K_p\) indicates that the products are favored, while a small \(K_p\) suggests that the reactants are more stable at equilibrium.
• This concept is essential for predicting how a system at equilibrium will respond to changes such as pressure, temperature, or concentration (Le Chatelier's Principle).

Understanding chemical equilibrium allows chemists to control and optimize reactions, ensuring desired products are obtained efficiently.

Partial Pressure

Partial pressure refers to the pressure exerted by a single gas in a mixture of gases.
Each gas in a mixture behaves independently and contributes to the total pressure according to its proportion in the mixture.
In a chemical reaction involving gases, partial pressures are crucial because they help to describe the system quantitatively.

• The total pressure in a container is the sum of the partial pressures of all gases present, according to Dalton's Law of Partial Pressures:

\[P_{\text{total}} = P_{\text{H}_2} + P_{\text{I}_2} + P_{\text{HI}}\]

• The partial pressure of each gas is related to the mole fraction of that gas and the total pressure of the mixture. This relationship is represented as:

\[P_i = \chi_i \times P_{\text{total}}\]

• In equilibrium reactions, the equilibrium constant \(K_p\) is expressed using the partial pressures of the gaseous components. Knowing the initial partial pressures allows us to track changes and calculate equilibrium states effectively.

Mastering the concept of partial pressure is key to understanding how gases behave and interact in different chemical scenarios.

ICE Table

An ICE table is a simple yet powerful tool used to track changes in concentrations or pressures of substances in a chemical reaction.
"ICE" stands for Initial, Change, and Equilibrium, representing different stages in the reaction process.

• Initial (I): Record the initial partial pressures or concentrations before the reaction begins.
• Change (C): Describe how much the concentrations or pressures change as the reaction progresses. This is represented by variables like \(x\), which indicates the amount reacting or forming based on stoichiometry.
• Equilibrium (E): Calculate the final concentrations or pressures using the initial and change values. These are used to solve equilibrium constant expressions.

Using an ICE table standardizes the process of solving equilibrium problems, ensuring all changes are accounted for clearly.
It helps visualize the relationship between substances in a reaction and provides a systematic way to plug values into the equilibrium expression. Mastering the ICE table approach allows students to handle complex equilibrium calculations with confidence.