Problem 102 - Chemistry: An Atoms-Focused Approach

Question

For the decomposition of \(\mathrm{HI}(g)\) into \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) at \(400^{\circ} \mathrm{C}, K_{c}=0.0183\) $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) $$ If \(80.0 \mathrm{g}\) of \(\mathrm{HI}(g)\) is placed in a 2.5 I. chamber at \(400^{\circ} \mathrm{C}\) what are the concentrations of all species when the system comes to cquilibrium?

Step-by-Step Solution

Answer

The equilibrium concentrations are approximately 0.1824 M for HI, 0.0338 M for H2, and 0.0338 M for I2.

Step 1: Find initial concentration of HI

To find the initial concentration of HI, first convert the mass of HI into moles using its molar mass. HI has one hydrogen and one iodine atom. The molar mass of hydrogen is 1.01 g/mol and that of iodine is 126.9 g/mol. Molar mass of HI = 1.01 g/mol + 126.9 g/mol = 127.91 g/mol Now, find the moles of HI: moles of HI = mass of HI / molar mass of HI = 80.0 g / 127.91 g/mol = 0.625 mol After that, we can find the concentration of HI by dividing the moles by the volume of the chamber: Initial concentration of HI = moles of HI / volume of chamber = 0.625 mol / 2.5 L = 0.25 M

Step 2: Set up ICE table

Now, we can set up our ICE table showing initial concentrations, changes to the concentrations and the final equilibrium concentrations. Since there is no H2 or I2 initially present, their initial concentrations are zero. | | HI | H2 | I2 | |------|---------|-----|-----| | I | 0.25 M | 0 M | 0 M | | C | -2x | +x | +x | | E | 0.25-2x | x | x | Here, 'x' is the change in molar concentration of each species during the reaction.

Step 3: Write the Kc expression and solve for x

Write the expression for Kc using the equilibrium concentrations: \(K_c = \frac{[H_2][I_2]}{[HI]^2}\) Substitute the equilibrium concentrations from the ICE table and the given Kc value: \(0.0183 = \frac{x \cdot x}{(0.25 - 2x)^2}\) Now we have a quadratic equation in terms of x. Solving for x can be a bit tricky, but we can make a reasonable assumption that since Kc is quite small compared to 1, the reaction does not proceed to a large extent, so 2x would be much smaller than 0.25. Therefore, the denominator can be approximated as \((0.25)^2\): \(0.0183 = \frac{x^2}{(0.25)^2}\) Now solve for x: \(x^2 = 0.0183 \cdot (0.25)^2\) \(x = \sqrt{0.001144} \approx 0.0338\)

Step 4: Find equilibrium concentrations

Now that we have the value of x, we can find the equilibrium concentrations using the ICE table: \([HI]_{eq} = 0.25 - 2x = 0.25 - 2(0.0338) \approx 0.1824 \: \text{M}\) \([H_2]_{eq} = x = 0.0338 \: \text{M}\) \([I_2]_{eq} = x = 0.0338 \: \text{M}\)

Step 5: Final answer

At equilibrium, the concentrations of HI, H2, and I2 are approximately 0.1824 M, 0.0338 M, and 0.0338 M, respectively.

Key Concepts

ICE Table

The ICE table is a helpful tool in chemical equilibrium problems. ICE stands for Initial, Change, and Equilibrium, and it's used to track concentrations of reactants and products. It starts by listing initial concentrations, which are the starting amounts before any reaction occurs.

Next, we record changes in concentration that happen as the reaction progresses. These changes are represented using the variable 'x', which indicates the extent of the reaction. The change for each species is determined by the stoichiometry of the balanced chemical equation. For reactions like the decomposition of HI into H₂ and I₂, the stoichiometric coefficients guide these changes:

• For HI: the change is (-2x) since 2 HI molecules decompose for every molecule of H₂ or I₂ formed.
• For H₂ and I₂: the change is (+x) as each is formed from the decomposition.

Finally, we calculate the equilibrium concentrations by adding the changes to the initial concentrations. The ICE table simplifies solving the equilibrium constant expression by organizing the necessary data efficiently.

Equilibrium Constant (Kc)

The equilibrium constant, denoted as \(K_c\), provides a quantifiable measure of the position of a chemical equilibrium. It is expressed in terms of concentrations of products and reactants raised to their respective stoichiometric coefficients. For the equilibrium involving the decomposition of HI:
\[K_c = \frac{[H_2][I_2]}{[HI]^2}\]
In this expression, the concentrations \([H_2]\), \([I_2]\), and \([HI]\)\ refer to the equilibrium concentrations as determined using the ICE table. The constant \(K_c\) is temperature-dependent and offers insight into the favorability of a reaction:

• A large \(K_c\) (greater than 1) implies the equilibrium is product-favored.
• A small \(K_c\) (less than 1), like in this case, indicates the equilibrium is reactant-favored, meaning the reaction forms fewer products relative to reactants at equilibrium.

Using \(K_c\), one can predict how changes in conditions (temperature, concentration) might shift the equilibrium, aiding in understanding how a reaction behaves under various circumstances.

Decomposition Reaction

Decomposition reactions involve the breakdown of a single compound into two or more simpler substances. They are generally represented as:
\[ ext{AB} ightarrow ext{A} + ext{B}\]
In this scenario, the decomposition of hydrogen iodide (HI) into hydrogen gas (H₂) and iodine gas (I₂) showcases a classic decomposition reaction. Such reactions can be spontaneous or require energy input, often in the form of heat, light, or electricity.

• For HI, decomposition occurs readily at elevated temperatures like 400°C, where the provided energy helps overcome the activation energy barrier.
• The result is the liberation of simpler diatomic molecules, such as H₂ and I₂, from the complex HI.

Understanding decomposition reactions is crucial in industries and laboratories where chemical breakdown is a part of processing or experimental procedures. It's important to recognize the role of temperature and energy in influencing the rate and extent of a decomposition reaction.