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The equilibrium constant for the dissociation of iodine molecules to iodine atoms $$ \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) $$ is \(3.76 \times 10^{-3}\) at \(1000 \mathrm{K}\). Suppose 0.105 mol of \(\mathrm{I}_{2}\) is placed in a \(12.3-\mathrm{L}\). flask at \(1000 \mathrm{K}\). What are the concentrations of I \(_{2}\) and I when the system comes to equilibrium?

Short Answer

Expert verified
At equilibrium, \([\mathrm{I}_2] = 0.00799 \text{ M}\) and \([\mathrm{I}] = 0.001106 \text{ M}\).

Step by step solution

01

Write the equilibrium expression

For the reaction \( \mathrm{I}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \), the equilibrium constant expression (\( K_c \)) is given by:\[K_c = \frac{[\mathrm{I}]^2}{[\mathrm{I}_2]}\]The value of \( K_c \) at \( 1000 \mathrm{K} \) is \( 3.76 \times 10^{-3} \).
02

Calculate initial concentrations

The initial concentration of \( \mathrm{I}_2 \) is calculated by dividing the moles by the volume of the flask:\[[\mathrm{I}_2]_0 = \frac{0.105 \text{ mol}}{12.3 \text{ L}} = 0.00854 \text{ M}\]Initially, \( [\mathrm{I}] = 0 \text{ M} \) because no iodine atoms are present.
03

Set up the expression for changes in concentration

Let the change in concentration of \( \mathrm{I}_2 \) be \( -x \). Thus, for \( \mathrm{I} \), the change will be \( +2x \) because two moles of \( \mathrm{I} \) are formed for each mole of \( \mathrm{I}_2 \) dissociated. At equilibrium, we have:- \( [\mathrm{I}_2] = 0.00854 - x \)- \( [\mathrm{I}] = 2x \)
04

Substitute into equilibrium expression

Substitute the expressions for equilibrium concentrations into the equilibrium expression:\[K_c = \frac{(2x)^2}{0.00854 - x} = 3.76 \times 10^{-3}\]
05

Solve the equation for x

Simplify and solve the equation \( \frac{4x^2}{0.00854 - x} = 3.76 \times 10^{-3} \) to find \( x \). Rearranging gives:\[4x^2 = 3.76 \times 10^{-3} \times (0.00854 - x)\]This simplifies to a quadratic equation. Solving the quadratic equation via factoring or the quadratic formula gives:\[ x \approx 0.000553 \text{ M} \]
06

Calculate equilibrium concentrations

Using the value of \( x \), calculate the equilibrium concentrations:- \( [\mathrm{I}_2] = 0.00854 - 0.000553 = 0.00799 \text{ M} \)- \( [\mathrm{I}] = 2 \times 0.000553 = 0.001106 \text{ M} \)

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Key Concepts

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

Equilibrium Constant
In chemistry, the equilibrium constant (\( K_c \)) plays a crucial role in understanding reactions at equilibrium. It provides a ratio of the concentrations of products to reactants for a reaction that has reached a stable state. Each equilibrium has its unique constant, indicating the extent of the reaction. For gas-phase reactions, like the dissociation of iodine molecules into atoms, the expression involves concentrations as \( M \) (molarity).

The formula is set up by placing the product concentrations in the numerator and the reactant concentrations in the denominator, each raised to the power of their stoichiometric coefficients from the balanced chemical equation. Essentially, \( K_c \) helps predict the direction and position of the equilibrium state:
  • If \( K_c \) > 1: Products are favored.
  • If \( K_c \) < 1: Reactants are favored.
Understanding \( K_c \) allows us to calculate unknown concentrations once the equilibrium state is established.
Dissociation
Dissociation describes the process by which a compound breaks down into simpler molecules or ions. For molecular iodine (\( \mathrm{I}_2 \)), dissociation involves splitting into individual iodine atoms—2 \( \mathrm{I} \).

When a substance dissociates, the relationship between its initial concentration and how much it changes is crucial. It's expressed in an ICE table (Initial, Change, Equilibrium) that tracks concentration shifts from the initial state to equilibrium.
  • Initial: Start concentration before reaction has occurred.
  • Change: Amount that dissociates, represented by variable \( x \).
  • Equilibrium: Result after dissociation reaches equilibrium.
For iodine molecules, the reaction \( \mathrm{I}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \) signals that dissociating one mole of \( \mathrm{I}_2 \) yields two moles of atoms, making understanding and calculating precise concentrations imperative for equilibrium predictions.
Iodine Molecules
Iodine molecules are diatomic, meaning they contain two iodine atoms bonded together (\( \mathrm{I}_2 \)). They can exist as gases under high temperatures or specific conditions, such as in a closed flask.

The concentration of iodine molecules initially is calculated by dividing the number of moles by the volume, using the formula:
  • \( [\mathrm{I}_2]_0 = \frac{\text{moles of } \mathrm{I}_2}{\text{volume of flask}} \)
This calculation sets the stage for understanding how much \( \mathrm{I}_2 \) will dissociate and how the reaction progresses within the closed system. It's the starting point from where changes due to dissociation are measured.
Concentration Calculations
Calculating concentrations involves a few simple mathematical operations. You start by determining initial concentrations and track changes that occur during the reaction to find equilibrium concentrations.

For reactions like \( \mathrm{I}_2(\mathrm{g}) \rightleftarrows 2 \mathrm{I}(\mathrm{g}) \), changes occur in accordance with the stoichiometry. Initially, only \( \mathrm{I}_2 \) might be present, hence:
  • Initial Concentration: \( [\mathrm{I}_2] = 0.00854 \text{ M} \); \( [\mathrm{I}] = 0 \text{ M} \)
  • Change: Represented with \( x \) for reactants and \( 2x \) for products due to stoichiometry.
  • Equilibrium: \( [\mathrm{I}_2] = 0.00854 - x \), \( [\mathrm{I}] = 2x \)
After setting up the equilibrium expression, algebraic manipulation (in this case solving a quadratic equation) allows us to find \( x \), eventually calculating the equilibrium concentrations of iodine molecules and atoms. This systematic approach ensures accuracy in dynamic systems like chemical equilibria.

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Most popular questions from this chapter

\(K_{\mathrm{p}}\) for the following reaction is 0.16 at \(25^{\circ} \mathrm{C}\) $$ 2 \mathrm{NOBr}(\mathrm{g}) \rightleftarrows 2 \mathrm{NO}(\mathrm{g})+\mathrm{Br}_{2}(\mathrm{g}) $$ The enthalpy change for the reaction at standard conditions is \(+16.3 \mathrm{kJ} .\) 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

\(K_{c}\) for the decomposition of ammonium hydrogen sulfide is \(1.8 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ (a) When the pure salt decomposes in a flask, what are the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S} ?\) (b) If \(\mathrm{NH}_{1} \mathrm{HS}\) is placed in a flask already containing \(0.020 \mathrm{mol} / \mathrm{L}\) of \(\mathrm{NH}_{3}\) and then the system is allowed to come to equilibrium, what are the equilibrium concentrations of \(\mathrm{NH}_{3}\) and \(\mathrm{H}_{2} \mathrm{S} ?\)

Ammonium hydrogen sulfide decomposes on heating. $$ \mathrm{NH}_{4} \mathrm{HS}(\mathrm{s}) \rightleftarrows \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{S}(\mathrm{g}) $$ If \(K_{p}\) for this reaction is 0.11 at \(25^{\circ} \mathrm{C}\) (when the partial pressures are measured in atmospheres), what is the total pressure in the flask at equilibrium?

Disturbing a Chemical Equilibrium Dinitrogen trioxide decomposes to NO and \(\mathrm{NO}_{2}\) in an endothermic process \((\Delta H=40.5 \mathrm{kJ} / \mathrm{mol})\) $$ \mathrm{N}_{2} \mathrm{O}_{3}(\mathrm{g}) \rightleftarrows \mathrm{NO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) $$ 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

Calculating an Equilibrium Constant The reaction $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftarrows \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ was examined at \(250^{\circ} \mathrm{C}\). At equilibrium, \(\left[\mathrm{PCl}_{5}\right]=4.2 \times\) \(10^{-5} \mathrm{mol} / \mathrm{L},\left[\mathrm{PCl}_{3}\right]=1.3 \times 10^{-2} \mathrm{mol} / \mathrm{L},\) and \(\left[\mathrm{Cl}_{2}\right]=\) \(3.9 \times 10^{-3} \mathrm{mol} / \mathrm{L} .\) Calculate \(K\) for the reaction.

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