Question

At a particular temperature, \(K=1.00 \times 10^{2}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)$$ In an experiment, 1.00 mole of \(\mathrm{H}_{2}, 1.00\) mole of \(\mathrm{I}_{2},\) and 1.00 mole of \(\mathrm{HI}\) are introduced into a \(1.00-\mathrm{L}\) container. Calculate the concentrations of all species when equilibrium is reached.

Short answer

The equilibrium concentrations of the species are approximately 0.755 M for H₂, 0.755 M for I₂, and 1.49 M for HI.

Step-by-step solution

Write equilibrium expression

To solve the problem, write the equilibrium expression using the definition of K given that: $$K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}$$

Set up Initial, Change, Equilibrium (ICE) table

In this problem, we are given initial concentrations of all species. Write an ICE table to keep track of the initial concentrations, change in concentrations, and equilibrium concentrations for each species. $$ \begin{array}{c|ccc} & [\mathrm{H}_{2}] & [\mathrm{I}_{2}] & [\mathrm{HI}] \\ \hline \text { Initial } & 1.00 & 1.00 & 1.00 \\ \text { Change } & -x & -x & +2 x \\ \text { Equilibrium } & 1.00-x & 1.00-x & 1.00+2 x \\ \end{array} $$

Substitute equilibrium concentrations into the equilibrium expression

Substitute the expressions for equilibrium concentrations in terms of x into the equilibrium expression from Step 1: $$K = \frac{(1.00+2x)^2}{(1.00-x)(1.00-x)}$$

Substitute the given value of K and solve for x

Substitute the given value of K, which is 1.00 × 10², into the equation from Step 3 and solve for x: $$1.00 × 10^{2} = \frac{(1.00+2x)^2}{(1.00-x)(1.00-x)}$$ You can solve the resulting quadratic equation either by factoring, using the quadratic formula, or using a graphical approach. Solving for x, we get x ≈ 0.245.

Calculate the equilibrium concentrations

Substitute x back into the expressions for equilibrium concentrations from the ICE table: $$[\mathrm{H}_{2}]_{eq} = 1.00 - x ≈ 1.00 - 0.245 = 0.755 \ \mathrm{M}$$ $$[\mathrm{I}_{2}]_{eq} = 1.00 - x ≈ 1.00 - 0.245 = 0.755 \ \mathrm{M}$$ $$[\mathrm{HI}]_{eq} = 1.00 + 2x ≈ 1.00 + 2(0.245) = 1.49 \ \mathrm{M}$$ Therefore, the equilibrium concentrations of the species are approximately 0.755 M for H₂, 0.755 M for I₂, and 1.49 M for HI.

Key concepts

equilibrium constant (K)

The equilibrium constant, often represented as \(K\), is an important factor in the study of chemical reactions that reach a state of equilibrium. It provides a ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficients in the balanced chemical equation.
For example, in the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g)\), the equilibrium expression is \(K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]}\).
This constant is temperature-dependent, meaning that changes in temperature can alter the value of \(K\).

• A large value of \(K\) (\(K > 1\)) indicates that, at equilibrium, the reaction mixture will contain more products than reactants.
• A small value of \(K\) (\(K < 1\)) suggests that the reactants are favored at equilibrium.

Understanding \(K\) helps predict the direction of the reaction and the concentrations of each chemical species at equilibrium.

ICE table

An ICE table is a useful tool for organizing the data surrounding the initial concentrations, the changes in concentration, and the equilibrium concentrations of reactants and products in a chemical reaction.
ICE stands for Initial, Change, Equilibrium. These tables are crucial when dealing with equilibrium problems, such as finding the concentrations of all species once equilibrium is reached.

Here's how you use an ICE table:

• Initial: Begin by noting the initial concentrations of all reactants and products before the reaction has progressed.
• Change: Determine the change in concentration for each species as the reaction moves towards equilibrium. Use variables (often \(x\)) to represent the change where needed.
• Equilibrium: Calculate the equilibrium concentrations by adding the changes to the initial values.

By inserting these equilibrium concentrations into the equilibrium expression, you can solve for \(x\), which gives the actual concentrations at equilibrium.

quadratic equation

When solving equilibrium problems, reaching a quadratic equation is common. This occurs when you have a substitution involving \(x\) in the equilibrium expression, which then must be solved to find the change in concentrations.
Quadratic equations typically look like \(ax^2 + bx + c = 0\). In our exercise, after substituting into the equilibrium expression, we arrive at:

\(1.00 \times 10^{2} = \frac{(1.00+2x)^2}{(1.00-x)^2}\)
This expands and rearranges into a quadratic form. There are different methods to solve quadratic equations:

• Factoring: If the equation is easily factorable.
• Quadratic Formula: Given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), suitable for any type of quadratic equation.
• Graphical Methods: Can also be used but are often less precise without computational tools.

Once you solve for \(x\), you can use its value to determine the equilibrium concentrations of the species involved.

concentration calculation

Calculating the concentration of chemical species at equilibrium involves using the value of \(x\) derived from solving the equilibrium equation.
In the given exercise, we identified \(x \approx 0.245\).
Here’s how you can calculate the equilibrium concentrations:

• For \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\), where the change is \(-x\), substitute as \([\mathrm{H}_{2}]_{eq} = 1.00 - x\) and \([\mathrm{I}_{2}]_{eq} \) similarly. After substituting, \([\mathrm{H}_{2}]_{eq} = [\mathrm{I}_{2}]_{eq} \approx 0.755 \ \mathrm{M}\).
• For \(\mathrm{HI}\), where the change is \(+2x\), substitute as \([\mathrm{HI}]_{eq} = 1.00 + 2x\). After substituting, \([\mathrm{HI}]_{eq} \approx 1.49 \ \mathrm{M}\).

By following these steps for each species, you obtain the equilibrium concentrations, giving insight into the state of the reaction at equilibrium.