Question

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

Short answer

The equilibrium concentrations of the species involved in the reaction \(\mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g)\) are found by solving the equation \(1.00 \times 10^{2} = \frac{(1.00 + 2x)^2}{(1.00 - x)^2}\) for x. In this case, we initially find an unrealistic negative value for the concentrations of H₂ and I₂, which implies the need to re-solve the equation using the quadratic formula for a more realistic value of x. The same procedure of finding x and then calculating equilibrium concentrations using x would still apply afterwards.

Step-by-step solution

Write out the balanced equation and the equilibrium expression

The balanced equation is given as: \[ \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2\mathrm{HI}(g) \] The expression for the equilibrium constant (K) is as follows: \[ K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}] [\mathrm{I}_{2}]} \]

Write down the initial moles of reactants and product that we have, set up a table to keep track of changes in moles and the associated concentrations

Let's use a table to keep track of the initial moles, the changes in moles (due to the reaction progressing towards its equilibrium), and the equilibrium moles. Species | Initial moles | Change in moles | Equilibrium moles --------|---------------|-----------------|------------------ H₂ | 1.00 | -x | 1.00 - x I₂ | 1.00 | -x | 1.00 - x HI | 1.00 | +2x | 1.00 + 2x The initial concentrations are calculated by dividing the initial moles by the volume of the container (1.00 L). Since the volume is 1.00 L, the moles of each species are numerically equal to their molar concentrations.

Write the equilibrium concentrations of all species and plug them into the equilibrium expression

Now let's write the equilibrium concentrations and substitute them into our K expression. \[ K = \frac{(1.00 + 2x)^2}{(1.00 - x)(1.00 - x)} = 1.00 \times 10^{2} \]

Solve the equation to find x

Now we need to solve this equation for x. We have: \[ 1.00 \times 10^{2} = \frac{(1.00 + 2x)^2}{(1.00 - x)^2} \] Rearranging gives: \[ (1.00 + 2x)^2 = 100(1.00 - x)^2 \] Taking the square root of both sides, we get: \[ 1.00 + 2x = 10(1.00 - x) \] Rearranging again gives: \[ 2x + x = 10 - 1.00 \] \[ 3x = 9 \] Now solve for x: \[ x = 3 \]

Calculate the equilibrium concentrations of all species using x

Now that we have x, we can find the equilibrium concentrations of all species: H₂: 1.00 - x = 1.00 - 3 = -2 (unrealistic value, so H₂ and I₂ cannot be completely consumed; use the quadratic formula in this case) I₂: 1.00 - x = 1.00 - 3 = -2 (unrealistic value) HI: 1.00 + 2x = 1.00 + 2(3) = 1.00 + 6 = 7.00 Since we got negative values for H₂ and I₂, we need to re-solve the equation in Step 4 using the quadratic formula and find a more realistic value for x. However, the process remains the same - find x, and then calculate the equilibrium concentrations of all species using x.

Key concepts

Equilibrium Constant (K)

The equilibrium constant, abbreviated as \( K \), is a vital component in understanding chemical reactions at equilibrium. It quantifies the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients under equilibrium conditions.
In the given exercise, the reaction is \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \), and the equilibrium constant \( K = 1.00 \times 10^{2} \). This means at equilibrium, the concentration of \( \mathrm{HI} \) is highly favored over \( \mathrm{H}_{2} \) and \( \mathrm{I}_{2} \).

When calculating \( K \), the equation is:

• \( K = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]} \)

This expression lets us solve for unknown concentrations at equilibrium when the value of \( K \) is known.
Overall, \( K \) is a constant that only changes with temperature, and it offers invaluable insight into where the equilibrium position lies.

Reaction Quotient

The reaction quotient, symbolized as \( Q \), is similar to the equilibrium constant \( K \), but is used for reactions that are not necessarily at equilibrium.
By comparing \( Q \) with \( K \), we can predict the direction in which a reaction will proceed to reach equilibrium.

For the reaction \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \),we calculate \( Q \) using initial concentrations:

• \( Q = \frac{[\mathrm{HI}]^2}{[\mathrm{H}_{2}][\mathrm{I}_{2}]} \)

Typically,

• If \( Q < K \), the forward reaction occurs to form more products.
• If \( Q = K \), the system is at equilibrium.
• If \( Q > K \), the reverse reaction is favored to form more reactants.

In the exercise, knowing that \( Q \) needs adjustment provides guidance on the shift needed in concentrations to achieve equilibrium.

Quadratic Formula

When solving equilibrium problems, you might encounter situations where solving for \( x \) (the change in concentration) leads to a quadratic equation.
The quadratic formula becomes an essential mathematical tool for finding exact solutions when other methods, such as simple factorization, are either impossible or impractical.

The standard quadratic equation is:

• \( ax^2 + bx + c = 0 \)

The quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our problem, when unrealistic values like negative concentrations emerge, the quadratic formula helps to provide accurate solutions.
By applying it, we can find the correct value of \( x \) that respects the physical constraints of the system.

ICE Table

An ICE table is a structured way to organize information for equilibrium problems, where 'ICE' stands for Initial, Change, and Equilibrium.
It is helpful for keeping track of concentrations as a reaction progresses toward equilibrium.

To use an ICE table, write down:

• Initial concentrations or moles.
• The change that occurs as the system moves to equilibrium.
• The equilibrium concentrations.

For the reaction \( \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \),we set initial moles in a table, and assume \( x \)changes as the reaction proceeds:

- Initial: \( 1.00 \) mol of each species.- Change: \(-x \) for reactants and \(+2x \) for products because of stoichiometry.- Equilibrium: Considered values using \( x \).The ICE table simplifies calculations and provides a clear picture of the relationships among reactants and products during the reaction's progression to equilibrium.