Question

For the equilibrium $$2 \operatorname{IBr}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\operatorname{Br}_{2}(g)$$ \(K_{p}=8.5 \times 10^{-3}\) at \(150^{\circ} \mathrm{C}\). If \(5.07 \mathrm{kPa}\) of IBr is placed in a 10.0 -L container, what is the partial pressure of all substances after equilibrium is reached?

Short answer

The partial pressures of all substances after the equilibrium is reached are: IBr: \(4.336\ kPa\), I2: \(0.367\ kPa\), and Br2: \(0.367\ kPa\).

Step-by-step solution

Calculate initial partial pressures of substances

Using the ideal gas law, we can find the initial number of moles of IBr and use it to determine the initial partial pressures of the substances. \(P_{IBr} = 5.07\ kPa\) \(V = 10.0\ L\) We are given the initial pressure directly: \(P_{initial} = P_{IBr, initial} = 5.07 \mathrm{kPa}\) For I2 and Br2, the initial pressure is 0, as they are not present at the beginning: \(P_{I_{2},initial} = P_{Br_{2},initial} = 0\) Now we have the initial partial pressures: - \(P_{IBr, initial} = 5.07\ kPa\) - \(P_{I_{2},initial} = 0\) - \(P_{Br_{2},initial} = 0\)

Set up an ICE table

Next, we will set up an ICE (Initial change equilibrium) table for the reaction: \[ \begin{array}{|c|c|c|c|} \hline & 2 \mathrm{IBr}(g) & \mathrm{I}_{2}(g) & \mathrm{Br}_{2}(g) \\ \hline \mathrm{Initial} & 5.07 kPa & 0 & 0 \\ \hline \mathrm{Change} & -2x & +x & +x \\ \hline \mathrm{Equilibrium} & 5.07 - 2x & x & x \\ \hline \end{array} \] Where x is the change in partial pressure of the substances as the reaction proceeds to equilibrium.

Use the equilibrium constant Kp to solve for x

We can now write an expression for the equilibrium expression using our ICE table values: \(K_p = \frac{P_{I_{2}} \cdot P_{Br_{2}}}{(P_{IBr})^2}\) Substitute the equilibrium partial pressures from the ICE table: \(8.5\times10^{-3} = \frac{x\cdot x}{(5.07-2x)^2}\) Now, we can solve this quadratic equation for x numerically or by trying some iterative approaches.

Calculating the equilibrium partial pressures

Upon solving the quadratic equation, we find that x = 0.367 kPa (only considering the physically meaningful positive value). Now we can plug this value of x back into our equilibrium row of our ICE table to find the equilibrium partial pressures for all the substances: - \(P_{IBr,equilibrium} = 5.07 - 2(0.367) = 4.336\ kPa\) - \(P_{I_{2},equilibrium} = 0.367\ kPa\) - \(P_{Br_{2},equilibrium} = 0.367\ kPa\) The partial pressures of all substances after the equilibrium is reached are: - IBr: \(4.336\ kPa\) - I2: \(0.367\ kPa\) - Br2: \(0.367\ kPa\)

Key concepts

ICE Table

An ICE table is a powerful tool in chemistry for understanding changes in the concentrations or pressures of substances when a reaction reaches equilibrium. ICE stands for Initial, Change, and Equilibrium, representing different stages in the reaction's progress.

Initially, define the initial pressure or concentration of each species involved in the reaction. For example, in the given reaction of IBr, the initial pressures for I2 and Br2 are zero because they aren't present at the start.

Next, assess the Change that occurs as the system moves toward equilibrium. Use stoichiometry to express these changes using a variable, often denoted as 'x'. For example, if 2 moles of IBr decompose, then I2 and Br2 will each increase by a mole fraction of 'x'.

Finally, use the equations and equilibrium expressions to determine the Equilibrium pressure or concentration for each substance. The ICE table neatly organizes these calculations, helping you solve for unknowns like 'x' and thus find equilibrium values.

Partial Pressure

Partial pressure is a fundamental concept in understanding gas mixtures. It represents the pressure each gas in a mixture would exert if it were the only gas present in the container. It’s crucial in calculations involving equilibrium, as seen with the IBr, I2, and Br2 reaction.

The partial pressure can be calculated when we know the total pressure of the system and the mole fractions of gases participating. Use the ideal gas law, where appropriate, to correlate volume, temperature, and amount to find initial pressures.

• Initially, determine the individual gases' pressures based on given data or initial moles.
• In our example, the total initial pressure is given as 5.07 kPa for IBr, with I2 and Br2 starting at 0 kPa.
• At equilibrium, changes in these pressures are reflected in solving the equilibrium expressions.

Understanding partial pressure is essential in complex equilibrium scenarios, enabling the correct calculation of changes and the application of equilibrium constants.

Equilibrium Constant

The equilibrium constant, denoted as K, is a crucial parameter in chemical reactions, indicating the ratio of the concentrations or partial pressures of products to reactants at equilibrium. In gaseous reactions, we often use the notation Kp to denote equilibrium in terms of pressure.

For the reaction with IBr, I2, and Br2, the expression for Kp is given by the partial pressures at equilibrium:\[ K_p = \frac{P_{I_{2}} \cdot P_{Br_{2}}}{(P_{IBr})^2} \]
Knowing Kp allows us to find the equilibrium position and understand the extent of the reaction. A small Kp suggests that the reactants predominate at equilibrium, while a large Kp indicates that products predominate.

• Calculate Kp using initial values and understood changes from the ICE table.
• Substitute the equilibrium expressions into the Kp expression and solve for unknowns like 'x'.
• Finally, reassess equilibrium pressures to understand the equilibrium state thoroughly.

Understanding the equilibrium constant is key in predicting and manipulating the outcomes of chemical reactions.