Question

Consider the hypothetical reaction \(\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})\). A flask is charged with \(0.75\) atm of pure A, after which it is allowed to reach equilibrium at \(0^{\circ} \mathrm{C}\). At equilibrium the partial pressure of \(\mathrm{A}\) is \(0.36 \mathrm{~atm} .\) (a) What is the total pressure in the flask at equilibrium? (b) What is the value of \(K_{p} ?\)

Short answer

The total pressure in the flask at equilibrium is \(1.14 atm\) and the value of \(K_p\) for the reaction is \(1.690\).

Step-by-step solution

Write down the balanced chemical equation

The balanced chemical equation for the given reaction is: \(A(g) \rightleftharpoons 2 B(g)\)

Set up an ICE (Initial, Change, Equilibrium) table

An ICE table is a convenient way to organize the information about the initial pressures (I), the changes in pressure (C), and the equilibrium pressures (E) during a chemical reaction. Let the initial pressure of A be \(P_A\) and the change in pressure be \(x\). Since A is converted into 2 moles of B for every mole of A that reacts, the change in pressure for B will be \(2x\). We are given the initial pressure \(P_A = 0.75 atm\) and equilibrium pressure \(P_{A,eq} = 0.36 atm\). ICE Table: | | A(g) | 2B(g) | |------|-----------------|---------| |Initial pressure (atm) | 0.75 | 0 | |Change in pressure (atm) | -x | +2x | |Equilibrium pressure (atm)| 0.36 | ? |

Determine the change in pressure (x)

We will use the given information about the equilibrium pressure of A to find the value of x: Initial pressure of A - Change in pressure of A = Equilibrium pressure of A \(0.75 - x = 0.36\) => \(x = 0.75 - 0.36 = 0.39 atm\) Now we can find the equilibrium pressure of B:

Calculate the equilibrium pressure of B

Use the change in pressure we found for A (x) to find the equilibrium pressure of B: Initial pressure of B + Change in pressure of B = Equilibrium pressure of B \(0 + 2x = 2(0.39)\) => Equilibrium pressure of \(B = 0.78 atm\)

Calculate the total pressure at equilibrium (Part a)

To find the total pressure at equilibrium, add the equilibrium pressures of gases A and B: Total pressure = Equilibrium pressure of A + Equilibrium pressure of B Total pressure = \(0.36 + 0.78= 1.14 atm\) So, the total pressure in the flask at equilibrium is \(1.14 atm\).

Calculate the value of Kp (Part b)

The expression for Kp is given by the ratio of the partial pressures of the products to the partial pressures of the reactants, raised to their stoichiometric coefficients: \(K_p = \frac{[B]^2}{[A]}\) Now, we can plug in the equilibrium pressures we found earlier: \(K_p = \frac{(0.78)^2}{(0.36)}\) \(K_p = \frac{0.6084}{0.36} = 1.690\) So, the value of Kp for the reaction is \(1.690\).

Key concepts

Partial Pressure

In gaseous chemical reactions, understanding partial pressure is crucial. It describes how the pressure exerted by each individual gas component in a mixture contributes to the total pressure.
For example, in the reaction \(\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{B}(\mathrm{g})\), each molecule of gas has its own partial pressure. Initially, gas A is pure, meaning it alone exerts the partial pressure of 0.75 atm.
At equilibrium, the partial pressure of A is given as 0.36 atm. From these values, we can compute changes in the system, leading us to better understand the behavior of gases within.

**How Partial Pressure Relates to Total Pressure**
Partial pressures help us to determine the total pressure within a system at equilibrium by simply adding them. This step is essential to solve for the total equilibrium pressure and also in calculating the equilibrium constant (Kp).
In the stated reaction, after formation of B, the contributions to total pressure from A and B are added: 0.36 atm from A, and 0.78 atm from 2B, yielding a total of 1.14 atm.

ICE Table

An essential method to solve equilibrium problems is by setting up an ICE (Initial, Change, Equilibrium) table. This table categorizes the changes in pressures of reactants and products over the course of the reaction.

**Setting Up an ICE Table**

• **Initial (I):** Begin by listing the initial pressures of reactants and products. In our example, A initially at 0.75 atm and B at 0 atm.
• **Change (C):** Denote the change in pressure during the reaction. For each mole of A that reacts (-x), two moles of B are produced (+2x).
• **Equilibrium (E):** Use the known equilibrium pressure. For A, the pressure drops to 0.36 atm, helping us solve for x (in this example x = 0.39 atm).

The ICE table provides a clear, visual method to track how pressures evolve and reach equilibrium.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as Kp, is a key value representing the ratio of the partial pressures of products to reactants raised by their respective coefficients.

**Expression for Kp**
For the reaction \(A(g) \rightleftharpoons 2B(g)\), the expression for Kp becomes:
\[K_p = \frac{(P_B)^2}{P_A}\] where \(P_B\) and \(P_A\) are the partial pressures of B and A, respectively.

We used equilibrium pressures from the ICE table: \(P_A = 0.36 \, \text{atm}\) and \(P_B = 0.78 \, \text{atm}\), leading to:
\[K_p = \frac{(0.78)^2}{0.36} = 1.690\]
This value, 1.690, indicates the reaction's tendency towards products or reactants at equilibrium. A large Kp suggests more products than reactants at equilibrium, guiding predictions about how the reaction behaves under different conditions.