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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.

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.

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.