6.3.2. Scenario 2: Calculating Equilibrium Concentrations/Partial Pressures from Initial Conditions and K (ICE Tables)

When you are given initial concentrations and the value of K, you need to determine how the concentrations will change to reach equilibrium. This is commonly done using an ICE (Initial, Change, Equilibrium) table.

ICE Tables are useful tools for calculating the changes in concentrations of reactants and products as a system approaches equilibrium. The table is structured into three rows: Initial (I), Change (C), and Equilibrium (E). In the Initial row, we note the starting concentrations of all chemicals involved. The Change row indicates how those concentrations will change as the reaction progresses towards equilibrium, often represented with a variable like 'x'. Finally, the Equilibrium row calculates the concentrations at equilibrium based on the initial amounts and the changes.

Think of an ICE table like a recipe for a cake. The initial ingredients represent the reactants you start with. As the mixing and baking process progresses, some ingredients change into cake (the products) until you have a delicious treat ready to serve (the equilibrium state).

Consider the decomposition of PCl₅: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At a certain temperature, Kc = 0.024. If 1.00 mol of PCl₅ is placed in a 10.0 dm³ container, calculate the equilibrium concentrations.

In this example, we start with 1.00 mol of PCl₅ in a 10 dm³ container, giving us an initial concentration of 0.100 mol dm⁻³ for PCl₅. The initial concentrations of PCl₃ and Cl₂ are both zero because they haven't formed yet. We set up our ICE table where we denote the change in the concentration of PCl₅ as '-x'. When PCl₅ reacts, it decreases in concentration while PCl₃ and Cl₂ increase by 'x'. This leads to a new equilibrium concentration for each substance. We then write the equilibrium expression based on the reaction and substitute our equilibrium concentrations into the expression to set it equal to K and solve for 'x.' Finally, we calculate the equilibrium concentrations for all substances.

Imagine a balloon filled with only air (PCl₅). When you start to let the air out, the balloon's volume decreases and the air inside it (which represents the gas in our reaction) finds a new balance. When you find balance again—with some air now outside the balloon (representing PCl₃ and Cl₂)—that illustrates reaching equilibrium.

1. Initial Concentrations: [PCl₅] = 1.00 mol / 10.0 dm³ = 0.100 mol dm⁻³ [PCl₃] = 0 mol dm⁻³ [Cl₂] = 0 mol dm⁻³

2. Set up the ICE Table: Let 'x' be the change in concentration of PCl₅. Concentration (mol dm⁻³) PCl₅ PCl₃ Cl₂ Initial (I) 0.100 0 0 Change (C) -x +x +x Equilibrium (E) 0.100-x x x

Initially, we calculate the concentration of PCl₅ as 0.100 mol dm⁻³ since we have 1.00 mol in a 10 dm³ container. For PCl₃ and Cl₂, their initial concentrations are zero as they haven't formed yet. We use a variable 'x' to represent how much of PCl₅ dissociates into PCl₃ and Cl₂. In the ICE table, for the Initial row, we place our starting concentrations. For the Change row, we mark a decrease of 'x' for PCl₅ and an increase of 'x' for PCl₃ and Cl₂. Consequently, the Equilibrium row reflects the concentrations after the reaction reaches equilibrium.

This process is similar to deciding how many slices of a cake (PCl₅ in our example) to eat at a party. Initially, you have an entire cake (0.100 mol of PCl₅). As you and your friends eat slices (the change), the remaining cake decreases while the consumed pieces can be equated to the number of happy faces (PCl₃ and Cl₂) around you, leading to an eventual calm party atmosphere (equilibrium).

3. Write the Kc expression: Kc = [PCl₃][Cl₂] / [PCl₅]

4. Substitute equilibrium expressions into Kc and solve for x: 0.024 = (x)(x) / (0.100 - x) 0.024 (0.100 - x) = x² 0.0024 - 0.024x = x² x² + 0.024x - 0.0024 = 0 This is a quadratic equation (ax² + bx + c = 0). Use the quadratic formula: x = [-b ± sqrt(b² - 4ac)] / 2a

The equilibrium constant expression relates the concentrations of products to those of reactants at equilibrium. For the decomposition of PCl₅, we express Kc in terms of PCl₃ and Cl₂ concentrations as they are the products, and PCl₅ is the reactant. After substituting the values from our ICE table into the Kc expression, we simplify and rearrange the equation, which ultimately leads us to a quadratic equation in the form ax² + bx + c = 0. We can solve this equation using the quadratic formula to find the value of 'x' which tells us how much of PCl₅ has converted into PCl₃ and Cl₂.

Imagine you're setting up a balance scale (the equilibrium constant) with weights (the concentrations). As you adjust the weights of cakes (PCl₃ and Cl₂), you want to find out how much cake has been transferred from one side (PCl₅) to balance it effectively. The quadratic equation acts as the detailed plan to find the right amount of cake needed to achieve perfect equilibrium on the scale.

x = [-0.024 ± sqrt((0.024)² - 4(1)(-0.0024))] / 2 x = [-0.024 ± sqrt(0.000576 + 0.0096)] / 2 x = [-0.024 ± sqrt(0.010176)] / 2 Two possible values for x: x₁ = (-0.024 + 0.10087) / 2 = 0.0384 (approx.) x₂ = (-0.024 - 0.10087) / 2 = -0.0624 (approx.) Since concentration cannot be negative, we choose x = 0.0384 mol dm⁻³.

5. Calculate Equilibrium Concentrations: [PCl₅] = 0.100 - 0.0384 = 0.0616 mol dm⁻³ [PCl₃] = 0.0384 mol dm⁻³ [Cl₂] = 0.0384 mol dm⁻³

When we solve the quadratic equation, we end up with two values for 'x': one positive and one negative. Since concentration cannot be negative, we discard the negative solution. We keep the positive solution, x = 0.0384 mol dm⁻³, which represents the amount of reactant converted to products. Next, we calculate the equilibrium concentrations by subtracting 'x' from the initial concentration of PCl₅ and keeping 'x' for the concentrations of PCl₃ and Cl₂. Thus, we find the equilibrium concentrations for all species involved in the reaction.

Returning to our cake analogy, when we're adjusting the recipe (solving for 'x'), we discard any impossible solutions (like negative cake amounts). The positive amount tells us the final slices shared at the party, giving us the final cake pieces present for everyone to enjoy, leading to that delightful moment when everyone relishes the cake (the equilibrium state).

Approximation Method: If the value of K is very small (typically K < 10⁻³ or 10⁻⁴) and the initial concentration of reactants is relatively large, you can often make the approximation that 'x' is negligible compared to the initial concentration. For example, in 0.100 - x, if x is very small, 0.100 - x ≈ 0.100. This simplifies the calculation by avoiding the quadratic formula. After solving for x with the approximation, you must check if x is less than 5% of the initial concentration. If it is, the approximation is valid. If not, the quadratic formula must be used. In the example above, x (0.0384) is not negligible compared to 0.100 (it's 38.4%), so the approximation would not be valid.

When K is small compared to the initial concentration of reactants, we can simplify the calculation process. This is because the amount of reactant that converts to products (x) will be very small relative to the initial amount. Therefore, we can assume that the initial concentration changes only very slightly, allowing us to approximate calculations without using complex equations. However, it’s important to check if this assumption holds true: if 'x' is less than 5% of the initial concentration, the approximation can be used safely. If 'x' turns out to be significant, we need to revert to using the quadratic formula to ensure accuracy.

Think about making a tiny adjustment to a large car engine (the reaction). If you're only making a small tweak to the throttle (the reaction shifting slightly), it's often easier to keep everything else the same, especially if it works well (ignoring that tiny adjustment). But if your tweak becomes significant, leading to a noticeable change in performance, you need to look closely at the entire engine (use the quadratic equation) to avoid any issues. This is how we approach calculations in chemistry conservatively.

--