Problem 32 - Chemistry The Central Science In Si Units

Question

Gaseous hydrogen iodide is placed in a closed container at \(450^{\circ} \mathrm{C}\), where it partially decomposes to hydrogen and iodine: \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g)\). At equilibrium it is found that \([\mathrm{HI}]=4.50 \times 10^{-3} \mathrm{M},\left[\mathrm{H}_{2}\right]=5.75 \times 10^{-4} \mathrm{M}\) and \(\left[\mathrm{I}_{2}\right]=5.75 \times 10^{-4} \mathrm{M}\). What is the value of \(K_{c}\) at this temperature?

Step-by-Step Solution

Answer

The value of \( K_{c} \) at this temperature is approximately 0.0163.

Step 1: Write the Expression for Kc

The equilibrium constant expression, \( K_c \), for the given reaction \( 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g) \), is used to determine the concentrations of the products and reactants at equilibrium. It is given by: \[ K_c = \frac{[\mathrm{H}_{2}][\mathrm{I}_{2}]}{[\mathrm{HI}]^2} \].

Step 2: Substitute Known Concentrations

Substitute the equilibrium concentrations into the expression for \( K_c \): \[ K_c = \frac{(5.75 \times 10^{-4} \mathrm{M})(5.75 \times 10^{-4} \mathrm{M})}{(4.50 \times 10^{-3} \mathrm{M})^2} \].

Step 3: Calculate the Numerator and Denominator

Calculate the numerator: \( (5.75 \times 10^{-4}) \times (5.75 \times 10^{-4}) = 3.30625 \times 10^{-7} \). Calculate the denominator: \( (4.50 \times 10^{-3})^2 = 2.025 \times 10^{-5} \).

Step 4: Solve for Kc

Divide the numerator by the denominator to find \( K_c \): \[ K_c = \frac{3.30625 \times 10^{-7}}{2.025 \times 10^{-5}} \approx 0.01633 \].

Key Concepts

Equilibrium Constant

When we talk about chemical equilibrium, one of the most significant concepts is the equilibrium constant, often denoted as \( K_c \) for reactions in terms of concentrations. The equilibrium constant provides valuable information about the ratio of the concentrations of products to reactants in a chemical reaction at equilibrium. For instance, in the reaction \(2 \mathrm{HI}(g) \rightleftharpoons \mathrm{H}_{2}(g) + \mathrm{I}_{2}(g)\), it helps us quantify how far the reaction has proceeded.

• If \( K_c \) is large (greater than 1), it implies that the products are favored at equilibrium.
• If \( K_c \) is small (less than 1), it suggests that the reactants are favored.

The expression for \( K_c \) in our example reaction is derived from the molar concentrations of the gases at equilibrium. It is calculated using the equation: \[ K_c = \frac{[\mathrm{H}_{2}][\mathrm{I}_{2}]}{[\mathrm{HI}]^2} \] This formula emphasizes that equilibrium constant values depend on the concentrations of substances raised to the power of their coefficients in the balanced equation.

Reaction Quotient

The reaction quotient, represented as \( Q \), is another tool that resembles the equilibrium constant but is used to gauge the reaction state at any given point—not necessarily at equilibrium.The expression for \( Q \) is structured the same way as \( K_c \):\[ Q = \frac{[\mathrm{H}_{2}][\mathrm{I}_{2}]}{[\mathrm{HI}]^2} \]The key difference between \( Q \) and \( K_c \) is that \( Q \) can be calculated using the initial concentrations or concentrations at any moment in time.

• If \( Q = K_c \), the system is at equilibrium.
• If \( Q < K_c \), the forward reaction is favored, meaning more products will form.
• If \( Q > K_c \), the reverse reaction is favored, leading to more reactants being produced.

Knowing this, one can predict the direction in which the reaction will shift to achieve equilibrium. This is particularly useful in predicting reaction behavior, especially when initial conditions change.

Concentration Calculation

Calculating concentrations is essential in solving equilibrium problems. For any chemical reaction, you must understand how to manipulate concentration values effectively.In our example, the concentrations at equilibrium are given as:- \([\mathrm{HI}] = 4.50 \times 10^{-3} \mathrm{M}\)- \([\mathrm{H}_{2}] = 5.75 \times 10^{-4} \mathrm{M}\)- \([\mathrm{I}_{2}] = 5.75 \times 10^{-4} \mathrm{M}\)Using these values to substitute into the equilibrium equation allows for the deduction of \( K_c \). Here's a step-by-step fracture:1. **Substitute:** Plug the values into the expression:\[ K_c = \frac{(5.75 \times 10^{-4}) \times (5.75 \times 10^{-4})}{(4.50 \times 10^{-3})^2} \]2. **Intermediate Calculation:** Calculate products and squares separately: - Numerator: \((5.75 \times 10^{-4}) \times (5.75 \times 10^{-4}) = 3.30625 \times 10^{-7}\) - Denominator: \((4.50 \times 10^{-3})^2 = 2.025 \times 10^{-5}\)3. **Result:** Divide the two to find \(K_c:\) \[ K_c = \frac{3.30625 \times 10^{-7}}{2.025 \times 10^{-5}} \approx 0.01633 \]Understanding these calculations is essential, as they allow you to confirm reaction behavior and equilibrium status, providing a robust prediction system in chemistry.