Consider the reaction IO4- (aq) + 2 H2O (l) ⇌ H4IO6- (aq); Kc = 3.5 * 10^-2. If you start with 25.0 mL of a 0.905 M solution of NaIO4 and then dilute it with water to 500.0 mL, what is the concentration of H4IO6- at equilibrium?

Verified step by step guidance

Calculate the initial concentration of IO4^- after dilution. Use the dilution formula: C1V1 = C2V2, where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.

Set up an ICE (Initial, Change, Equilibrium) table to determine the changes in concentration for the reaction IO4^- + 2 H2O ⇌ H4IO6^- at equilibrium. Start with the initial concentration of IO4^- and assume the initial concentration of H4IO6^- is zero.

Define the change in concentration of IO4^- as -x and the change in concentration of H4IO6^- as +x, since the stoichiometry of the reaction shows a 1:1 ratio between IO4^- and H4IO6^-.

Write the expression for the equilibrium constant Kc in terms of the concentrations at equilibrium: Kc = [H4IO6^-] / [IO4^-]. Substitute the equilibrium concentrations from the ICE table into this expression.

Solve the equation for x, which represents the equilibrium concentration of H4IO6^-. This will involve substituting the known value of Kc and solving for x algebraically.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. In this case, Kc = 3.5 * 10^-2 indicates that at equilibrium, the concentration of H4IO6- relative to IO4- and H2O is low, suggesting that the reaction favors the reactants.

Dilution and Concentration Calculations

Dilution involves reducing the concentration of a solute in a solution, typically by adding more solvent. The dilution formula, C1V1 = C2V2, allows us to calculate the new concentration (C2) after dilution. In this scenario, the initial concentration of NaIO4 is diluted from 0.905 M in 25.0 mL to a final volume of 500.0 mL, affecting the concentrations of all species involved in the equilibrium.

ICE Table (Initial, Change, Equilibrium)

An ICE table is a tool used to organize the initial concentrations, changes in concentrations, and equilibrium concentrations of reactants and products in a chemical reaction. By setting up an ICE table for the given reaction, we can systematically determine the equilibrium concentrations of H4IO6- after accounting for the dilution and the reaction's stoichiometry.

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Related Practice

Textbook Question

Consider the hypothetical reaction A(𝑔) + 2 B(𝑔) ⇌ 2 C(𝑔), for which 𝐾_𝑐= 0.25 at a certain temperature. A 1.00-L reaction vessel is loaded with 1.00 mol of compound C, which is allowed to reach equilibrium. Let the variable x represent the number of mol/L of compound A present at equilibrium.

(d) The equation from part (c) is a cubic equation (one that has the form ax3 + bx2 + cx + d = 0). In general, cubic equations cannot be solved in closed form. However, you can estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at which the cubic equation crosses the x-axis is the solution.

(e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of your answer by substituting these concentrations into the equilibrium expression.)

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Textbook Question

At a temperature of 700 K, the forward and reverse rate constants for the reaction 2 HI(g) ⇌ H₂(g) + I₂(g) are k_f = 1.8×10⁻³⁰ M⁻¹s⁻¹ and k_r = 0.063 M⁻¹s⁻¹.

(a) What is the value of the equilibrium constant K_c at 700 K?

(b) Is the forward reaction endothermic or exothermic if the rate constants for the same reaction have values of k_f= 0.097M⁻¹s⁻¹ and k_r= 2.6 M⁻¹s⁻¹ at 800 K?