Question

In a closed tube, \(\mathrm{HI}(\mathrm{g})\) is heated at \(440^{\circ} \mathrm{C}\) up to establishment of equilibrium. If it dissociates into \(\mathrm{H}_{2}(\mathrm{~g})\) and \(\mathrm{I}_{2}(\mathrm{~g})\) up to \(22 \%\), the dissociation constant is (a) \(0.282\) (b) \(0.0796\) (c) \(0.0199\) (d) \(1.99\)

Short answer

The dissociation constant is approximately \(0.0199\), which corresponds to option (c).

Step-by-step solution

Understand the Reaction and Define the Equilibrium Constant

Firstly, write the balanced chemical equation for the dissociation of hydrogen iodide: \(\mathrm{2HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)}\).Then, define the equilibrium constant expression based on the balanced equation:\(K_c = \frac{[\mathrm{H_2}][\mathrm{I_2}]}{[\mathrm{HI}]^2}\).

Set up the Initial and Equilibrium Concentrations

Assuming the initial concentration of \(\mathrm{HI}\) is 'a' mol/L, after 22% dissociation, the concentration of \(\mathrm{HI}\) will be \(0.78a\) mol/L at equilibrium. The concentration of \(\mathrm{H_2}\) and \(\mathrm{I_2}\) at equilibrium will be \(0.22a/2\) mol/L each, given that the stoichiometry of the reaction is 2:1:1.

Write the Equilibrium Concentrations in Terms of Initial Concentration

The equilibrium concentrations can now be represented as follows:\([\mathrm{HI}] = 0.78a\ [\mathrm{H_2}] = 0.11a\ [\mathrm{I_2}] = 0.11a\).

Substitute the Equilibrium Concentrations into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium constant expression:\(K_c = \frac{(0.11a)(0.11a)}{(0.78a)^2}\).

Calculate the Dissociation Constant

Calculate the value of \(K_c\):\(K_c = \frac{(0.11)^2}{(0.78)^2} \approx \frac{0.0121}{0.6084} \approx 0.0199\).

Key concepts

Chemical Equilibrium

When a chemical reaction occurs in a closed system and the reactants and products reach a state where their concentrations no longer change with time, the system is said to have reached chemical equilibrium. This does not mean the reaction has stopped; rather, it continues with forward and reverse reactions occurring at the same rate, leading to a constant ratio of product and reactant concentrations. This dynamic balance is crucial because it helps predict the concentrations of substances involved in reactions under constant conditions.

It's essential to understand that equilibrium can be reached from either direction of a reversible reaction, and once achieved, the system is in a state of balance. However, if any external condition such as temperature or pressure is altered, the position of equilibrium may shift, favoring either the forward or reverse reaction according to Le Chatelier's principle.

Equilibrium Constant Expression

The equilibrium constant expression (K_c) quantifies the position of equilibrium for a given chemical reaction. For the dissociation of hydrogen iodide, the expression is derived from the balanced chemical equation. It is important to note that each concentration is raised to the power of its respective coefficient in the balanced equation. The products are in the numerator, while the reactants are in the denominator.

To calculate K_c, plug in the equilibrium concentrations of the products and reactants into the expression. Keep in mind that the equilibrium constant is dimensionless and only includes substances that are gases or aqueous solutions. Pure solids and liquids do not appear in the expression, as their concentrations do not change.

Dissociation of Hydrogen Iodide

Hydrogen iodide (HI) is a gas that can undergo a reversible reaction to form hydrogen gas (H₂) and iodine gas (I₂). The degree of dissociation refers to the fraction of the original substance that has been transformed into products at equilibrium. This exercise explores what happens when 22% of HI dissociates.

In this case, the initial amount of HI decreases while the amounts of H₂ and I₂ increase proportionally, as they are produced from the dissociation of HI. The exercise demonstrates how these changes in concentration impact the calculation of the dissociation constant, K_c, which is a special case of the equilibrium constant for this particular reaction.

A clear understanding of this problem would also involve considering the effect of temperature, equilibrium shifts, and the stochiometry of the reaction on the dissociation constant. Remember that the amount of dissociation and the dissociation constant are intimately related but are not the same thing – the former is a percentage, while the latter is a ratio of product and reactant concentrations at equilibrium.