Question

At a certain temperature, \(K_{\text {eq }}\) is \(2.2 \times 10^{-3}\) for the reaction $$ 2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Now calculate the \(K_{\text {eq }}\) value for the reaction $$ \mathrm{I}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{ICl}(g) $$

Short answer

\( K_{eq} \) for the reverse reaction is \( \frac{1}{2.2 \times 10^{-3}} \).

Step-by-step solution

Understand the Relationship Between the Two Reactions

Identify that the second reaction given is the reverse of the first reaction. In equilibrium, if the direct reaction has an equilibrium constant of \( K_{eq} \), the reverse reaction will have an equilibrium constant that is the reciprocal of that value.

Calculate the Reverse Reaction's Equilibrium Constant

Take the reciprocal of the given equilibrium constant \( K_{eq} \) for the direct reaction to find the equilibrium constant for the reverse reaction. The calculation will be \( K_{eq \text{ reverse}} = \frac{1}{K_{eq \text{ direct}}} \).

Key concepts

Equilibrium Constant Calculation

Understanding how to calculate the chemical equilibrium constant, denoted as \(K_{eq}\), is fundamental to the study of reversible chemical reactions. The equilibrium constant represents the ratio of the product concentrations to the reactant concentrations, each raised to the power of their respective coefficients in the balanced chemical equation, when the reaction has reached equilibrium.

For instance, consider the reaction for which the equilibrium constant \(K_{eq}\) is given as \(2.2 \times 10^{-3}\):
\[2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g) + \mathrm{Cl}_{2}(g)\]
Calculating \(K_{eq}\) involves using the concentrations of the products (here, \(\mathrm{I}_{2}(g)\) and \(\mathrm{Cl}_{2}(g)\)) divided by the concentrations of the reactants (\(\mathrm{ICl}(g)\)), with each concentration raised to the power of the coefficient from the balanced equation.

Reciprocal of Equilibrium Constant

The relationship between forward and reverse reactions and their corresponding equilibrium constants is critical for fully grasping chemical equilibria. When you are given the equilibrium constant for a forward reaction and need to find the equilibrium constant for the reverse reaction, you simply take the reciprocal of the forward equilibrium constant. The reciprocal denotes the inverse value of the given constant.

For the reaction:
\[2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g) + \mathrm{Cl}_{2}(g)\]
with \(K_{eq} = 2.2 \times 10^{-3}\), the equilibrium constant for the reverse reaction is calculated by taking the reciprocal of this value:
\[K_{eq \text{ reverse}} = \frac{1}{2.2 \times 10^{-3}}\]
As such, calculating the reciprocal provides the magnitude of \(K_{eq}\) for the process proceeding in the opposite direction.

Reversible Chemical Reactions

Reversible chemical reactions are processes that can occur in both the forward and reverse directions, achieving a state of dynamic equilibrium where the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of the reactants and products remain constant over time, though both reactions continue to occur.

In reversible chemical reactions, such as:
\[2 \mathrm{ICl}(g) \rightleftharpoons \mathrm{I}_{2}(g) + \mathrm{Cl}_{2}(g)\]
the concept of chemical equilibrium plays a vital role, and the equilibrium constant \(K_{eq}\) quantifies the extent to which a reaction prefers the formation of products or reactants under a given set of conditions. Higher values of \(K_{eq}\) suggest a greater proportion of products under equilibrium, while lower values indicate a reaction that favors reactants.