Q.At 500 K, equilibrium constant, Kc, for the following reaction is 5.
(1/2) H2

Concept: Equilibrium Constant Calculation

The equilibrium constant for a reaction changes predictably when the reaction equation is manipulated.

1. The given reaction is $\frac{1}{2} \text{H}_2 (g) + \frac{1}{2} \text{I}_2 (g) \rightleftharpoons \text{HI} (g)$, with $K_{c1} = 5$.
2. The target reaction, $2\text{HI} (g) \rightleftharpoons \text{H}_2 (g) + \text{I}_2 (g)$, is the reverse of the given reaction, and then multiplied by a factor of 2.
3. When a reaction is reversed, its equilibrium constant becomes the reciprocal of the original constant. So, for $\text{HI} (g) \rightleftharpoons \frac{1}{2} \text{H}_2 (g) + \frac{1}{2} \text{I}_2 (g)$, the constant is $K'_{c1} = \frac{1}{K_{c1}} = \frac{1}{5}$. …

The equilibrium constant, $K_c$, quantifies the ratio of products to reactants at equilibrium for a given reaction at a specific temperature. Its value depends on the stoichiometry of the balanced chemical equation. Understanding how $K_c$ changes when a reaction is manipulated (reversed, multiplied by a factor, or added to another reaction) is crucial for solving such problems.

In this problem, we are given the equilibrium constant for one reaction and asked to find the equilibrium constant for a related reaction. The key is to identify the relationship between the two reactions and apply the corresponding rules for manipulating $K_c$.

Here's how we approach this:

1. Identify the given reaction and its equilibrium constant.

   We are given the reaction:

   $(1/2) H_2 (g) + (1/2) I_2 (g) \rightleftharpoons HI (g)$

   The equilibrium constant for this reaction at 500 K is $K_{c1} = 5$.

2. Identify the target reaction for which we need to find the equilibrium constant.

   The target reaction is:

   $2HI (g) \rightleftharpoons H_2 (g) + I_2 (g)$

   Let's call its equilibrium constant $K_{c2}$.

3. Compare the target reaction with the given reaction to determine the necessary manipulations.

   Observe the two reactions:

   • The products of the given reaction ($HI$) are the reactants of the target reaction.
   • The reactants of the given reaction ($H_2$ and $I_2$) are the products of the target reaction.
   • The coefficients in the target reaction are twice the coefficients in the given reaction. For example, $HI$ has a coefficient of 1 in the given reaction and 2 in the target reaction. $H_2$ and $I_2$ have coefficients of $1/2$ in the given reaction and 1 in the target reaction.

   This means we need to perform two operations on the given reaction to obtain the target reaction:

   a. Reverse the reaction.

   b. Multiply the entire reaction by a factor of 2.

4. Apply the rule for reversing a chemical reaction.

   When a chemical reaction is reversed, the new equilibrium constant is the reciprocal of the original equilibrium constant.

   If $A \rightleftharpoons B$ has equilibrium constant $K$, then $B \rightleftharpoons A$ has equilibrium constant $1/K$.

   Let's reverse the given reaction:

   $HI (g) \rightleftharpoons (1/2) H_2 (g) + (1/2) I_2 (g)$

   The equilibrium constant for this reversed reaction, let's call it $K_{c}'$, will be:

   $K_{c}' = \frac{1}{K_{c1}} = \frac{1}{5}$

5. Apply the rule for multiplying a chemical reaction by a factor. …