Problem 40 - Chemistry: An Atoms-Focused Approach

Question

Is the numerical value of \(K_{p}\) for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) $$ greater than, equal to, or less than the value of the equilibrium constant for the following reaction? $$ \frac{1}{2} \mathrm{H}_{2}(z)+\frac{1}{2} \mathrm{I}_{2}(g) \rightleftharpoons \mathrm{HI}(g) $$

Step-by-Step Solution

Answer

The numerical value of the equilibrium constant (\(K_p\)) for the first reaction (\(K_{p1}\)) is less than the value for the second reaction (\(K_{p2}\)) as denoted by the following relationship: $$K_{p1} = K_{p2}^{(-1/2)}$$

Step 1: Write given information

The given reactions are: 1. \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2 HI(g)\), with equilibrium constant \(K_{p1}\) 2. \(\frac{1}{2} H_{2}(g) + \frac{1}{2} I_{2}(g) \rightleftharpoons HI(g)\), with equilibrium constant \(K_{p2}\) We need to compare \(K_{p1}\) and \(K_{p2}\).

Step 2: Rewrite second reaction with whole number coefficients

To ease comparison, let's rewrite the second reaction as $$H_{2}(g) + I_{2}(g) \rightleftharpoons 2 HI(g) - H_{2}(g) - I_{2}(g) + HI(g)$$ which simplifies to $$HI(g) \rightleftharpoons H_{2}(g) + I_{2}(g)$$ with the equilibrium constant \(K'_{p2}\).

Step 3: Determine the relationship between \(K_{p2}\) and \(K'_{p2}\)

The second and third equations are related by a factor of 2, such that the third equation is half the second equation. The relationship between the equilibrium constants is given by: $$K'_{p2} = \sqrt[n]{K_p}$$ where \(n\) is the factor by which one equation is multiplied or divided to form the other. For this case, we have: $$K'_{p2} = \sqrt[2]{K_{p2}}$$

Step 4: Relate \(K_{p1}\) and \(K'_{p2}\)

Now, we relate the equilibrium constants for the first and third equations: $$\frac{1}{K_{p1}} = K'_{p2}$$ which can be rewritten as: $$K_{p1} = \frac{1}{K'_{p2}}$$

Step 5: Determine the relationship between \(K_{p1}\) and \(K_{p2}\)

Using the relationships derived in step 3 and 4, we substitute \(K'_{p2}\): $$K_{p1} = \frac{1}{\sqrt[2]{K_{p2}}}$$ or $$K_{p1} = K_{p2}^{(-1/2)}$$

Step 6: Compare \(K_{p1}\) and \(K_{p2}\)

Since the exponent in the above relationship is negative, the value of \(K_{p1}\) will be smaller if the value of \(K_{p2}\) is greater than 1, and vice versa. Therefore, we can conclude that the numerical value of \(K_{p1}\) is less than the value of \(K_{p2}\) in this case.

Key Concepts

Equilibrium Constant

The equilibrium constant, represented by the symbol \(K\), is a fundamental concept in chemical equilibrium. It reflects the ratio of the concentrations of products to reactants at equilibrium for a given reaction. For reactions involving gases, the equilibrium constant in terms of partial pressures is denoted as \(K_p\). This value is a fixed quantity at a specific temperature for any given reaction. However, it varies if the reaction is modified, like in the case of different stoichiometric coefficients.
For example, consider the reactions:

• \(H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)\) - equilibrium constant \(K_{p1}\)
• \(\frac{1}{2} H_2(g) + \frac{1}{2} I_2(g) \rightleftharpoons HI(g)\) - equilibrium constant \(K_{p2}\)

When stoichiometry changes, like in the second reaction which is half of the first, it affects the equilibrium constant. The mathematical relationship adjusts accordingly, showing how equilibrium constants can transform with the equation's manipulation.

Reaction Quotients

Reaction quotients, denoted as \(Q\), are used to determine the direction in which a reaction will proceed before reaching equilibrium. They are calculated similarly to equilibrium constants but with initial concentrations of the reactants and products. When a reaction is at equilibrium, \(Q = K\).
Here’s a comparison of \(Q\) and \(K\):

• If \(Q < K\), the reaction will proceed forward, converting reactants into products until equilibrium is attained.
• If \(Q > K\), the reaction will proceed in reverse, converting products back into reactants.
• If \(Q = K\), the system is already at equilibrium, and no net change will occur.

Understanding \(Q\) helps predict the shift required to reach equilibrium, offering a snapshot of the present state relative to equilibrium.

Le Chatelier's Principle

Le Chatelier's Principle provides insight into how a chemical system at equilibrium responds to disturbances. The principle states that if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust to counteract the effect of the disturbance.
Key considerations include:

• Adding or removing reactants or products: The system shifts in the direction that uses up the added component or replenishes the removed one.
• Changing temperature: If the reaction is exothermic, increasing temperature will shift the equilibrium to favor the reactants. If endothermic, it will favor the products.
• Changing pressure: For gaseous reactions, increasing pressure will shift equilibrium towards the side with fewer gas molecules, and vice versa.

Understanding Le Chatelier's Principle is crucial for predicting how changes will affect a reaction at equilibrium, ensuring balance maintains within the system.