Problem 41 - Chemistry: An Atoms-Focused Approach

Question

\(K_{c}=120\) for the following reaction at \(425 \mathrm{K}\) $$ \mathrm{I}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{IBr}(g) $$ What is the value of \(K_{e}\) for the following reaction, also at \(425 \mathrm{K} ?\) $$ \frac{1}{2} \mathrm{I}_{2}(g)+\frac{1}{2} \mathrm{B} \mathrm{r}_{2}(g) \rightleftharpoons \operatorname{IBr}(g) $$

Step-by-Step Solution

Answer

Answer: The value of the equilibrium constant (\(K_e\)) for the second reaction at 425 K is approximately 10.95.

Step 1: Write down the given reaction and its equilibrium constant

First, let's write down the given reaction and its equilibrium constant: $$ I_2(g) + Br_2(g) \rightleftharpoons 2 IBr(g) \qquad K_c = 120 $$

Step 2: Determine the relationship between the two reactions

Next, let's analyze the relationship between the given reaction and the reaction for which we need to calculate the equilibrium constant \(K_e\). Clearly, the second reaction has half the number of reactants and products compared to the first reaction: $$ \frac{1}{2}I_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons IBr(g) $$ We can see that this reaction is just a half of the initial reaction.

Step 3: Determine the formula to relate the equilibrium constants

The relationship between equilibrium constants for two reactions can be determined by examining the stoichiometry of the reactions. In this case, the second reaction is just half of the first reaction. Therefore, we can write the relation between \(K_c\) and \(K_e\) as: $$ K_e = (K_c)^\frac{1}{2} $$

Step 4: Calculate the value of \(K_e\)

Finally, we can use the formula derived in Step 3 and the given value of \(K_c\) to calculate the value of \(K_e\): $$ K_e = (120)^\frac{1}{2} $$ $$ K_e = \sqrt{120} \approx 10.95 $$ So the value of \(K_e\) for the second reaction at 425 K is approximately 10.95.

Key Concepts

Equilibrium Constant

In chemistry, the equilibrium constant, denoted by the symbol \( K \), is a crucial parameter used to describe the state of a reversible reaction at equilibrium. It provides insight into the extent to which products and reactants are present in the reaction mixture when the reaction has reached a point of balance. Understanding \( K \) involves:

• Knowing that it depends on the concentrations of the reactants and products.
• Recognizing that it is specific to a given reaction at a particular temperature.

For a general reaction of the form \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_c \) is expressed as:\[K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
Here, \([C]\), \([D]\), \([A]\), and \([B]\) are the equilibrium concentrations of the products and reactants respectively, and \(a\), \(b\), \(c\), and \(d\) are their stoichiometric coefficients in the balanced equation.In the case of our problem, the given \( K_c = 120 \) reflects the ratio of formation of the new compound from iodine and bromine gases. By comparing other reactions to this one, we can use mathematical relationships based on stoichiometry to find new equilibrium constants.

Stoichiometry

Stoichiometry is the foundation of chemical calculations and involves the quantitative relationships between the reactants and products in a chemical reaction. It uses the coefficients from a balanced chemical equation to allow us to calculate the amounts of substances involved.In our exercise:

• The initial reaction is \( I_2(g) + Br_2(g) \rightleftharpoons 2 IBr(g) \), where the stoichiometric coefficients are 1 for each reactant and 2 for the product.
• The modified reaction is \( \frac{1}{2}I_2(g) + \frac{1}{2}Br_2(g) \rightleftharpoons IBr(g) \) with coefficients each halved.

This means the second reaction is derived by taking half of each part of the initial reaction. Mathematically, this alters the equilibrium constant because changing stoichiometry affects how concentrations of substances are represented. By understanding stoichiometry, students can see why the relation \( K_e = (K_c)^{1/2} \) is used for the new reaction, indicating how stoichiometric changes require adjusting the equilibrium constant.

Reaction Quotient

The reaction quotient, denoted by \( Q \), is a crucial tool for determining the direction in which a reaction mixture will proceed to reach equilibrium. It is calculated using the same expression as the equilibrium constant \( K \), but at any point in time before the system reaches equilibrium.So, for a reaction of the form \( aA + bB \rightleftharpoons cC + dD \), \( Q \) is given by:\[Q = \frac{[C]^c [D]^d}{[A]^a [B]^b}\]
Depending on its value, \( Q \) can inform us about the reaction’s progress:

• If \( Q < K \), the reaction will proceed in the forward direction to produce more products.
• If \( Q > K \), the reaction will proceed in the reverse direction to produce more reactants.
• If \( Q = K \), the reaction is at equilibrium, and no net change occurs in the concentrations of reactants and products.

Although the exercise does not explicitly calculate \( Q \), understanding this concept helps differentiate between conditions at equilibrium \( (Q = K) \) and non-equilibrium states, underpinning why equilibrium constants like \( K_c \) are used to describe specific reactions and their extents under equilibrium conditions.