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Given: (1) \(2 \mathrm{BrCl}(g) \leftrightharpoons \mathrm{Cl}_{2}(g)+\mathrm{Br}_{2}(g) \quad K_{\mathrm{P}_{1}}=0.45\) (2) \(2 \operatorname{IBr}(g) \leftrightharpoons \mathrm{Br}_{2}(g)+\mathrm{I}_{2}(g) \quad K_{\mathrm{P}_{2}}=21.0\) Determine the value of \(K_{\mathrm{p}}\) for the reaction equation $$ \mathrm{BrCl}(g)+\frac{1}{2} \mathrm{I}_{2}(g) \leftrightharpoons \operatorname{IBr}(g)+\frac{1}{2} \mathrm{Cl}_{2}(g) $$

Short Answer

Expert verified
\( K_p = \sqrt{0.45} \times \sqrt{\frac{1}{21}} \)

Step by step solution

01

Understanding the Task

We have been provided with two equilibrium reactions and their equilibrium constants. We are asked to find the equilibrium constant, \( K_p \), for a third reaction. The given reactions are: \( 2 \mathrm{BrCl}(g) \leftrightharpoons \mathrm{Cl}_2(g)+\mathrm{Br}_2(g) \) with \( K_{P_1} = 0.45 \), and \( 2 \operatorname{IBr}(g) \leftrightharpoons \mathrm{Br}_2(g)+\mathrm{I}_{2}(g) \) with \( K_{P_2} = 21.0 \). We need to express the desired reaction using combinations of the given reactions.
02

Analyze the Desired Reaction

The reaction for which we have to find \( K_p \) is \( \mathrm{BrCl}(g) + \frac{1}{2} \mathrm{I}_{2}(g) \leftrightharpoons \operatorname{IBr}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \). This reaction involves \( \mathrm{BrCl}, \mathrm{I}_2, \mathrm{IBr} \), and \( \mathrm{Cl}_2 \). Our goal is to manipulate the given reactions to achieve this reaction.
03

Manipulating Reaction (1)

To use Reaction (1), we divide it by 2: \( \mathrm{BrCl}(g) \leftrightharpoons \frac{1}{2}\mathrm{Cl}_2(g) + \frac{1}{2}\mathrm{Br}_2(g) \). When a reaction is multiplied or divided by a factor, we raise \( K_p \) to the power of that factor, so \( K_{p1}' = (0.45)^{1/2} \).
04

Manipulating Reaction (2)

To form the desired reaction, reverse Reaction (2) and divide by 2: \( \mathrm{Br}_2(g) + \frac{1}{2} \mathrm{I}_{2}(g) \leftrightharpoons \operatorname{IBr}(g) \). When a reaction is reversed, we take the reciprocal of \( K_p \), and dividing by 2 involves taking the square root, so \( K_{p2}' = (1/21.0)^{1/2} \).
05

Combining the Modified Reactions

Adding the two modified reactions from Step 3 and Step 4 gives us the desired reaction:\[ \mathrm{BrCl}(g) + \frac{1}{2} \mathrm{I}_{2}(g) \leftrightharpoons \operatorname{IBr}(g) + \frac{1}{2} \mathrm{Cl}_{2}(g) \]. The equilibrium constant for this reaction is \( K_p \), which is the product of \( K_{p1}' \) and \( K_{p2}' \).
06

Calculate the Final Equilibrium Constant

Thus, \( K_p = (0.45)^{1/2} \times (1/21.0)^{1/2} \). Calculate \( K_p \) using these values.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant
The concept of an equilibrium constant, denoted as \( K \), is pivotal in understanding chemical reactions at equilibrium. It provides a quantitative measure of the concentrations of reactants and products in a reaction mixture where the rates of the forward and reverse reactions are equal. For a generic reaction \( aA + bB \leftrightharpoons cC + dD \), the equilibrium constant expression is represented as:
  • \( K = \frac{{[C]^c[D]^d}}{{[A]^a[B]^b}} \)
In this expression, square brackets indicate the concentration of substances, and the coefficients \( a, b, c, \) and \( d \) are the stoichiometric coefficients from the balanced equation.

In the case of gases, the equilibrium constant is often expressed in terms of partial pressures and denoted as \( K_p \). It's crucial for predicting reaction direction and extent.

A reaction with a large \( K \) favours product formation, indicating equilibrium lies to the right, while a small \( K \) suggests equilibrium favours the reactants. Understanding \( K \) allows chemists to manipulate conditions to drive reactions in the desired direction.
Manipulation of Equilibrium Expressions
When solving problems involving multiple reactions, understanding how to manipulate equilibrium expressions is essential. This involves performing operations such as multiplication, division, and reversal of reactions, each of which have specific effects on the equilibrium constant.
  • **Multiplying or Dividing**: If a reaction is multiplied or divided by a factor, the equilibrium constant is raised to the power of that factor. For example, if you have \( K \) for a reaction, dividing by 2 would change \( K \) to \( K^{1/2} \).
  • **Reversing**: Reversing a reaction inverts the equilibrium constant, changing it to its reciprocal (\( 1/K \)).
In the provided exercise, we modified two given reactions to achieve a target reaction. This was done by dividing the first by 2 and reversing and dividing the second by 2. The resultant equilibrium constants for these modified reactions were \( K'_{p1} = (0.45)^{1/2} \) and \( K'_{p2} = (1/21.0)^{1/2} \).

Combining these operations allows for the calculation of the overall \( K_p \) for the desired reaction by multiplying the modified constants:

  • \( K_p = K'_{p1} \times K'_{p2} \)
This method is highly useful in multi-step synthesis and analysis of complex chemical reactions.
Reaction Mechanism
Reaction mechanisms are sequences of elementary steps that make up a complex reaction, offering insights into the pathway a reaction follows. Each step in the mechanism is an individual reaction, and the combination of these steps accounts for the overall transformation observed in a chemical reaction.

In analyzing equilibrium reactions, it is important to note that no matter how complex or how many steps are involved, the principle of equilibrium applies to each step, and the equilibrium constant remains a defining feature.

Understanding reaction mechanisms is critical for determining the rate-determining step, or the slowest step in the sequence that limits the overall reaction rate. While the provided exercise focused primarily on manipulating equilibrium expressions, discerning reaction mechanisms can provide additional layers of understanding.
  • By knowing the individual steps, chemists can identify intermediates and transition states, allowing for more refined control over reaction conditions.
  • Mechanisms help predict the effects of changing concentration, pressure, or temperature on the reaction rate and equilibrium position.
Mastering these concepts enhances the ability to optimize chemical reactions for industrial applications and research.

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Most popular questions from this chapter

Consider the chemical equilibrium described by the equation $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-74.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ Predict the way in which the equilibrium will shift in response to each of the following changes in conditions (if the equilibrium is unaffected by the change, then write no change): (a) decrease in temperature (b) decrease in reaction volume (c) decrease in \(\bar{P}_{\mathrm{H}_{2}}\) (d) increase in \(\bar{P}_{\mathrm{CH}_{4}}\) (e) addition of \(\mathrm{C}(s)\)

Prior to learning about equilibrium states, we solved stoichiometric problems using the concept of "limiting reactants." Under what conditions does the method of limiting reactants apply?

For the reaction described by $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \leftrightharpoons 2 \mathrm{HI}(g) $$ the value of \(K_{\mathrm{p}}\) is \(54.4\) at \(355^{\circ} \mathrm{C}\). What percentage of \(\mathrm{I}_{2}(g)\) will be converted to \(\mathrm{HI}(g)\) if \(0.200\) moles each of \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) are mixed and allowed to come to equilibrium at \(355^{\circ} \mathrm{C}\) in a \(1.00\) -liter container?

An equilibrium mixture of \(\mathrm{CO}(g), \mathrm{Cl}_{2}(g)\), and \(\mathrm{COCl}_{2}(g)\) has partial pressures \(P_{\mathrm{co}}=P_{\mathrm{Cl}_{2}}=1.09 \mathrm{bar}\) and \(P_{\mathrm{Cocl}_{2}}=0.144\) bar. A quantity of \(\mathrm{CO}(g)\) is suddenly injected into the reaction vessel and the total pressure jumps to \(3.31\) bar. Calculate the total pressure after equilibrium is reestablished. The relevant chemical equation is $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \leftrightharpoons \operatorname{COCl}_{2}(g) $$

Consider the methanation reaction described by the chemical equation $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ It was found that \(0.618\) moles of \(\mathrm{CO}(g), 0.387\) moles of \(\mathrm{H}_{2}(\mathrm{~g}), 0.387 \mathrm{moles}\) of \(\mathrm{CH}_{4}(g)\), and \(0.387\) moles of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) were present in an equilibrium mixture in a 1.00-liter container. All the water vapor was removed and the system allowed to come to equilibrium again. Calculate the concentration of all gases in the new equilibrium system. (You must solve the resulting equation by trial and error.)

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