Problem 28 - Chemistry The Central Science In Si Units

Question

Consider the equilibrium $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) $$ Calculate the equilibrium constant \(K_{p}\) for this reaction, given the following information (at \(298 \mathrm{~K}\) ): $$ \begin{array}{lr} 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) & K_{c}=2.0 \\ 2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & K_{c}=2.1 \times 10^{30} \end{array} $$

Step-by-Step Solution

Answer

The equilibrium constant \(K_p\) is \(3.91 \times 10^{-32}\).

Step 1: Write the Given Reactions and Their Equilibrium Constants

We have two separate reactions: \(2\mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_{c1} = 2.0\) and \(2\mathrm{NO}(g) \rightleftharpoons \mathrm{N}_2(g) + \mathrm{O}_2(g)\) with \(K_{c2} = 2.1 \times 10^{30}\). Our goal is to find the equilibrium constant \(K_p\) for the new reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\).

Step 2: Derive Target Reaction

To derive the target reaction, we need to manipulate the given reactions. If we reverse reaction 2, it becomes \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\). This changes the equilibrium constant to \(K_{c2}' = \frac{1}{2.1 \times 10^{30}} = 4.76 \times 10^{-31}\).

Step 3: Combine Reactions

Combine the reversed reaction 2 and reaction 1:- \(\mathrm{N}_2(g) + \mathrm{O}_2(g) \rightleftharpoons 2\mathrm{NO}(g)\) with \(K_{c2}' = 4.76 \times 10^{-31}\)- \(2\mathrm{NO}(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_{c1} = 2.0\)The sum gives \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) with \(K_c = K_{c1} \times K_{c2}'\).

Step 4: Calculate \(K_c\) for Target Reaction

Calculate \(K_c\) for the target reaction using the formula:\[ K_c = K_{c1} \times K_{c2}' = 2.0 \times 4.76 \times 10^{-31} = 9.52 \times 10^{-31} \]

Step 5: Convert \(K_c\) to \(K_p\)

The relationship between \(K_c\) and \(K_p\) is given by \(K_p = K_c \times (RT)^{\Delta n}\), where \(\Delta n\) is the change in moles of gas. For our reaction, \(\Delta n = (2 - (1+1+1)) = -1\).Using \(R = 0.0821 \text{ L atm mol}^{-1} \text{K}^{-1}\) and \(T = 298 \text{ K}\), we find:\[ K_p = 9.52 \times 10^{-31} \times (0.0821 \times 298)^{-1} = 9.52 \times 10^{-31} \times 0.041 \]\[ K_p = 3.91 \times 10^{-32} \]

Step 6: Conclusion

The equilibrium constant \(K_p\) for the reaction \(\mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g)\) at 298 K is \(3.91 \times 10^{-32}\).

Key Concepts

Chemical Equilibrium

In chemistry, chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate. At this point, the concentrations of the reactants and products remain constant over time. Chemical equilibrium is crucial because it helps to predict the behavior of reactions in different conditions.
The equilibrium constant, represented as either \( K_c \) for concentrations or \( K_p \) for partial pressures, quantitatively describes the ratio of product and reactant concentrations at equilibrium. This is determined by the nature of the reaction and temperature.
For example, in our target reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g) \), we need to find the equilibrium constant to understand how much \( \mathrm{NOBr} \) is formed and how much \( \mathrm{N}_2, \mathrm{O}_2, \) and \( \mathrm{Br}_2 \) remain.

Reaction Manipulation

Reaction manipulation is a technique used to modify chemical equations to derive a desired target equation. In solving equilibrium problems, we often manipulate given reactions to find information about a related reaction.
There are several common operations used in reaction manipulation:

• Reversing a reaction, which involves swapping the reactants and products. When this is done, the equilibrium constant becomes the reciprocal of the original.
• Combining reactions, adding two or more reactions to get a net reaction. The equilibrium constant for the net reaction is the product of the constants for the individual steps.
• Multiplying a reaction by a coefficient scales the equilibrium constant exponentially.

In our example, reversing and combining given reactions allowed us to derive the reaction \( \mathrm{N}_2(g) + \mathrm{O}_2(g) + \mathrm{Br}_2(g) \rightleftharpoons 2\mathrm{NOBr}(g) \) and find its \( K_c \) value, which is crucial for further calculations.

Thermodynamics

Thermodynamics plays a vital role in understanding chemical equilibrium, especially when predicting how a reaction shifts with changes in temperature and pressure. The first and second laws of thermodynamics lay the foundation for calculating equilibrium constants.
One of the key insights is the relationship between \( K_c \) and \( K_p \), which represents concentrations and pressures, respectively. This relationship is given by: \[ K_p = K_c \times (RT)^{\Delta n} \]where \( R \) is the gas constant, \( T \) is the temperature in Kelvin, and \( \Delta n \) is the change in moles of gas from reactants to products.
In our reaction, \( \Delta n = -1 \) because we have fewer moles of gas on the product side than the reactant side. Using the values of \( R \) and \( T \), we can convert between \( K_c \) and \( K_p \), aiding in our comprehension of how pressure affects the reaction outcome.
This understanding is invaluable in many chemical processes, from industrial synthesis to biological systems.