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 \(298 \mathrm{~K}\) ): \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g) \quad K_{c}=2.0\) \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \quad K_{c}=2.1 \times 10^{30}\)

Short answer

The short answer for calculating the equilibrium constant \(K_{p}\) for the given reaction is: 1. Reverse Reaction 2 and add it with Reaction 1 to get the Target Reaction 2. Multiply the equilibrium constants Kc of the two reactions used: \(K_c (Target Reaction) = (1/(2.1\times10^{30}))\times2.0=9.52\times10^{-31}\) 3. Convert Kc to Kp: \(K_p=(9.52\times10^{-31})(0.0821\times298)^{-1}=16.4\times10^{-29}\) Therefore, the equilibrium constant \(K_{p}\) for the given reaction is approximately \(1.64\times10^{-28}\).

Step-by-step solution

Identify the related reactions

We have been given two reactions: Reaction 1: \(2 NO(g) + Br_2(g) \rightleftharpoons 2 NOBr(g) \quad K_{c1}=2.0\) Reaction 2: \(2 NO(g) \rightleftharpoons N_2(g) + O_2(g) \quad K_{c2}=2.1 \times 10^{30}\) We need to use these two reactions to come up with the given reaction, which is: Target Reaction: \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\)

Combine the given reactions to get the target reaction

Reverse Reaction 2, and then add it with Reaction 1, to arrive at the Target Reaction: Reverse Reaction 2: \(N_2(g) + O_2(g) \rightleftharpoons 2 NO(g) \quad K_{c2}^{-1}= 1 / (2.1 \times 10^{30})\) Add: Reverse Reaction 2 + Reaction 1 \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\)

Calculate Kc for target reaction from given Kc values

Multiply the equilibrium constants (Kc) of the two reactions used: Kc (Target Reaction) = Kc (Reverse Reaction 2) × Kc (Reaction 1) Kc (Target Reaction) = \( (1 / (2.1 \times 10^{30})) × 2.0 \) Kc (Target Reaction) = \(9.52 \times 10^{-31}\)

Convert Kc to Kp for the target reaction

Use the following formula to convert Kc to Kp for the target reaction: \[K_p = K_c (RT)^{\Delta n}\] R is the ideal gas constant (0.0821 L atm / (mol K)), T is the temperature (298 K), and Δn is the change in moles of gas in the reaction (calculated as the difference between the moles of products and reactants). Δn = moles of products - moles of reactants = 2 (NOBr) - (1 N2 + 1 O2 + 1 Br2) = 2 - 3 = -1 Thus, \(K_p = (9.52 \times 10^{-31}) (0.0821 \times 298)^{-1}\) \(K_p = 16.4 \times 10^{-29}\) The equilibrium constant Kp for the given reaction is approximately \(1.64 \times 10^{-28}\).

Key concepts

Reaction Mechanism

In chemistry, understanding the path that a reaction takes, from reactants to products, involves exploring the reaction mechanism. It is essentially the series of elementary steps or processes that make up the overall reaction. The exercise we are discussing involves combining and reversing reactions to reach the target equation, showcasing an application of reaction mechanisms.
To derive the target reaction:

• Reverse Reaction 2, which initially converts nitrogen (\(N_2\)) and oxygen (\(O_2\)) gases into 2 moles of nitric oxide (\(NO\)).
• Add this reversed reaction to Reaction 1, which already includes bromine (\(Br_2\)) and nitric oxide.

By reversing and combining these reactions, we align our approach to fit the original mechanism that forms the target reaction: \(N_2(g) + O_2(g) + Br_2(g) \rightleftharpoons 2 NOBr(g)\). This demonstrates how individual steps can be adjusted and joined to explain a chemical process.

Chemical Equilibrium

Chemical equilibrium represents a state where the forward and reverse reactions occur at equal rates, leading to constant concentrations of products and reactants. This concept is essential for understanding how equilibrium constants like \(K_c\) and \(K_p\) are calculated and used.
In the problem, the exercise focuses on reaching equilibrium for the given target reaction, using known equilibrium constants from related reactions. When reactions are combined, the equilibrium constant for the overall process is found by multiplying the constants from each step, considering any necessary inversions.

• The steps involve reversing Reaction 2, creating an inverse \(K_c\) value.
• Multiplying the resulting \(K_c\) from the reversed reaction with the \(K_c\) from Reaction 1 gives the overall \(K_c\) for the target reaction.

Therefore, chemical equilibrium simplifies finding the overall \(K_c\), crucial for converting to \(K_p\) and understanding the dynamics of gaseous reactions at a given temperature.

Gas Laws

Gas laws provide a foundation for understanding how gases behave under different conditions, and they are closely related to calculating equilibrium constants in gaseous reactions. In the exercise, converting \(K_c\) to \(K_p\) for the target reaction involves application of these laws.
The conversion from \(K_c\) to \(K_p\) requires the equation:\[K_p = K_c \times (RT)^{\Delta n}\]where \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. The change in moles \(\Delta n\) reflects the difference in the mole quantity of gaseous products and reactants.

• From the balanced equation, \(\Delta n\) for the target reaction is \(-1\), indicating one less mole of gas results after reaction compared to before.
• This negative \(\Delta n\) implies a conversion factor less than one, affecting \(K_p\).

Understanding these principles helps accurately calculate the equilibrium constant \(K_p\), reflecting the gas behavior defined by the laws, especially under conditions specified by temperature and volume changes.