Question

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

Short answer

The equilibrium constant \(K_p\) for the given reaction is approximately \(3.85 \times 10^{-34}\).

Step-by-step solution

Manipulate given reactions

We have the following two given reactions: 1. \(2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)\) with \(K_{c1}=2.0\) 2. \(2 \mathrm{NO}(g) \rightleftharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)\) with \(K_{c2}=2.1 \times 10^{30}\) We will reverse the second reaction and divide it by 2 to obtain: 2'. \(\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) with \(K_{c2'}=\dfrac{1}{K_{c2}}=\dfrac{1}{2.1 \times 10^{30}}\) Now, let's add the first reaction and 2' reaction: \[ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+ \cancel{2NO(g)}+ \mathrm{Br}_{2}(g) \rightleftharpoons \cancel{2 \mathrm{NO}(g)}+2 \operatorname{NOBr}(g)\] This results in the desired reaction: \[\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{NOBr}(g) \]

Calculate the equilibrium constant for the desired reaction

Since the reactions were combined by addition, the equilibrium constants are multiplied: \[K_{c} = K_{c1} \times K_{c2'} = 2.0 \times \dfrac{1}{2.1 \times 10^{30}} \] Now calculate the value of \(K_c\): \[K_{c} = \dfrac{2.0}{2.1 \times 10^{30}} = \dfrac{1}{1.05 \times 10^{30}} \]

Convert \(K_c\) to \(K_p\)

To convert \(K_c\) to \(K_p\), we can use the following relation: \[K_p = K_c(RT)^{\Delta n}\] where R: Gas constant = \(8.314~\mathrm{J~mol^{-1}K^{-1}}\) T: Temperature = 298K \(\Delta n\): Change in the number of moles of gas between reactants and products The change in the number of moles of gas: \[\Delta n = (\mathrm{n_{NOBr}} \cdot 2) - (\mathrm{n_{N_2}} + \mathrm{n_{O_2}} + \mathrm{n_{Br_2}}) = (2 -1 - 1 - 1) = -1\] Now calculate \(K_p\): \[K_p = K_c(RT)^{\Delta n} = \dfrac{1}{1.05 \times 10^{30}} \times (8.314 \times 298)^{-1} \] \[K_p = \dfrac{1}{1.05 \times 10^{30}} \times (2.474 \times 10^{3})^{-1}\] \[K_p = \dfrac{1}{1.05 \times 10^{30} \times 2.474 \times 10^{3}} \] \[K_p = 3.85 \times 10^{-34}\] Hence, the equilibrium constant \(K_p\) for the given reaction is approximately \(3.85 \times 10^{-34}\).

Key concepts

Understanding Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, allowing the concentrations of reactants and products to remain constant over time. This dynamic state does not mean the reactions have stopped, but that they are occurring at a rate that balances each other out.

For example, in the synthesis of NOBr from N2, O2, and Br2, the system reaches equilibrium when the formation rate of NOBr equals its decomposition rate back to N2, O2, and NO. An important thing to note is that equilibrium can be achieved from either direction of a reversible reaction and the concentrations of reactants and products do not have to be equal, but are fixed at a particular ratio.

Le Chatelier's Principle

Le Chatelier's principle provides insight into how a system at equilibrium responds to disturbances. It states that if an external change is applied to a system at equilibrium, the system adjusts in such a way as to counteract the imposed change and a new equilibrium is established.

These changes can include variations in concentration, pressure, volume, or temperature. In the context of our reaction example, if the concentration of one of the reactants is increased, the system will respond by shifting the equilibrium to the right, producing more NOBr to restore equilibrium. Conversely, decreasing the concentration of one of the products would cause the system to shift to the left, to produce more reactants.

Reaction Quotient and its Role

The reaction quotient, Q, predicts the direction in which a reaction mixture will shift to reach equilibrium. It is calculated using the same expression as the equilibrium constant (K), but with the initial concentrations of the reactants and products instead of the equilibrium concentrations.

When Q is compared to K, it tells us if the reaction is at equilibrium (Q = K), or if it will shift to the right (Q < K) to produce more products, or to the left (Q > K) to produce more reactants. In the given exercise, calculating Q would determine whether the reaction needs to shift towards the formation of NOBr or the formation of N2, O2, and Br2 prior to reaching equilibrium.

Partial Pressure in Equilibrium

Partial pressure, an important concept for gases, is the pressure which a gas would exert if it alone occupied the volume of the mixture at the same temperature. In the context of equilibrium, the equilibrium constant expressed in terms of partial pressure is denoted as Kp.

For gas-phase equilibria, the relationship between Kc (equilibrium constant in terms of concentration) and Kp is given by the equation Kp = Kc(RT)^(Δn), where R is the ideal gas constant, T is the temperature, and Δn is the change in moles of gas. In the given solution, we saw how Kc was converted to Kp by considering the change in the number of moles of gas when proceeding from reactants to products.