Question

Consider the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If nitrosyl bromide, NOBr, is 34 percent dissociated at \(25^{\circ} \mathrm{C}\) and the total pressure is \(0.25 \mathrm{~atm},\) calculate \(K_{P}\) and \(K_{\mathrm{c}}\) for the dissociation at this temperature.

Short answer

The equilibrium constants \(K_c\) and \(K_P\) for the dissociation of NOBr gas are both equal to 0.3341

Step-by-step solution

Calculate concentrations at equilibrium

Initially, let's say, there is 'a' moles of NOBr. Thus, at equilibrium, there is \(a - 2x\) moles of NOBr, \(2x\) moles of NO, and \(x\) moles of Br2 present. Given that NOBr is 34% dissociated, \(x = 0.34a\). Let's denote our total concentration at equilibrium, \(C = a/V\), where \(V\) is the volume.

Calculate equilibrium constant \(K_c\)

The equilibrium constant \(K_c\) is defined as \[K_c = \frac{[NO]^2[Br2]}{[NOBr]^2}\] Substituting the expressions we found for the equilibrium concentrations of the substances in above equilibrium equation, we find \[K_c = \frac{(0.34C)^2 * 0.34C}{(0.66C)^2} = 0.3341\]

Calculate partial pressures at equilibrium

Partial pressures can be calculated using the formula: \(P = \frac{n}{V}RT\), where n is the number of moles, V is the volume, R is the universal gas constant and T is the temperature. Using the total pressure \(P_{total} = 0.25\) atm, the partial pressures of NO, Br2, and NOBr are \(P_{NO} = 2xP_{total}\), \(P_{Br2} = xP_{total}\) and \(P_{NOBr} = 2(1-x)P_{total}\) respectively.

Calculate \(K_P\)

The equilibrium constant in terms of pressure, \(K_P\), can be calculated as \(K_P = \frac {P_{NO}^2 * P_{Br2}}{P_{NOBr}^2} \). Substituting the partial pressures we found into the equation gives \(K_P = \frac {(0.68P_{total})^2 * 0.34P_{total}}{(1.32P_{total})^2} = 0.3341\).

Comparing \(K_P\) and \(K_c\)

In this case, because the total moles of gas don't change from reactants to products, \(K_P\) is equal to \(K_c\).

Key concepts

Equilibrium Constant (Kc)

The equilibrium constant (Kc) is a fundamental concept in chemistry that represents the ratio of product concentrations to reactant concentrations for a chemical reaction at equilibrium. Specifically, it applies to reactions taking place in a liquid or solution, where concentrations are measured in moles per liter. The value of Kc is temperature-dependent and does not change unless the temperature does.

For the balanced reaction \[2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)\], the Kc expression would be: \[K_c = \frac{[NO]^2[Br_2]}{[NOBr]^2}\]
It's important to plug in the equilibrium concentrations of the reactants and products to calculate Kc. As seen in the step-by-step solution, we can utilize the degree of dissociation to determine these concentrations and find that Kc is a fixed value, showing the ratio of products to reactants at equilibrium.

Partial Pressure (Kp)

In gases, the concept of equilibrium also applies, and the equilibrium constant can be expressed in terms of partial pressures, known as Kp. The partial pressure of a gas in a mixture is the pressure it would exert if it occupied the entire volume alone. In the step-by-step solution, the partial pressure is calculated using the equation \(P = \frac{n}{V}RT\), where n is the number of moles and V is volume.

To calculate the equilibrium constant for gas-phase reactions (Kp), we use partial pressures instead of concentrations. The equation for our example is: \[K_P = \frac {P_{NO}^2 \cdot P_{Br2}}{P_{NOBr}^2} \]
The solution shows that Kp and Kc are equal in this particular problem because the change in the number of moles of gas during the reaction is zero. However, this is not always the case; generally, these two constants are related by the equation: \[K_P = K_c(RT)^{\Delta n}\], where \(\Delta n\) is the change in moles of gas and R is the gas constant.

Dissociation in Chemistry

Dissociation is the process where a compound breaks down into its constituent parts, which can be atoms, ions, or smaller molecules. In the context of the given exercise and their solution, dissociation refers to the partial breakdown of nitrosyl bromide (NOBr) into nitric oxide (NO) and bromine (Br2).

We are given that 34% of NOBr is dissociated, which is crucial information needed to calculate the equilibrium concentrations of all species involved. This percentage allows us to establish a relationship between the initial amount of NOBr and the amount present at equilibrium along with the amount of products formed. Understanding how to calculate the degree of dissociation aids in solving for Kc and Kp.

Le Chatelier's Principle

Le Chatelier's principle is a qualitative tool in chemistry that helps predict how a system at equilibrium reacts to external changes. The principle states that if a system in a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract the change and a new equilibrium will be established.

Changes can include alterations in concentration, pressure, or temperature. For example, if the pressure is increased on a system where gases are in equilibrium, the system shifts towards the side with fewer moles of gas to reduce pressure. Conversely, if the temperature increases, the system will favor the endothermic reaction to absorb excess heat.

In the exercise presented, if one were to change the temperature or the total pressure, Le Chatelier's principle would guide us in predicting whether the dissociation would increase or decrease. However, since we are dealing with a constant temperature and pressure scenario, the principle tells us that the equilibrium constants (Kc and Kp) will solely depend on temperature and will remain constant without external perturbations.