Problem 76 - Chemistry: The Central Science in SI Units, Global Edition

Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g).$$ An equilibrium mixture in a 5.00-L vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of NOBr, \(3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) of \(\mathrm{Br}_{2}\). (a) Calculate \(K_{c}\). (b) What is the total pressure exerted by the mixture of gases? (c) What was the mass of the original sample of NOBr?

Step-by-Step Solution

Answer

The equilibrium constant Kc for the decomposition of nitrosyl bromide (NOBr) into nitric oxide (NO) and bromine gas (Br₂) in a 5.00-L vessel at 100°C is 12.3. The total pressure exerted by this mixture of gases is 1.091 atm, and the original mass of the NOBr sample was 14.51 g.

Step 1: Convert grams to moles for each component

First, let's determine the number of moles for NOBr, NO, and Br₂ by using their respective molar masses. Molar masses: NOBr: 14.01(g/mol for N) + 15.999(g/mol for O) + 79.904(g/mol for Br) = 109.923 g/mol NO: 14.01(g/mol for N) + 15.999(g/mol for O) = 30.009 g/mol Br₂: 2 × 79.904(g/mol for Br) = 159.808 g/mol Moles of each component: n_NOBr = 3.22 g / 109.923 g/mol = 0.0293 mol n_NO = 3.08 g / 30.009 g/mol = 0.1027 mol n_Br₂ = 4.19 g / 159.808 g/mol = 0.0262 mol

Step 2: Calculate the concentrations of each component at equilibrium

Since the volume of the reaction is given as 5.00 L, we can determine the concentrations of each component at equilibrium by dividing the number of moles by the volume. [NOBr] = n_NOBr / 5.00 L = 0.0293 mol / 5.00 L = 0.00586 M [NO] = n_NO / 5.00 L = 0.1027 mol / 5.00 L = 0.02054 M [Br₂] = n_Br₂ / 5.00 L = 0.0262 mol / 5.00 L = 0.00524 M

Step 3: Calculate the equilibrium constant Kₚ

Now that we have the concentrations at equilibrium, we can find the equilibrium constant using the balanced chemical reaction: 2 NOBr(g) ⇌ 2 NO(g) + Br₂(g) $$K_{c} = \frac{[\mathrm{NO}]^{2}[\mathrm{Br}_{2}]}{[\mathrm{NOBr}]^{2}}$$ Plugging in the concentration values: $$K_{c} = \frac{(0.02054)^{2}(0.00524)}{(0.00586)^{2}}$$ Kc = 12.3 So, the equilibrium constant Kc = 12.3.

Step 4: Find the total pressure exerted by the mixture

To find the total pressure exerted by the mixture, we can use the ideal gas law, which is given by: PV = nRT The values of R (gas constant) and T (temperature) are given as R = 0.08206 L atm / mol K and T = 100 °C = 373 K. Let's start by finding the total number of moles of gas in the mixture: n_total = n_NOBr + n_NO + n_Br₂ = 0.0293 mol + 0.1027 mol + 0.0262 mol = 0.1582 mol Now we can substitute the values into the ideal gas law equation. P * 5.00 L = 0.1582 mol * 0.08206 L atm / mol K * 373 K Solving this equation, we get: P = 1.091 atm Thus, the total pressure exerted by the mixture is 1.091 atm.

Step 5: Calculate the original mass of NOBr

To find the original mass of NOBr, we need to determine the number of moles of NOBr that dissociated during the reaction. Since the balanced equation shows that for every 2 moles of NOBr that decompose, 2 moles of NO are formed and 1 mole of Br₂ is formed, we can compare the number of moles to determine the initial number of moles of NOBr. From the balanced equation, the decomposition ratio of NOBr to NO is 1:1. So, the change in the number of moles of NO will be equal to the change in the number of moles of NOBr decomposed. Δn_NOBr = Δn_NO = n_NO - n_NO (initial) Since n_NO (initial) = 0, we have: Δn_NOBr = 0.1027 mol Now, let's find the initial number of moles of NOBr: n_NOBr (initial) = n_NOBr + Δn_NOBr = 0.0293 mol + 0.1027 mol = 0.132 mol Finally, let's convert the moles into grams using the molar mass of NOBr: Initial mass of NOBr = n_NOBr (initial) * molar mass of NOBr = 0.132 mol * 109.923 g/mol = 14.51 g Thus, the initial mass of the nitrosyl bromide sample was 14.51 g.

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \( K_c \), is a crucial concept in chemical equilibrium. It provides insight into the relative concentrations of reactants and products at equilibrium in a chemical reaction. For a given reaction, the equilibrium constant is expressed in terms of the concentrations of these species.

In the exercise, we've examined a reaction where nitrosyl bromide decomposes into nitrogen monoxide and bromine. The equation for the equilibrium constant \( K_c \) is drawn from the balanced chemical equation:

• For the reaction: \( 2 \, \text{NOBr} (g) \rightleftharpoons 2 \, \text{NO} (g) + \text{Br}_2 (g) \)
• The expression for \( K_c \) is: \( K_c = \frac{[\text{NO}]^2[\text{Br}_2]}{[\text{NOBr}]^2} \)

The equilibrium constant can provide a sense of whether reactants or products are favored at equilibrium:

• A large \( K_c \) (much greater than 1) indicates that products are predominant at equilibrium.
• A small \( K_c \) (much smaller than 1) suggests that reactants are favored.

For the reaction in this exercise, we found \( K_c \) to be 12.3. This tells us that, at equilibrium, products are preferred over reactants, but the relationship is moderate given that the value isn't excessively large.

Ideal Gas Law

The ideal gas law provides a relationship between pressure, volume, temperature, and the number of moles in a gaseous system. The formula for the ideal gas law is:

• \( PV = nRT \)

Where:

• \( P \) is the pressure
• \( V \) is the volume
• \( n \) is the number of moles
• \( R \) is the universal gas constant, 0.08206 L atm / mol K
• \( T \) is the temperature in Kelvin

For problems involving gases at equilibrium, the ideal gas law is a vital tool. It allows us to calculate unknown variables when the other properties are known. In this exercise, we used it to find the total pressure exerted by the gaseous mixture in the reaction vessel.

• By summing up the moles of all gases, we determined the total pressure, resulting in 1.091 atm.
• This was achieved by rearranging the ideal gas law equation to solve for the pressure \( P \), and inputting the values for the total number of moles, volume, gas constant, and temperature.

Using the ideal gas law in chemical equilibrium problems allows us to better understand the behavior of gases in the system.

Reaction Stoichiometry

Reaction stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. The balanced chemical equation provides the ratio in which reactants react and products are form. This information is essential for calculating the amounts of substances involved.

In our example, we handled the reaction \( 2 \, \text{NOBr} (g) \rightleftharpoons 2 \, \text{NO} (g) + \text{Br}_2 (g) \) to find information about the initial and equilibrium masses of nitrosyl bromide. Here's how stoichiometry was used:

• The balanced equation tells us 2 moles of NOBr produce 2 moles of NO and 1 mole of \( \text{Br}_2 \).
• The decomposition of NOBr matches 1:1 with the formation of NO; for each mole of NOBr decomposed, one mole of NO is formed.

In the solution, stoichiometry helped determine the change in moles of NOBr by comparing it to the increase in NO moles. We calculated the initial moles of NOBr as the sum of equilibrium moles and those that decomposed. This was then converted back to grams to find the initial mass.

Understanding stoichiometry is vital for any calculation involving chemical reactions, especially in predicting the amounts of reactants or products and ensuring equations are balanced.