Problem 74 - Chemistry: The Central Science

Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$ 2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ An equilibrium mixture in a \(5.00-\mathrm{L}\) vessel at \(100^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO}\), and \(4.19 \mathrm{~g}^{2}\) 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

(a) The equilibrium constant, \(K_{c}\), for the given reaction, is 2.64. (b) The total pressure exerted by the mixture of gases at the equilibrium position is 4.72 atm. (c) The mass of the original sample of NOBr before it started to decompose is 18.87 g.

Step 1: Calculate the moles of species at equilibrium

First, let's find the moles of each gas species at equilibrium: Moles of NOBr = (mass of NOBr) / (molar mass of NOBr) = \( \frac{3.22\,g}{102.90\,g/mol} \) = 0.0313 mol Moles of NO = (mass of NO) / (molar mass of NO) = \( \frac{3.08\,g}{30.01\,g/mol} \) = 0.1026 mol Moles of Br2 = (mass of Br2) / (molar mass of Br2) = \( \frac{4.19\,g}{159.8\,g/mol} \) = 0.0262 mol

Step 2: Calculate molar concentrations

Next, we need to find the molar concentrations of the substances at equilibrium. We can do this by dividing the moles of each substance by the volume of the container: [NOBr] = moles of NOBr / volume = \( \frac{0.0313\,mol}{5.00\,L} \) = 0.00626 M [NO] = moles of NO / volume = \( \frac{0.1026\,mol}{5.00\,L} \) = 0.0205 M [Br2] = moles of Br2 / volume = \( \frac{0.0262\,mol}{5.00\,L} \) = 0.00524 M

Step 3: Find Kc and total pressure

Using the balanced chemical equation and the molar concentrations, we can calculate the equilibrium constant Kc: Kc = \( \frac{[\mathrm{NO}]^2[\mathrm{Br}_{2}]}{[\mathrm{NOBr}]^2} \) = \( \frac{(0.0205)^2(0.00524)}{(0.00626)^2} \) = 2.64 Now we can find the total pressure exerted by the mixture of gases. The sum of the partial pressures at equilibrium equals the total pressure: Total pressure = partial pressure of NOBr + partial pressure of NO + partial pressure of Br2 Total pressure = (0.00626)RT/(1 atm/(0.0821 L/mol K)) + (0.0205)RT/(1 atm/(0.0821 L/mol K)) + (0.00524)RT/(1 atm/(0.0821 L/mol K)) Total pressure = \(0.00626 \times 8.314 \times (273+100)/(0.0821)+0.0205 \times 8.314 \times (273+100)/(0.0821)+0.00524 \times 8.314 \times (273+100)/(0.0821)\) K = 4.72 atm

Step 4: Find the initial mass of NOBr

We can find the initial moles of NOBr by using the stoichiometry of the balanced chemical equation. For every 2 moles of NOBr consumed, 2 moles of NO and 1 mole of Br2 are produced. Initial moles of NOBr = moles of NOBr + 2( moles of NO - moles of Br2 ) = 0.0313 + 2(0.1026 - 0.0262) = 0.1834 mol Initial mass of NOBr = initial moles of NOBr × molar mass of NOBr = 0.1834 × 102.90 = 18.87 g To summarize our results: (a) The equilibrium constant, Kc, is 2.64. (b) The total pressure exerted by the mixture of gases is 4.72 atm. (c) The mass of the original sample of NOBr is 18.87 g.

Key Concepts

Equilibrium Constant

Understanding the equilibrium constant, denoted as K_c, is crucial to mastering chemical equilibrium. It is a parameter that quantifies the ratio of the concentrations of products to reactants at equilibrium for a reversible chemical reaction, each raised to the power of their respective stoichiometric coefficients. In the given problem, for the decomposition of nitrosyl bromide (NOBr), the equilibrium constant K_c is calculated using the molar concentrations of the gases once the system has reached equilibrium. The concentrations are obtained by dividing the number of moles by the volume of the container in which the reaction takes place.

The formula for the equilibrium constant looks like this:
\[ K_c = \frac{{[\text{{Product 1}}]^{{n_1}}[\text{{Product 2}}]^{{n_2}}...}}{{[\text{{Reactant 1}}]^{{m_1}}[\text{{Reactant 2}}]^{{m_2}}...}} \]
where n and m are the stoichiometric coefficients and the square brackets signify molar concentrations. For our reaction:
\[ 2\,\text{{NOBr}}(g) \rightleftharpoons 2\,\text{{NO}}(g) + \text{{Br}}_2(g) \]
The equilibrium constant is expressed as:\[ K_c = \frac{{[\text{{NO}}]^2[\text{{Br}}_2]}{{[\text{{NOBr}}]^2}} \]
It implies that at equilibrium, the concentration of NOBr squared in the denominator is matched by the squared concentration of NO and concentration of Br₂ in the numerator. For students, it's key to remember that K_c is a reflection of a particular reaction's tendency to proceed in a given direction at a specific temperature. A higher K_c means the reaction favors the formation of products, while a lower value indicates a preference for reactants. Understanding this concept will enhance problem-solving skills related to reaction equilibrium.

Molar Mass

The concept of molar mass is a fundamental component in the study of chemistry, serving as a bridge between the microscale (atoms and molecules) and the macroscale (grams and kilograms). Simply put, molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

Here's how you calculate it:
\[ \text{{Molar Mass}} = \frac{{\text{{mass of the substance in grams}}}}{{\text{{number of moles of the substance}}}} \]
In the exercise, to find the moles of NOBr, NO, and Br₂, we first need their respective molar masses: 102.90 g/mol for NOBr, 30.01 g/mol for NO, and 159.8 g/mol for Br₂. With their masses provided, we divide by their molar masses to get the number of moles present. Knowing the molar mass allows students to convert between mass and number of moles, which is a vital step in any stoichiometric calculation. It's essential to use accurate molar masses to ensure the correctness of subsequent calculations, like those determining the equilibrium constant or partial pressures. By grasping this concept, students become more proficient in their ability to manipulate and calculate various chemical quantities in both theoretical and practical scenarios.

Partial Pressure

The concept of partial pressure is a centerpiece in understanding gas mixtures. It defines the pressure that a particular gas in a mixture would exert if it alone occupied the entire volume of the container. In a mixture of gases, each gas contributes to the total pressure, in accordance with Dalton's Law of Partial Pressures.

To calculate partial pressures, you would use the equation:
\[ P_\text{{gas}} = \frac{{\text{{moles of gas}} \times RT}}{{V}} \]
Where P_gas is the partial pressure of the gas, R is the ideal gas constant, T is the temperature in Kelvin, and V is the volume of the container. In our exercise problem, the total pressure exerted by the gas mixture at equilibrium can be determined by adding the partial pressures of NOBr, NO, and Br₂. By knowing the individual partial pressures and their contributions, we not only characterize the system under equilibrium but also move closer to deducing other valuable properties such as the original mass of NOBr present before the reaction reached equilibrium. This knowledge is particularly important when dealing with gas laws and understanding the behavior of gases in diverse chemical contexts.