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 -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 \(\mathrm{g}\) of 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 \(\mathrm{NOBr}\) ?

Short answer

In summary, (a) the equilibrium constant, \(K_c\), is 0.0257, (b) the total pressure exerted by the mixture of gases is 1.45 atm, and (c) the initial mass of the NOBr sample was 6.28 g.

Step-by-step solution

Convert the given amounts of substances into moles

: First, we need to convert the given masses of NOBr, NO, and Br2 into moles. We can do this using the molar masses of the substances: Molar mass of NOBr = 30.01 g/mol (14.01 g/mol for N + 16.00 g/mol for O + 79.90 g/mol for Br) Molar mass of NO = 30.01 g/mol (14.01 g/mol for N + 16.00 g/mol for O) Molar mass of Br2 = 159.80 g/mol (79.90 g/mol * 2) Now, let's use these molar masses to convert the given masses into moles: moles_NOBr = 3.22 g / 30.01 g/mol = 0.107 mol moles_NO = 3.08 g / 30.01 g/mol = 0.102 mol moles_Br2 = 4.19 g / 159.80 g/mol = 0.0262 mol

Calculate the equilibrium concentrations

: Now we calculate the concentrations of those species at equilibrium: We have been given the volume of the vessel, 5.00 L. Concentration_NOBr = 0.107 mol / 5.00 L = 0.0214 M Concentration_NO = 0.102 mol / 5.00 L = 0.0204 M Concentration_Br2 = 0.0262 mol / 5.00 L = 0.00524 M

Determine the equilibrium constant, Kc

: Using the equation provided, we can calculate Kc using the equilibrium concentrations: \[K_c = \frac{[\mathrm{NO}]^2[\mathrm{Br}_2]}{[\mathrm{NOBr}]^2}\] Therefore, Kc = (0.0204^2 * 0.00524)/(0.0214^2) = 0.0257

Calculate the total pressure exerted

: To calculate the total pressure, we first need to determine the total number of moles of gas present in the equilibrium mixture: Total moles = moles_NOBr + moles_NO + moles_Br2 = 0.107 + 0.102 + 0.0262 = 0.2352 mol Now we can use the ideal gas law to determine the total pressure. The temperature is given as 100°C, which is equal to 373.15 K. PV = nRT Let the total pressure be P. Then, P * 5.00 L = 0.2352 mol * 0.0821 L atm/mol K * 373.15 K Now, solve for the pressure: P = (0.2352 * 0.0821 * 373.15) / 5 = 1.45 atm

Calculate the initial mass of NOBr

: To find the initial mass of NOBr, we first need to determine how many moles of NOBr decomposed. From the balanced equation, it is clear that the number of moles of NO produced equals the number of moles of NOBr that decomposed. We have calculated the moles of NO present at equilibrium as 0.102 mol. So, the initial moles of NOBr = moles_NOBr + moles_NO = 0.107 + 0.102 = 0.209 mol Now, we can convert this value back to grams using the molar mass of NOBr: Initial mass of NOBr = 0.209 mol * 30.01 g/mol = 6.28 g Therefore, the initial mass of the NOBr sample was 6.28 g.

Key concepts

Equilibrium Constant (Kc)

When we delve into the concept of chemical equilibrium, we enter a realm where reactions occur in such a way that the rate of the forward reaction equals the rate of the reverse reaction, leading to a state where the concentrations of reactants and products remain constant over time.

At this point, we can define the equilibrium constant, denoted as \(K_c\), which quantifies the ratio of the concentrations of products raised to their respective coefficients in the balanced equation to the concentrations of reactants raised to their own coefficients. In the supplied exercise, the equilibrium constant for the decomposing of nitrosyl bromide (NOBr) into nitrogen monoxide (NO) and bromine (Br\(_2\)) is calculated using the formula:

\[K_c = \frac{[\mathrm{NO}]^2[\mathrm{Br}_2]}{[\mathrm{NOBr}]^2}\]

where [NO], [Br\(_2\)], and [NOBr] represent the concentrations of the respective substances in moles per liter (M). This calculation assumes that the reaction has reached a state of equilibrium and that the concentrations remain stable.

To further aid understanding, envision \(K_c\) as a mathematical expression of the position of equilibrium; a larger \(K_c\) indicates a disposition towards products, while a smaller \(K_c\) denotes a reaction that favors the reactants. It's pivotal to recognize that \(K_c\) is only influenced by temperature changes and is not affected by other factors such as pressure or concentration adjustments.

Ideal Gas Law

Another core concept linked to chemical equilibrium, especially in gaseous systems, is the ideal gas law. This law relates the pressure (P), volume (V), temperature (T), and the number of moles (n) of a gas with the equation:

\[PV = nRT\]

In this formula, R represents the universal gas constant with a value of 0.0821 L atm/mol K. The ideal gas law is an excellent approximation for the behavior of real gases under many conditions, albeit with some limitations when gases are at very high pressure or low temperature.

Application in Equilibrium Problems

In the context of the given problem, we apply the ideal gas law to determine the total pressure exerted by the gases in equilibrium within the 5.00-liter vessel. By inserting the total number of moles of gas present and the temperature (converted to Kelvin) into the formula, we can solve for the total pressure. This process connects the macroscopic properties of gases, such as pressure, to their microscopic properties, such as the amount in moles, providing a crucial link between the two realms.

The ideal gas law is fundamental in various fields, ranging from engineering and meteorology to medicine and chemistry. It is particularly important in our problem because it helps to establish the total pressure in the vessel, a necessary step in analyzing the system's overall condition.

Mole Concept

A mole is a unit of measurement used in chemistry to express amounts of a chemical substance. One mole corresponds to the number of atoms in 12 grams of carbon-12 and is approximately \(6.022 \times 10^{23}\) entities (Avogadro’s number).

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we can measure, essentially serving as a conversion factor between these two realms.

From Mass to Moles and Back

In the problem at hand, the concept is used to convert the mass of nitrosyl bromide (NOBr), nitrogen monoxide (NO), and bromine (Br\(_2\)) into moles before calculating the equilibrium constant \(K_c\). Subsequently, the moles of decomposed NOBr are used to determine the initial mass of NOBr.

Understanding the mole concept is essential in chemistry. It allows for the calculation of quantities that participate in chemical reactions, evaluation of yield, and comprehension of stoichiometry, which is the quantitative relationship between reactants and products in a chemical reaction.

Emphasizing the mole concept in equilibrium problems is crucial, as it offers a common ground for comparing different substances and their behavior in reactions, and it provides insights into how different substances interconvert during a chemical process.