Problem 74 - Chemistry The Central Science

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 \(5.00-\mathrm{L}\) vessel at \(100{ }^{\circ} \mathrm{C}\) contains \(3.22 \mathrm{~g}\) of \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) of \(\mathrm{NO},\) and \(4.19 \mathrm{~g}\) 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 \(K_c\) for the given reaction is 0.157, the total pressure exerted by the equilibrium mixture is 1.021 atm, and the mass of the original sample of nitrosyl bromide (NOBr) was 14.51 g.

Step 1: Calculate the moles of each species at equilibrium

To calculate the moles of each species at equilibrium, we will use their given masses and molar masses: Molar mass of NOBr = 14.0067 g/mol (N) + 15.999 g/mol (O) + 79.904 g/mol (Br) = 109.910 g/mol Moles of NOBr = 3.22 g / 109.910 g/mol = 0.0293 mol Molar mass of NO = 14.0067 g/mol (N) + 15.999 g/mol (O) = 30.006 g/mol Moles of NO = 3.08 g / 30.006 g/mol = 0.1027 mol Molar mass of Br2 = 2 * 79.904 g/mol (Br) = 159.808 g/mol Moles of Br2 = 4.19 g / 159.808 g/mol = 0.0262 mol

Step 2: Calculate the concentrations of each species at equilibrium

To obtain the equilibrium concentrations for each species, we will divide their moles by the volume of the container (5.00 L): [NOBr] = 0.0293 mol / 5.00 L = 0.00586 M [NO] = 0.1027 mol / 5.00 L = 0.02054 M [Br2] = 0.0262 mol / 5.00 L = 0.00524 M

Step 3: Calculate the equilibrium constant Kc

Using the balanced chemical equation and the concentrations at equilibrium, we can calculate the equilibrium constant Kc: Kc = [NO]^2[Br2] / [NOBr]^2 = (0.02054)^2 * (0.00524) / (0.00586)^2 = 0.157

Step 4: Find the total pressure exerted by the gases

The total moles of gas in the container at equilibrium are: Total moles = Moles of NOBr + Moles of NO + Moles of Br2 = 0.0293 + 0.1027 + 0.0262 = 0.1582 mol Now, we can use the ideal gas law to find the total pressure: PV = nRT Total pressure P = nRT / V = (0.1582 mol) * (0.0821 L * atm / mol * K) * (373 K) / (5.00 L) = 1.021 atm

Step 5: Find the mass of the original sample of NOBr

The balanced chemical equation tells us that, for each mole of NO formed, one mole of NOBr decomposes. Therefore, the number of moles of NOBr that decomposed is equal to the number of moles of NO at equilibrium: Moles of NOBr decomposed = Moles of NO at equilibrium = 0.1027 mol So the original number of moles of NOBr was: Original moles of NOBr = Moles of NOBr at equilibrium + Moles of NOBr decomposed = 0.0293 mol + 0.1027 mol = 0.1320 mol Now we can convert this back to mass: Mass of the original sample of NOBr = 0.1320 mol * 109.910 g/mol = 14.51 g #Conclusion# The equilibrium constant Kc is 0.157, the total pressure exerted by the equilibrium mixture is 1.021 atm, and the mass of the original sample of nitrosyl bromide was 14.51 g.

Key Concepts

Equilibrium Constant

Chemical equilibrium represents a state in which the concentrations of reactants and products no longer change with time. For reactions that reach equilibrium, we can quantify their chemical balance using the equilibrium constant, denoted as \( K_c \).
In the reaction \( 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_2(g) \), the equilibrium constant expression is determined from the concentrations of the products and reactants at equilibrium. The formula for \( K_c \) is:

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

The brackets "[]" indicate the molar concentrations of the reactants and products.
The value of \( K_c \) indicates the position of equilibrium. A small \( K_c \) (less than 1) suggests that the concentration of the reactants is higher than that of the products at equilibrium, which was the case here with a \( K_c \) of 0.157.
Understanding these balances is crucial for predicting how changes in conditions, like temperature and pressure, can shift equilibria according to Le Chatelier's principle.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry and physics that relates the pressure, volume, temperature, and number of moles of a gas. It is utilized to predict the behavior of gases under various conditions. The equation for the Ideal Gas Law is:

• \( PV = nRT \)

Where:

• \( P \) stands for pressure in atmospheres,
• \( V \) is the volume in liters,
• \( n \) represents the number of moles,
• \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and
• \( T \) is the temperature in Kelvin.

In the exercise, the Ideal Gas Law was applied to find the total pressure exerted by the gases in the container. Knowing the total moles of gases, the volume, and temperature allowed us to solve for \( P \), yielding approximately 1.021 atm.
This relation is key to understanding gas mixtures and is essential for solving many chemistry problems involving gases. Always ensure that you properly convert temperature to Kelvin and use consistent units for accurate calculations.

Molar Mass Calculation

Calculating molar masses correctly is vital when doing any quantitative chemical analysis. The molar mass of a compound is the mass of one mole of its molecules. It's determined by summing the atomic masses of all the atoms in a formula.
For example, to find the molar mass of \( \mathrm{NOBr} \):

• Nitrogen (N) = 14.0067 g/mol
• Oxygen (O) = 15.999 g/mol
• Bromine (Br) = 79.904 g/mol

Add these values to get 109.910 g/mol for \( \mathrm{NOBr} \).
These calculations help determine moles from a given mass, which was crucial in assessing the composition of the equilibrium mixture.
To find moles, you rearrange the formula:

• \( \text{Moles} = \frac{\text{Mass}}{\text{Molar Mass}} \)

By computing correct moles for each component, you can derive important equilibrium and pressure data, making molar mass assessment a foundational skill in chemistry.