Problem 76 - Chemistry The Central Science In Si Units

Question

A sample of nitrosyl bromide (NOBr) decomposes according to the equation $$ 2 \operatorname{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\operatorname{Br}_{2}(g) $$ 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 \(\mathrm{Br}_{2}\). (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 \(\mathrm{NOBr}\) ?

Step-by-Step Solution

Answer

(a) \(K_c \approx 0.356\). (b) Total pressure is \(0.969\, \text{atm}\). (c) Original mass of \(\mathrm{NOBr}\) was \(11.29\, \text{g}\).

Step 1: Convert Mass to Moles

To start, we need to convert the given masses of each gas into moles using their molecular weights. The molecular weights are: \( \mathrm{NOBr} = 109.91\, \text{g/mol}, \mathrm{NO} = 30.01\, \text{g/mol}, \mathrm{Br}_2 = 159.81\, \text{g/mol} \).\[\text{Moles of NOBr} = \frac{3.22\, \text{g}}{109.91\, \text{g/mol}} \approx 0.0293\, \text{mol} \\text{Moles of NO} = \frac{3.08\, \text{g}}{30.01\, \text{g/mol}} \approx 0.1027\, \text{mol} \\text{Moles of Br}_2 = \frac{4.19\, \text{g}}{159.81\, \text{g/mol}} \approx 0.0262\, \text{mol} \]

Step 2: Calculate Concentrations

Next, calculate the concentration of each species since all are in a 5.00 L vessel: \[\left[\mathrm{NOBr}\right] = \frac{0.0293\, \text{mol}}{5.00\, \text{L}} = 0.00586\, \text{M} \\left[\mathrm{NO}\right] = \frac{0.1027\, \text{mol}}{5.00\, \text{L}} = 0.02054\, \text{M} \\left[\mathrm{Br}_2\right] = \frac{0.0262\, \text{mol}}{5.00\, \text{L}} = 0.00524\, \text{M} \]

Step 3: Calculate Equilibrium Constant \(K_c\)

Using the equilibrium concentrations, find the equilibrium constant \(K_c\) for the reaction: \[\ K_c = \frac{[\mathrm{NO}]^2 \cdot [\mathrm{Br}_2]}{[\mathrm{NOBr}]^2} = \frac{(0.02054)^2 \cdot (0.00524)}{(0.00586)^2} \approx 0.356\]

Step 4: Calculate Total Pressure

Using the ideal gas law, calculate the total pressure exerted by the gas mixture. First, find the total number of moles in the vessel: \[ \text{Total moles} = 0.0293 + 0.1027 + 0.0262 = 0.1582\, \text{mol} \]Then apply the ideal gas law: \(PV = nRT\), where \(R = 0.0821\, \text{L·atm/mol·K}, T = 373\, \text{K}\).\[P = \frac{nRT}{V} = \frac{0.1582 \times 0.0821 \times 373}{5.00} \approx 0.969\, \text{atm} \]

Step 5: Calculate Original Mass of \(\mathrm{NOBr}\)

The composition of NOBr itself is affected by the reaction; assume the initial moles of NO added becomes the moles of NO formed, thus\: 0.1027 moles NO. For every 2 mols of NOBr decomposed, 2 mols of NO and 1 mol Br2 are formed. Thus, initial moles NOBr = (final NO - initial NO). \[\text{Initial moles of NOBr} = 0.1027\, \text{mol NO} = 0.1027 \}\ \text{Mass of initial NOBr} = 0.1027 \times 109.91 \approx 11.29\, \text{g}. \]

Key Concepts

Equilibrium Constant (Kc)

Chemical equilibrium plays a significant role in understanding reactions in a closed system. The equilibrium constant, \( K_c \), is a fundamental aspect when analyzing the concentration of reactants and products at equilibrium in gaseous reactions. The equilibrium constant can be expressed using the balanced chemical equation where the concentrations of the products are divided by the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. For the decomposition reaction of nitrosyl bromide (NOBr), the equilibrium constant \( K_c \) is given by:\[K_c = \frac{[\mathrm{NO}]^2 \cdot [\mathrm{Br}_2]}{[\mathrm{NOBr}]^2}\]This formula allows us to determine the proportions of substances at equilibrium, highlighting how far a reaction progresses before reaching that state. In this reaction setup, it is essential to accurately calculate the concentration of each gaseous component using their mass and the vessel's volume.

Ideal Gas Law

The Ideal Gas Law connects several essential aspects, like pressure, volume, temperature, and the number of moles of a gas in a system. The formula is expressed as:\[PV = nRT\]Where:

• \( P \) is the pressure of the gas
• \( V \) is the volume of the gas
• \( n \) is the number of moles
• \( R \) is the ideal gas constant (0.0821 L·atm/mol·K)
• \( T \) is the temperature in Kelvin

In this case, calculating the total pressure of the gas mixture involves adding up the moles of each component gas and plugging them into the Ideal Gas Law. Hence, understanding these relationships is crucial for accurately gauging the behavior of gases as they interact within their container.

Stoichiometry

Stoichiometry is about the quantitative relationships between reactants and products within a chemical reaction, which dictates how much product is formed from a certain amount of reactant. In the decomposition of NOBr:\[2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g) + \mathrm{Br}_2(g)\]The stoichiometric coefficients tell us that two moles of NOBr decompose to form two moles of NO and one mole of \( \mathrm{Br}_2 \). This balance provides a blueprint for how substances transform during the reaction. Understanding stoichiometry is critical for predicting the outcomes of chemical reactions, helping you convert moles of a starting material directly into moles of the desired product or vice versa.

Mole Calculations

Converting between mass and moles is a fundamental skill when dealing with chemical reactions. Each substance has a specific molar mass (molecular weight), which enables the conversion from mass to moles. For instance, to find how many moles are present in a certain mass of nitrosyl bromide (NOBr), you use its molar mass, which is 109.91 g/mol:\[\text{Moles of NOBr} = \frac{\text{Mass of NOBr}}{\text{Molar Mass of NOBr}}\]This principle applies to any chemical species and is pivotal in achieving accurate reaction calculations. Once you have the moles, these can be employed to find concentrations, which are essential for further calculations involving equilibrium constants or pressures using the Ideal Gas Law. Understanding how to switch between these quantities is crucial for mastering chemical equations.