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 \(\mathrm{NOBr}, 3.08 \mathrm{~g}\) 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?

Short answer

\(K_c = 1.55\) and the total pressure exerted by the mixture of gases is 1.83 atm.

Step-by-step solution

Convert mass to moles

Using the molar mass of each substance, convert the given mass of each substance into moles: - Molar mass of NOBr = 30.01 g/mol - Molar mass of NO = 30.01 g/mol - Molar mass of Br2 = 159.8 g/mol So, Moles of NOBr = (3.22 g) / (30.01 g/mol) = 0.107 mol Moles of NO = (3.08 g) / (30.01 g/mol) = 0.103 mol Moles of Br2 = (4.19 g) / (159.8 g/mol) = 0.0262 mol

Find the initial and equilibrium concentrations

We are given that the volume of the vessel is 5.00 L. So, we can calculate the initial and equilibrium concentrations as follows: Initial Concentrations: [NOBr]0 = Moles of NOBr / Volume = (0.107 mol) / (5.00 L) = 0.0214 mol/L [NO]0 = 0 (as no initial moles) [Br2]0 = 0 (as no initial moles) Equilibrium Concentrations: [NOBr] = (0.107 - 2x) mol / 5.00 L = (0.0214 - 0.4x) mol/L [NO] = (0.103 + 2x) mol / 5.00 L = (0.0206 + 0.4x) mol/L [Br2]= (0.0262 + x) mol / 5.00 L = (0.00524 + 0.2x) mol/L

Set up the reaction quotient

The equation for the reaction quotient (Qc) based on the balanced reaction: $$ Q_{c}=\dfrac{[\mathrm{NO}]^{2}[\mathrm{Br}_{2}]}{[\mathrm{NOBr}]^{2}} $$ Plug the equilibrium concentrations into the equation: $$ K_{c}=\dfrac{(0.0206+0.4x)^{2}(0.00524+0.2x)}{(0.0214-0.4x)^{2}} $$ Since our reaction is at equilibrium state, and \(K_c\) is unknown, we cannot use the equation and must rely on stoichiometry as we know the moles of the reactants and products at equilibrium conditions: \(0.0214-0.4x = 0.0206+0.4x\) \(0.4x+0.4x = 0.0206-0.0214\) \(0.8x = -0.0008\) \(x = -0.0008/0.8 = -0.001\) This indicates that the reaction is already at the equilibrium state and there is actually no additional change that occurs. Therefore, we can use the initial concentrations as the equilibrium concentrations and calculate \(K_c\).

Calculate \(K_c\)

We will now calculate the equilibrium constant using the initial concentrations as the equilibrium concentrations: $$ K_{c}=\dfrac{(0.0206)^{2}(0.00524)}{(0.0214)^{2}} $$ $$ K_{c}=1.550 $$

Calculate the total pressure

To calculate the total pressure, we will use the ideal gas law for the mixture: $$ P_{total}=\dfrac{n_{total}RT}{V} $$ The total number of moles of the gases present in the reaction mixture: \(n_{total} = n_{NOBr} + n_{NO} + n_{Br2} = 0.107 + 0.103 + 0.0262 = 0.2362 \, moles\) The temperature is given as \(100^{\circ} C\), we need to convert it to Kelvin: \(T = (100+273.15)K = 373.15K\) R, the gas constant, is given in L·atm/(mol·K), so R = 0.0821 L·atm/(mol·K) Now, we plug in the values: \(P_{total} = \dfrac{(0.2362 \, mol) (0.0821 \, L·atm/(mol·K))(373.15 \, K)}{5.00 \, L}\) \(P_{total} = 1.830 \, atm\) So the equilibrium constant, \(K_c\) is 1.55 and the total pressure exerted by the mixture of gases is 1.83 atm.

Key concepts

Equilibrium Constant

In chemistry, equilibrium is a state where the concentrations of reactants and products remain constant over time. This state is reached when the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant, denoted as \(K_c\), is a measure of how far a reaction proceeds towards products at equilibrium. It is expressed as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their respective coefficients in the balanced equation.

For the decomposition of nitrosyl bromide (NOBr) into NO and Bri.e. \(2\mathrm{NOBr(g)} \rightleftharpoons 2\mathrm{NO(g) + Br_2(g)}\), the equilibrium constant expression would be:

\[ K_{c} = \dfrac{[\mathrm{NO}]^2[\mathrm{Br}_2]}{[\mathrm{NOBr}]^2}\]

A high \(K_c\) value indicates a reaction with more products than reactants at equilibrium, whereas a low \(K_c\) value indicates more reactants than products. In this exercise, the equilibrium constant was found to be 1.55, suggesting a balanced creation of products and reactants at equilibrium.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. The concept is rooted in the balanced chemical equation, which provides a ratio of moles of each substance involved. This is critical when determining concentrations in equilibrium calculations.

In our exercise, the stoichiometry of the reaction \(2 \mathrm{NOBr(g)} \rightleftharpoons 2 \mathrm{NO(g) + Br_2(g)}\) means that decomposing 2 moles of \(\mathrm{NOBr}\) produces 2 moles of \(\mathrm{NO}\) and 1 mole of \(\mathrm{Br_2}\). The calculations involve:

• Converting mass to moles to find the initial quantity of each substance.
• Using the stoichiometric coefficients to understand how these moles distribute at equilibrium.

Understanding stoichiometry is essential when calculating the equilibrium constant or any related measures in the reaction.

Gas Laws

Gas laws describe the behavior of gases, and the Ideal Gas Law is particularly useful for determining the total pressure of a gas mixture. The Ideal Gas Law is represented as \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.

In the exercise, to find the total pressure exerted by the mixture of gases, the given temperature was first converted to Kelvin. Then, the total moles of gases \((n_{total})\) was calculated by summing the moles of NOBr, NO, and Br₂. These values were plugged into the Ideal Gas Law equation:

\[ P_{total} = \dfrac{n_{total}RT}{V}\]

This allowed us to find that the total pressure in the vessel was 1.83 atm, demonstrating an application of the gas laws in determining properties of gas mixtures.