Question

Assume 3.60 mol of ammonia is placed in a 2.00 - \(L\) vessel and allowed to decompose to the elements at \(723 \mathrm{K}\) $$ 2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftharpoons \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If the experimental value of \(K_{r}\) is 6.3 for this reaction at the temperature in the reactor, calculate the equilibrium concentration of each reagent. What is the total pressure in the flask?

Short answer

Equilibrium concentrations are \([\text{NH}_3] = 0.82\, \text{M}\), \([\text{N}_2] = 0.49\, \text{M}\), \([\text{H}_2] = 1.47\, \text{M}\). Total pressure is 164.8 atm.

Step-by-step solution

Write the Initial Concentrations

Given 3.60 moles of \( \text{NH}_3 \) in a 2.00 L vessel, the initial concentration of \( \text{NH}_3 \) is \( \frac{3.60}{2.00} = 1.80 \text{ M} \). The initial concentrations of \( \text{N}_2 \) and \( \text{H}_2 \) are both 0 \text{ M}.

Set Up the Equilibrium Expression

The reaction is \( 2 \text{NH}_3 \rightleftharpoons \text{N}_2 + 3 \text{H}_2 \). The equilibrium expression for \( K_r = 6.3 \) is:\[ K_r = \frac{\left[ \text{N}_2 \right] \left[ \text{H}_2 \right]^3}{\left[ \text{NH}_3 \right]^2} = 6.3 \]

Define Changes in Concentrations

Let \( x \) be the change in moles of \( \text{N}_2 \) formed. Then, \( 2x \) moles of \( \text{NH}_3 \) decompose, and \( 3x \) moles of \( \text{H}_2 \) are formed.- \( \left[ \text{NH}_3 \right]_e = 1.80 - 2x \)- \( \left[ \text{N}_2 \right]_e = x \)- \( \left[ \text{H}_2 \right]_e = 3x \)

Substitute and Solve for x

Substitute the equilibrium concentrations into the equilibrium expression:\[ 6.3 = \frac{x \cdot (3x)^3}{(1.80 - 2x)^2} \]Solve this equation to find \( x \).

Simplify and Solve Equation

This simplifies to:\[ 6.3 = \frac{27x^4}{(1.80 - 2x)^2} \]After solving numerically or using approximation methods, we find:\( x \approx 0.49 \).

Calculate Equilibrium Concentrations

Substitute \( x = 0.49 \) back into the expressions:- \( \left[ \text{NH}_3 \right]_e = 1.80 - 2(0.49) = 0.82 \text{ M} \)- \( \left[ \text{N}_2 \right]_e = 0.49 \text{ M} \)- \( \left[ \text{H}_2 \right]_e = 3(0.49) = 1.47 \text{ M} \)

Calculate Total Pressure

Using ideal gas law and given total concentration of gases:\[ P = CRT \]where \( R = 0.0821 \text{ L atm/mol K} \), and \( T = 723 \text{ K} \):Total concentration = \( \left[ \text{NH}_3 \right]_e + \left[ \text{N}_2 \right]_e + \left[ \text{H}_2 \right]_e = 0.82 + 0.49 + 1.47 = 2.78 \text{ M} \)\[ P = 2.78 \times 0.0821 \times 723 \approx 164.8 \text{ atm} \]

Key concepts

Ammonia Decomposition

Ammonia decomposition involves the chemical process where ammonia (\( \text{NH}_3 \)) breaks down into nitrogen (\( \text{N}_2 \)) and hydrogen (\( \text{H}_2 \)). This reaction is reversible, which means it can proceed both forwards and backwards. The equation for ammonia decomposition is given by: \ \[ 2 \text{NH}_3(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \] This balanced equation shows that two molecules of ammonia decompose to form one molecule of nitrogen and three molecules of hydrogen. Understanding this stoichiometry is crucial for determining how the concentration of each species changes as the reaction reaches equilibrium. Initially, we start with a certain amount of ammonia, which decomposes until a balance of all substances is reached. Knowing the initial concentrations allows us to use the stoichiometric relationship to express how the concentrations of \( \text{N}_2 \) and \( \text{H}_2 \) change relative to \( \text{NH}_3 \).

Equilibrium Constant

The equilibrium constant (\( K_r \)) is a fundamental concept in the context of chemical reactions. It quantitatively describes the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a particular temperature. For the decomposition of ammonia, the equilibrium constant expression is: \ \[ K_r = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \] Here, \( [\text{N}_2] \), \( [\text{H}_2] \), and \( [\text{NH}_3] \) represent the molar concentrations of nitrogen, hydrogen, and ammonia at equilibrium, respectively. In this problem, the given \( K_r \) value is 6.3, which provides crucial information about how far the reaction proceeds towards products. When \( K_r \) is greater than 1, as in this case, it suggests that at equilibrium, products are favored over reactants. By knowing \( K_r \) and using initial concentrations, we can calculate the specific equilibrium concentrations of all substances involved.

Ideal Gas Law

The Ideal Gas Law is a critical equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as: \ \[ PV = nRT \] Where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. In this exercise, once the equilibrium concentrations of \( \text{NH}_3 \), \( \text{N}_2 \), and \( \text{H}_2 \) are known, the Ideal Gas Law can be used to determine the total pressure in the vessel. Since the gas law combines concentration (which is moles per volume) with temperature to estimate pressure, it becomes a straightforward calculation to find: \ \[ P = CRT \] Here, \( C \) is the total concentration of gases, and by substituting the known values, the pressure is calculated to be approximately 164.8 atm. This highlights how macroscopic properties such as pressure can be derived from molecular-scale activities.

Reaction Stoichiometry

Reaction stoichiometry involves using balanced chemical equations to determine the relationships between reactants and products in a chemical reaction. For the ammonia decomposition reaction, the balanced equation: \ \[ 2 \text{NH}_3(\text{g}) \rightleftharpoons \text{N}_2(\text{g}) + 3 \text{H}_2(\text{g}) \] provides a roadmap for understanding how the quantities change as the reaction proceeds. Stoichiometry tells us that for every 2 moles of \( \text{NH}_3 \) that decompose, 1 mole of \( \text{N}_2 \) and 3 moles of \( \text{H}_2 \) are produced. This means the changes in concentrations will follow these stoichiometric coefficients. For example, if \( x \) moles of \( \text{N}_2 \) is formed, then \( 3x \) moles of \( \text{H}_2 \) are formed, and \( 2x \) moles of \( \text{NH}_3 \) is decomposed. Using these relationships allows chemists to relate changes in concentration to one another and to solve for unknown values, such as equilibrium concentrations.