Problem 47 - Chemistry and Chemical Reactivity

Question

Assume 3.60 mol of ammonia is placed in a \(2.00-\mathrm{L}\) vessel and allowed to decompose to the elements. $$ 2 \mathrm{NH}_{3}(\mathrm{g}) \rightleftarrows \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) $$ If the experimental value of \(K_{\varepsilon}\) 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?

Step-by-Step Solution

Answer

Determine values by solving the equilibrium equation and use ideal gas law for pressure.

Step 1: Write the Expression for K_e

For the reaction \(2\, \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\, \text{H}_2(g)\), the equilibrium constant \(K_\varepsilon\) is given by the expression:\[ K_\varepsilon = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \]where \([\text{NH}_3]\), \([\text{N}_2]\), and \([\text{H}_2]\) are the equilibrium concentrations of ammonia, nitrogen, and hydrogen respectively.

Step 2: Initial Concentration Calculation

Calculate the initial concentration of \(\text{NH}_3\) using the formula \(C = \frac{n}{V}\).\[ [\text{NH}_3]_0 = \frac{3.60\, \text{mol}}{2.00\, \text{L}} = 1.80\, M \]

Step 3: Set Up the ICE Table

- **Initial Concentrations:** - \([\text{NH}_3] = 1.80\, M\), \([\text{N}_2] = 0\, M\), \([\text{H}_2] = 0\, M\)- **Change:** - Let \(x\) be the change in concentration for \(\text{N}_2\). Then, \([\text{H}_2] = 3x\) and \([\text{NH}_3] = 1.80 - 2x\) at equilibrium.- **Equilibrium Concentrations:** - \([\text{NH}_3] = 1.80 - 2x\) - \([\text{N}_2] = x\) - \([\text{H}_2] = 3x\)

Step 4: Solve For x Using K_e

Substitute the equilibrium concentrations into the equilibrium expression:\[ K_\varepsilon = \frac{x(3x)^3}{(1.80 - 2x)^2} = 6.3 \]This simplifies to:\[ \frac{x(27x^3)}{(1.80 - 2x)^2} = 6.3 \]\[ 27x^4 = 6.3 (1.80 - 2x)^2 \]Solve this quartic equation to find the value of \(x\).

Step 5: Calculate Equilibrium Concentrations

Using the value of \(x\) from the previous step, calculate each concentration:- \([\text{NH}_3] = 1.80 - 2x\)- \([\text{N}_2] = x\)- \([\text{H}_2] = 3x\)

Step 6: Calculate the Total Pressure

To find the total pressure, use the ideal gas law where the total number of moles at equilibrium is considered:- Total moles = \([\text{NH}_3] + [\text{N}_2] + [\text{H}_2]\) at equilibrium.- Use \(P = \frac{nRT}{V}\) to find the total pressure (assuming \(R = 0.0821\, L\, atm\, mol^{-1}\, K^{-1}\) and \(T\) is known).

Key Concepts

Equilibrium Constant

The equilibrium constant, often denoted as \(K_c\) or \(K_p\), is a crucial concept in understanding how chemical reactions reach a state of balance, or equilibrium. It is defined as the ratio of the concentrations of products to the concentrations of reactants, each raised to the power of their respective coefficients in the balanced chemical equation. For the decomposition of ammonia: \[ 2\, \text{NH}_3(g) \rightleftharpoons \text{N}_2(g) + 3\, \text{H}_2(g) \] The equilibrium constant \(K_c\) is given by: \[ K_c = \frac{[\text{N}_2][\text{H}_2]^3}{[\text{NH}_3]^2} \] **Why is it important?**

• Predicting Direction: If \(Q\), the reaction quotient, is less than \(K_c\), the reaction will move forward to reach equilibrium.
• Calculating Concentrations: Knowing \(K_c\) allows us to determine the unknown concentrations of reactants or products at equilibrium.
• Temperature Dependence: The value of \(K_c\) is temperature-dependent, thus providing insight into reaction favorability under different conditions.

ICE Table

The ICE table stands for Initial, Change, and Equilibrium and is a methodical way to track concentration changes in reactions reaching equilibrium. This tool is particularly powerful when dealing with reactions where the changes in concentration are not immediately obvious. For the ammonia decomposition reaction:

• Initial: Start by noting the initial concentrations of all species. For ammonia, it is 1.80 M, while nitrogen and hydrogen start at 0 M.
• Change: Represent changes due to reaction stoichiometry using variables like \(x\). For every mole of \(\text{N}_2\) formed, 2 moles of \(\text{NH}_3\) are decomposed, and 3 moles of \(\text{H}_2\) are released.
• Equilibrium: Express equilibrium concentrations in terms of the initial concentrations and the changes, such as \([\text{NH}_3] = 1.80 - 2x\).

By filling out an ICE table, you are systematically setting up a plan to insert values into the equilibrium constant expression.

Chemical Equilibrium

Chemical equilibrium occurs when the forward and reverse reactions occur at the same rate, resulting in constant concentrations of reactants and products. This balance is dynamic, meaning the reactions are still occurring, just without net change. In the ammonia decomposition example:

• The system reaches equilibrium once the concentrations of \(\text{NH}_3\), \(\text{N}_2\), and \(\text{H}_2\) remain unchanged over time.
• The equilibrium constant reflects the ratio of these concentrations at equilibrium.

**Le Chatelier's Principle:** Explains how a system at equilibrium responds to disturbances:- If more ammonia is added, the system will shift to produce more nitrogen and hydrogen until equilibrium is re-established.- Similarly, removing \(\text{H}_2\) will push the reaction forward to balance the loss.

Ideal Gas Law

The ideal gas law combines various principles to relate pressure, volume, temperature, and the amount of gas. Its formula is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is temperature. For calculating the total pressure in systems at chemical equilibrium:

• Sum all different moles of gases present after the reaction reaches equilibrium.
• Consider that each gas contributes to the overall pressure proportionally to its mole fraction.

Using this law, one can compute the total pressure by substituting the total moles of gas at equilibrium into the gas law equation, factoring in known conditions like the temperature and volume of the vessel used.