Problem 2 - General Chemistry: Principles and Modern Applications

Question

Based on these descriptions, write a balanced equation and the corresponding \(K_{\mathrm{p}}\) expression for each reversible reaction. (a) Oxygen gas oxidizes gaseous ammonia to gaseous nitrogen and water vapor. (b) Hydrogen gas reduces gaseous nitrogen dioxide to gaseous ammonia and water vapor. (c) Nitrogen gas reacts with the solid sodium carbonate and carbon to produce solid sodium cyanide and carbon monoxide gas.

Step-by-Step Solution

Answer

The balanced chemical equations are: \n(a) \(4NH_3 + 5O_2 \rightarrow 4N_2 + 6H_2O\) with \(K_{\mathrm{p}} = \frac{(P_{N2})^4 (P_{H2O})^6}{(P_{NH3})^4 (P_{O2})^5}\) \n(b) \(3H_2 + 2NO_2 \rightarrow 2NH_3 + 2H_2O\) with \(K_{\mathrm{p}} = \frac{(P_{NH3})^2 (P_{H2O})^2}{(P_{H2})^3 (P_{NO2})^2}\) \n(c) \(N_2 + 4Na_2CO_3 + C \rightarrow 4NaCN + 3CO\) with \(K_{\mathrm{p}} = (P_{CO})^3/\(P_{N2} \(P_{C}\)

Step 1: Problem a - Balancing Chemical Equation

First, let's write down the unbalanced equation using the given elements and compounds: \(O_2 + NH_3 \rightarrow N_2 + H_2O\). Now let's balance this equation. The balanced chemical equation is: \(4NH_3 + 5O_2 \rightarrow 4N_2 + 6H_2O\).

Step 2: Problem a - \(K_{\mathrm{p}}\) expression

The \(K_{\mathrm{p}}\) expression for a reaction at equilibrium is the ratio of the products to the reactants, each raised to the power of its stoichiometric coefficient in the balanced chemical equation. Therefore, the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = \frac{(P_{N2})^4 (P_{H2O})^6}{(P_{NH3})^4 (P_{O2})^5}\).

Step 3: Problem b - Balancing Chemical Equation

Similarly, let’s write down the unbalanced equation for the reaction: \(H_2 + NO_2 \rightarrow NH_3 + H_2O\). Then, we balance the equation, which gives: \(3H_2 + 2NO_2 \rightarrow 2NH_3 + 2H_2O\).

Step 4: Problem b - \(K_{\mathrm{p}}\) expression

Following the same process as in part a, the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = \frac{(P_{NH3})^2 (P_{H2O})^2}{(P_{H2})^3 (P_{NO2})^2}\).

Step 5: Problem c - Balancing Chemical Equation

Let’s write down the unbalanced equation for the reaction: \(N_2 + Na_2CO_3 + C \rightarrow NaCN + CO\). Then, balance this equation, which results in: \(N_2 + 4Na_2CO_3 + C \rightarrow 4NaCN + 3CO\).

Step 6: Problem c - \(K_{\mathrm{p}}\) expression

The \(K_{\mathrm{p}}\) expression is a bit different in this case because a solid (Na_2CO_3 and NaCN) is involved. The concentrations of solids are not included in the equilibrium expressions, so the \(K_{\mathrm{p}}\) expression for this reaction is: \(K_{\mathrm{p}} = (P_{CO})^3/\(P_{N2} \(P_{C}\).

Key Concepts

Balanced Chemical Equations

Balancing chemical equations is an essential part of understanding chemical reactions. This process ensures that the same number of each type of atom is present on both sides of the equation, adhering to the law of conservation of mass. In simple terms, if you start with a specific number of atoms of an element in the reactants, the same number must be in the products.

To balance a chemical equation, you can follow these steps:

• Write down the unbalanced chemical equation using chemical symbols and formulas from the problem statement.
• Count the number of atoms of each element in both reactants and products. Adjust coefficients, which are numbers placed before compounds, to balance the atoms.
• Ensure the final equation has the same number of each type of atom on both sides.

In the exercise, we balanced equations by adjusting coefficients to ensure that oxygen and nitrogen atoms matched on both sides of the equation. For example, in the reaction of oxygen gas with ammonia, we ended up with the balanced equation: \[4NH_3 + 5O_2 \rightarrow 4N_2 + 6H_2O \] This shows 4 ammonia, 5 oxygen, producing 4 nitrogen and 6 water molecules, balancing nitrogen and hydrogen atoms correctly.

Equilibrium Constant (Kp)

The equilibrium constant, denoted as \(K_{\mathrm{p}}\), gives us insight into the balance between products and reactants at chemical equilibrium, specifically in gas-phase reactions. It is a value that describes the ratio of the concentrations of products to reactants, each raised to the power of their respective stoichiometric coefficients.

Here's how you can derive the \(K_{\mathrm{p}}\):

• Write the balanced equation for the chemical reaction.
• Use the partial pressures of the gases involved to create a fraction. The numerator contains the products, and the denominator contains the reactants.
• Each term in the \(K_{\mathrm{p}}\) expression is raised to the power of its stoichiometric coefficient from the balanced equation.

For example, the \(K_{\mathrm{p}}\) expression for the oxidation of ammonia is:\[K_{\mathrm{p}} = \frac{(P_{N2})^4 (P_{H2O})^6}{(P_{NH3})^4 (P_{O2})^5}\]This indicates a high product/reactant ratio might prefer product formation, while a low value suggests the opposite.

Stoichiometry

Stoichiometry is the study of the quantitative relationships between reactants and products in a chemical reaction. It helps us predict the amounts of substances consumed and produced during a reaction, ensuring the reaction is balanced accurately.

Key concepts in stoichiometry include:

• The mole concept, facilitating the conversion between mass and number of particles.
• Using stoichiometric coefficients from the balanced equation to understand the proportions of reactants to products.
• Limiting reactants, which are substances that restrict the amount of product formed.

In the exercise provided, stoichiometry guided the balancing of chemical equations and the formulation of \(K_{\mathrm{p}}\) expressions. By knowing that for every 4 molecules of ammonia, 5 molecules of oxygen are needed (as in problem a), we can predict amounts needed and produced in more complex industrial processes. This establishes a clear relationship between theoretical and practical chemical reactions in the real world.