Question

Given two reactions at equilibrium: $$ \begin{array}{l}{\text { a. } \mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})} \\\ {\text { b. } \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{HCl}(\mathrm{g})}\end{array} $$

Short answer

In summary, we have two reactions at equilibrium: a. Formation of Ammonia (NH₃): \(N_2(g) + 3H_2(g) ⇌ 2NH_3(g)\) with the equilibrium constant expression \(K_1 = \frac{[NH_3]^2}{[N_2] \times [H_2]^3}\) b. Formation of Hydrogen Chloride (HCl): \(H_2(g) + Cl_2(g) ⇌ 2HCl(g)\) with the equilibrium constant expression \(K_2 = \frac{[HCl]^2}{[H_2] \times [Cl_2]}\)

Step-by-step solution

Introduction to Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant. In other words, the system is in a state of balance, and no net change is observed. The reactions are represented by the symbol ⇌ (double arrow) which implies that the reaction can proceed in both directions (forward and backward).

Equilibrium Constant (K)

For a general reaction at equilibrium: \(aA + bB ⇌ cC + dD\), where \(A\), \(B\), \(C\), and \(D\) are species, and \(a\), \(b\), \(c\), and \(d\) are their stoichiometric coefficients. The equilibrium constant, K, is given by: \[K = \frac{[C]^{ c} \times [D]^{d}}{[A]^{a} \times [B]^{b}}\] where [A], [B], [C], and [D] are the molar concentrations of the respective species at equilibrium. Now, let's analyze and break down each given reaction.

Reaction a: Formation of Ammonia (NH₃)

The reaction can be represented as: \(N_2(g) + 3H_2(g) ⇌ 2NH_3(g)\) To write the equilibrium constant expression for this reaction, we can follow the general formula of the equilibrium constant: \[K_1 = \frac{[NH_3]^2}{[N_2] \times [H_2]^3}\]

Reaction b: Formation of Hydrogen Chloride (HCl)

The reaction can be represented as: \(H_2(g) + Cl_2(g) ⇌ 2HCl(g)\) The equilibrium constant expression for this reaction is: \[K_2 = \frac{[HCl]^2}{[H_2] \times [Cl_2]}\] Now we have the equilibrium constant expressions for both reactions. They can be used to solve problems related to equilibrium concentrations and the extent of reactions.

Key concepts

Equilibrium Constant

When a chemical reaction reaches equilibrium, it is governed by a vital parameter known as the equilibrium constant, denoted as \(K\). This constant provides significant insight into the balance between reactants and products at equilibrium.
In a general reaction: \[ aA + bB \rightleftharpoons cC + dD \] where \(A\), \(B\), \(C\), and \(D\) represent chemical species and \(a, b, c,\) and \(d\) are their stoichiometric coefficients, the equilibrium constant \(K\) is expressed as:\[ K = \frac{[C]^{c} \times [D]^{d}}{[A]^{a} \times [B]^{b}} \] This equation allows us to calculate the concentrations of products and reactants at equilibrium.
- **Significance of \(K\):** A large \(K\) value suggests that, at equilibrium, products predominate, indicating a reaction that proceeds to completion.
- Conversely, a small \(K\) value indicates that reactants are favored, suggesting a reaction that hardly moves forward.
Knowing \(K\) is crucial as it helps predict how a change in conditions such as concentration, temperature, or pressure might shift the equilibrium position.

Reversible Reactions

Reversible reactions are unique because they can move forward and backward simultaneously. At equilibrium, these reactions do not stop—they achieve a balance where the forward and backward reactions occur at the same rate.
In a reversible reaction, the system can shift its path based on external conditions, such as temperature and concentration.
- **Dynamic Nature:** Even at equilibrium, both the forward and the reverse reactions continue to occur at equal rates. This dynamic characteristic ensures the relationship between reactants and products reaches a steady state.
- **Le Chatelier's Principle:** This principle states that if a change in concentration, temperature, or pressure is applied to a system at equilibrium, the system will adjust to counteract that change. For instance, increasing the concentration of a reactant encourages the forward reaction to restore equilibrium.
Understanding reversible reactions helps us grasp why reactions do not always produce 100% yields and how we can manipulate conditions to favor the production of either reactants or products.

Reaction Stoichiometry

Reaction stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction.
The balanced chemical equation shows us how many moles of reactants are required to form a certain number of moles of products.
For example, in the ammonia formation reaction
\[ N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \] - **Proportional Moles:** For every 1 mole of \(N_2\), 3 moles of \(H_2\) are consumed to produce 2 moles of \(NH_3\).
- These stoichiometric coefficients are critical when calculating the yield or understanding the progression of a reaction.
In equilibrium expressions, these coefficients become exponents, reflecting the proportionate use of reactants and formation of products. This stoichiometric balance ensures the law of conservation of mass applies even at equilibrium.
Being familiar with stoichiometry is vital both in optimizing reaction conditions and in adjusting quantities of reactants to steer reactions in a desired direction.