Problem 1 - Chemistry Matter and Change

Question

Write equilibrium constant expressions for these equilibria. \begin{equation} \begin{array}{l}{\text { a. } N_{2} O_{4}(g) \rightleftharpoons 2 N O_{2}(g)} \\\ {\text { b. } 2 \mathrm{H}_{2} \mathrm{S}(9) \rightleftharpoons 2 \mathrm{H}_{2}(\mathrm{g})+\mathrm{S}_{2}(\mathrm{g}) \\ \text { c. } \mathrm{CO}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{4}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})} \\ {\text { d. } 4 \mathrm{NH}_{3}(\mathrm{g})+5 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})} \\ {\text { e. } \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{S}(\mathrm{g}) \rightleftharpoons \mathrm{CS}_{2}(\mathrm{g})+4 \mathrm{H}_{2}(\mathrm{g}) }\end{array} \end{equation}

Step-by-Step Solution

Answer

a: \( \frac{[NO_2]^2}{[N_2O_4]} \), b: \( \frac{[H_2]^2[S_2]}{[H_2S]^2} \), c: \( \frac{[CH_4][H_2O]}{[CO][H_2]^3} \), d: \( \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5} \), e: \( \frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2} \).

Step 1: Understanding the Equilibrium Constant Expression

The equilibrium constant expression for a reaction is given by the ratio of the concentrations of the products raised to their stoichiometric coefficients to the concentrations of the reactants raised to their stoichiometric coefficients. For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K \) is expressed as: \[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \] where \([X]\) denotes the molarity of substance \( X \).

Step 2: Write Equilibrium Expression for Reaction a

For the reaction \( N_{2}O_{4}(g) \rightleftharpoons 2 NO_{2}(g) \), apply the formula. The products are \(NO_2\) and the reactants are \(N_2O_4\). Thus, the expression is: \[ K = \frac{[NO_2]^2}{[N_2O_4]} \]

Step 3: Write Equilibrium Expression for Reaction b

The reaction \( 2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \) has products \(H_2\) and \(S_2\), and reactant \(H_2S\). Therefore, the expression is: \[ K = \frac{[H_2]^2[S_2]}{[H_2S]^2} \]

Step 4: Write Equilibrium Expression for Reaction c

For the reaction \( CO(g) + 3H_2(g) \rightleftharpoons CH_4(g) + H_2O(g) \), the expression is formulated by identifying products \(CH_4\) and \(H_2O\), and reactants \(CO\) and \(H_2\): \[ K = \frac{[CH_4][H_2O]}{[CO][H_2]^3} \]

Step 5: Write Equilibrium Expression for Reaction d

The reaction \( 4NH_3(g) + 5O_2(g) \rightleftharpoons 4NO(g) + 6H_2O(g) \) gives the expression: \[ K = \frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5} \] by following the stoichiometry of each reactant and product.

Step 6: Write Equilibrium Expression for Reaction e

For the equilibrium \( CH_4(g) + 2H_2S(g) \rightleftharpoons CS_2(g) + 4H_2(g) \), we have: \[ K = \frac{[CS_2][H_2]^4}{[CH_4][H_2S]^2} \] based on the concentrations and their respective coefficients.

Key Concepts

Chemical Equilibria

Chemical equilibria represent a state in a chemical reaction where the forward and backward reactions happen at the same rate. At this point, the concentrations of the reactants and products remain constant over time, even though they might not be equal.

The equilibrium is characterized by the equilibrium constant, denoted as \(K\), which provides a measure of the relative concentrations of products and reactants at equilibrium. This concept is paramount for predicting how a reaction will behave under different conditions, such as changes in concentration, pressure, or temperature.

The position of equilibrium tells us if the products or reactants are favored. A high \(K\) value indicates a large concentration of products at equilibrium, whereas a low \(K\) value suggests that reactants are more prevalent.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It uses the balanced chemical equation to determine the relative amounts of substances involved.

When writing equilibrium constant expressions, stoichiometry plays a critical role as it determines the powers to which the concentrations are raised in the expression. For example, in the equilibrium reaction: \[2H_2S(g) \rightleftharpoons 2H_2(g) + S_2(g) \]The coefficients from the balanced chemical equation become exponents in the equilibrium expression: \[K = \frac{[H_2]^2[S_2]}{[H_2S]^2}.\]

By following the stoichiometry of each reactant and product properly, one can ensure accurate calculations of the equilibrium positions.

Reaction Quotient

The reaction quotient, \(Q\), is a valuable concept that offers insight into a chemical reaction's direction compared to its equilibrium. The formula for \(Q\) looks exactly like that for the equilibrium constant \(K\), but it uses the initial concentrations rather than those at equilibrium.

By calculating \(Q\) and comparing it to \(K\), one can predict whether a reaction needs to shift to the left (towards reactants) or to the right (towards products) to reach equilibrium:

• If \(Q < K\), the reaction will proceed forward, towards the products, to achieve equilibrium.
• If \(Q > K\), the reaction will shift backwards, favoring the formation of reactants.
• If \(Q = K\), the reaction is at equilibrium.

This tool is vital for chemists in determining the status of a reaction at any point before equilibrium is actually reached.

Le Chatelier's Principle

Le Chatelier's Principle predicts how a chemical equilibrium will respond to disturbances or changes in conditions, such as concentration, temperature, and pressure. When a system at equilibrium experiences such a change, it will adjust itself to partially counteract the effect of the change:

• Increase in concentration: The system will shift to the side that reduces the added concentration.
• Temperature change: For endothermic reactions, increasing temperature favors products, while for exothermic reactions, it favors reactants.
• Pressure change: Increasing pressure will favor the side with fewer moles of gas.

Understanding Le Chatelier's Principle helps to predict the behavior of a system and allows chemists to control reaction conditions to yield more products efficiently.