Problem 1 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

Write the expressions for the thermodynamic equilibrium constant \(K \text { for the following reactions. (Section } 15.1)\) (a) \(\quad 4 \mathrm{NH}_{3}(\mathrm{g})+7 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 4 \mathrm{NO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{f})\) (b) \(\mathrm{HCN}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{CN}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\) (c) \(\quad \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) (d) \(3 \mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{O}_{3}(\mathrm{g})\) (e) \(2 \mathrm{H}_{2} \mathrm{O}() \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{*}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq})\) (f) \(3 \operatorname{Zn}(s)+2 F e^{3+}(a q) \rightleftharpoons 2 F e(s)+3 \operatorname{Zn}^{2}+(a q)\)

Step-by-Step Solution

Answer

K_a = \frac{[\text{NO}_2]^4}{[\text{NH}_3]^4 [\text{O}_2]^7}, K_b = \frac{[\text{CN}^-][\text{H}_3\text{O}^+]}{[\text{HCN}]}, K_c = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}, K_d = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}, K_e = [\text{H}_3\text{O}^+][\text{OH}^-], K_f = \frac{[\text{Zn}^{2+}]^3}{[\text{Fe}^{3+}]^2}.

Step 1: Understanding the Thermodynamic Equilibrium Constant

The equilibrium constant \(K\) for a reaction at a given temperature is determined by the concentrations (or partial pressures for gases) of the reactants and products. For a general reaction \(aA + bB \rightleftharpoons cC + dD\), \(K\) is given by \(\frac{[C]^c [D]^d}{[A]^a [B]^b}\), where \([X]\) represents the concentration or pressure of species X. For pure solids and liquids, the concentration is factored as 1.

Step 2: Calculate K for Reaction (a)

The reaction is \(4 \text{NH}_3(\text{g}) + 7 \text{O}_2(\text{g}) \rightleftharpoons 4 \text{NO}_2(\text{g}) + 6 \text{H}_2\text{O}(\text{f})\). Since H2O is in the form of a liquid (factored as 1), the equilibrium constant expression is \[K = \frac{[\text{NO}_2]^4}{[\text{NH}_3]^4 [\text{O}_2]^7}.\]

Step 3: Calculate K for Reaction (b)

The reaction is \(\text{HCN}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightleftharpoons \text{CN}^- (\text{aq})+ \text{H}_3\text{O}^+ (\text{aq})\). As H2O is a liquid, it is not included in the \(K\) expression. Thus, \[K = \frac{[\text{CN}^-][\text{H}_3\text{O}^+]}{[\text{HCN}]}.\]

Step 4: Calculate K for Reaction (c)

The reaction is \(\text{PCl}_5(\text{g}) \rightleftharpoons \text{PCl}_3(\text{g}) + \text{Cl}_2(\text{g})\). Since all species are gases, \(K\) is expressed in terms of partial pressures: \[K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]}.\]

Step 5: Calculate K for Reaction (d)

The reaction is \(3 \text{O}_2(\text{g}) \rightleftharpoons 2 \text{O}_3(\text{g})\). Since all species are gases, \[K = \frac{[\text{O}_3]^2}{[\text{O}_2]^3}.\]

Step 6: Calculate K for Reaction (e)

The reaction is \(2 \text{H}_2\text{O}() \rightleftharpoons \text{H}_3\text{O}^+(\text{aq}) + \text{OH}^-(\text{aq})\). Here, H2O is a liquid and thus not included in the expression: \[K = [\text{H}_3\text{O}^+][\text{OH}^-].\]

Step 7: Calculate K for Reaction (f)

The reaction is \(3 \text{Zn}(\text{s}) + 2 \text{Fe}^{3+}(\text{aq}) \rightleftharpoons 2 \text{Fe}(\text{s}) + 3 \text{Zn}^{2+}(\text{aq})\). \(K\) expression excludes solids: \[K = \frac{[\text{Zn}^{2+}]^3}{[\text{Fe}^{3+}]^2}.\]

Key Concepts

Reaction Equilibrium Expressions

In the study of chemistry, understanding equilibrium is crucial. Equilibrium expressions allow chemists to predict how a reaction behaves. It represents the state where the rate of the forward reaction equals the rate of the backward reaction. In simpler terms, the amounts of reactants and products no longer change over time, despite both reactions happening continuously. The general form of an equilibrium expression for a reaction \( aA + bB \rightleftharpoons cC + dD \) is written as: \[ K = \frac{[C]^c [D]^d}{[A]^a [B]^b} \].
Here, \([X]\) denotes the concentration of chemical species \(X\). This expression reveals the ratio of the product concentrations to the reactant concentrations. It is essential because it quantifies the extent of a reaction at equilibrium once the concentrations are known.

Equilibrium Constant Calculations

Calculating the equilibrium constant \( K \) helps in predicting the position of equilibrium for a chemical reaction. If \( K \) is large, products are favored, representing a reaction that goes mostly to completion. Conversely, a small \( K \) suggests that reactants are favored, with minimal product formation. Calculations involve substituting the concentrations or partial pressures of the substances in the equilibrium expression.
Consider the reaction \( 4 \, \text{NH}_3( ext{g}) + 7 \, \text{O}_2( ext{g}) \rightleftharpoons 4 \, \text{NO}_2( ext{g}) + 6 \, \text{H}_2\text{O}( ext{f}) \). The equilibrium constant expression, given that water is in liquid form, is \[ K = \frac{[\text{NO}_2]^4}{[\text{NH}_3]^4 [\text{O}_2]^7} \].
Through similar steps, you can determine \( K \) for any balanced reaction. Remember that the coefficients from the balanced equation become the powers for each species in the expression.

Influence of Phases on Equilibrium Constant

The phase of each reactant and product in a reaction influences the equilibrium constant expression. Only gases and aqueous solutions are included in the equilibrium expression, while pure solids and liquids are not. This simplification occurs because the concentration of pure solids and liquids is constant.
For example, in the reaction \( \text{HCN}( ext{aq}) + \text{H}_2\text{O}( ext{l}) \rightleftharpoons \text{CN}^- ( ext{aq}) + \text{H}_3\text{O}^+ ( ext{aq}) \), water is a liquid. Hence, it is excluded from the equilibrium expression: \[ K = \frac{[\text{CN}^-][\text{H}_3\text{O}^+]}{[\text{HCN}]} \]. This principle simplifies the calculation and focuses on the species whose concentrations change significantly during the reaction.

Use of Concentrations and Pressures in K Expressions

For an equilibrium expression, you might use concentrations if the substances are in solution, or pressures if the substances are gases. This dual approach allows chemists to adapt the expression to the physical state of reactants and products. For gas-phase reactions, it’s often convenient to express the equilibrium in terms of partial pressures.
Consider the reaction \( \text{PCl}_5( ext{g}) \rightleftharpoons \text{PCl}_3( ext{g}) + \text{Cl}_2( ext{g}) \). Here, all species are gases, so the equilibrium constant is given in terms of partial pressures: \[ K = \frac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} \].
Using partial pressures is analogous to using concentrations for solutions but is better suited for reactions where gases are involved.