Question

For which of the following reactions is \(K_{\mathrm{c}}\) equal to \(K_{P}\) ? For which can we not write a \(K_{P}\) expression? (a) \(4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \rightleftarrows 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{CaCO}_{3}(s) \rightleftarrows \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(\mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \rightleftarrows \mathrm{Zn}^{2+}(a q)+\mathrm{H}_{2}(g)\) (d) \(\mathrm{PCl}_{3}(g)+3 \mathrm{NH}_{3}(g) \rightleftarrows 3 \mathrm{HCl}(g)+\mathrm{P}\left(\mathrm{NH}_{2}\right)_{3}(g)\) (e) \(\mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \rightleftarrows \mathrm{NH}_{4} \mathrm{Cl}(s)\) (f) \(\mathrm{NaHCO}_{3}(s)+\mathrm{H}^{+}(a q) \rightleftarrows \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)+\) \(\mathrm{Na}^{+}(a q)\) (g) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightleftarrows 2 \mathrm{HF}(g)\) (h) \(\mathrm{C}(\) graphite \()+\mathrm{CO}_{2}(g) \rightleftarrows 2 \mathrm{CO}(g)\)

Short answer

Reactions (d) and (g) have \(K_c = K_p\). Reactions (b), (e), and (f) don't have \(K_p\) expressions.

Step-by-step solution

Understanding the Relationship Between Kc and Kp

The equilibrium constants, \(K_c\) and \(K_p\), are related by the equation \(K_p = K_c(RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of gas, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. If \(\Delta n = 0\), then \(K_p = K_c\).

Calculate Δn for each reaction

For each reaction, count the number of moles of gaseous products and reactants to determine \(\Delta n\). If \(\Delta n = 0\), then \(K_c = K_p\). If \(\Delta n eq 0\), then \(K_c eq K_p\).- (a) \(\Delta n = (4+6) - (4+5) = 1\)- (b) \(\Delta n = 1 - 0 = 1\)- (c) \(\Delta n = 1 - 0 = 1\)- (d) \(\Delta n = (3+1) - (1+3) = 0\)- (e) \(\Delta n = 0 - (1+1) = -2\)- (f) \(\Delta n = 1 - 0 = 1\)- (g) \(\Delta n = 2 - (1+1) = 0\)- (h) \(\Delta n = 2 - 1 = 1\)

Identify Reactions where Kc = Kp

From the calculated \(\Delta n\), Reactions (d) and (g) both have \(\Delta n = 0\), so for these reactions \(K_c = K_p\).

Identify Reactions where Kp Expression is Not Possible

A \(K_p\) expression can only be written for reactions involving gases. For solids and liquids, \(\Delta n\) is not applicable, and therefore \(K_p\) cannot be calculated:- (b) Includes solids only.- (e) Product is a solid; reactants are gases.- (f) Includes solid and liquid states in the products.

Key concepts

Kc and Kp relationship

Understanding the relationship between the equilibrium constants, \(K_c\) and \(K_p\), is crucial when discussing gas-phase reactions. These constants represent the concentrations of reactants and products at equilibrium but in different units. \(K_c\) relates to concentrations (moles per liter), while \(K_p\) relates to partial pressures. The relationship can be expressed via the equation:\[ K_p = K_c(RT)^{\Delta n} \]Here, \(R\) is the ideal gas constant, \(T\)is temperature in Kelvin, and \(\Delta n\) is the change in the moles of gases from reactants to products. If \(\Delta n = 0\), indicating no net change in the number of moles of gas in the reaction, then \(K_p = K_c\). This shows that the equilibrium constant in terms of pressure will equal the one in terms of concentration.

Equilibrium expressions

Equilibrium expressions are mathematical representations of chemical equilibrium. They allow you to predict the concentrations or pressures of reactants and products at equilibrium. The general form for a reaction \(aA + bB \rightarrow cC + dD\) would be expressed as:\[ K = \frac{[C]^c[D]^d}{[A]^a[B]^b} \]Here, \([A]\), \([B]\), \([C]\), and \([D]\) are the concentrations or pressures of the chemical species involved. In gas-phase reactions, partial pressures are typically used, leading to the equilibrium expression written as \(K_p\). Always remember that pure solids and liquids are included in calculations with a value of one and do not appear in the expression. This means that for reactions with solids like CaCO₃ or liquids like H₂O(l), they do not impact \(K_p\) directly.

Gas phase reactions

Gas-phase reactions are a critical category for studying equilibrium constants, as they often involve changes in the number of moles of gas, which directly affects \(\Delta n\). When analyzing these reactions, it's important to focus on the gaseous state reactants and products: - Changes in the moles of gases can significantly alter the equilibrium positions.- Reactions \((a)\), \((d)\), \((e)\), and \((g)\) are predominantly gas-phase and, thus, suitable for \(K_p\) analysis if \(\Delta n\) is calculated.- Non-gaseous components, such as solids or liquids, are omitted when setting up expressions for \(K_p\).These factors make it vital to properly determine \(\Delta n\) for each reaction, ensuring accurate application of \(K_p = K_c(RT)^{\Delta n}\).

Delta n calculation

The calculation of \(\Delta n\), which represents the net change in moles of gas between products and reactants, is a fundamental step in equating \(K_c\) to \(K_p\) or determining their relationship. You calculate \(\Delta n\) as follows:- Sum the moles of gaseous products.- Subtract the moles of gaseous reactants.Illustratively, for reaction \(a\), \(\Delta n = (4+6) - (4+5) = 1\). If \(\Delta n = 0\), it implies an equal number of gas moles on both sides, resulting in \(K_c = K_p\), like in reactions \((d)\) and \((g)\). For cases where \(\Delta n eq 0\), adjustments using the expression \(K_p = K_c(RT)^{\Delta n}\) are necessary to equate the constants at equilibrium.

Equilibrium constant limitations

Equilibrium constants like \(K_p\) have specific limitations, particularly in terms of the phases of matter involved in the reaction. 1. **Solid and Liquid States**: These states do not appear in \(K_p\) expressions as they are considered to have a constant activity of one.2. **Reactions Without Gases**: In reactions where neither the reactants nor the products are gases, such as in components like CaCO₃(s) (Reaction (b)), \(K_p\) cannot be calculated.3. **Mixtures of Phases**: In mixed reactions containing gases along with solids or liquids, only the gaseous components are included in calculating \(K_p\).This is why reactions \((b)\), \((e)\), and \((f)\) cannot have a valid \(K_p\) expression. Always consider the state of each reactant and product when analyzing equilibrium systems to determine the applicability of \(K_p\).