Question

In which of the following reaction is \(K_{p}\) less than \(K_{c} ?\) (a) \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g})\) (b) \(\mathrm{PCl}_{5}(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g})\) (c) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})\) (d) \(2 \mathrm{SO}_{3}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})\)

Short answer

Among the provided reactions, \(K_p\) is less than \(K_c\) only for reaction (a), \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\).

Step-by-step solution

Analyzing reaction (a)

For the reaction \(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\), the moles of gaseous reactants are 1 (for \(N_2\)) + 3 (for \(H_2\)) = 4, whereas moles of gaseous products are 2 (for \(NH_3\)). So, \(\Delta n = n_{products} - n_{reactants} = 2 - 4 = -2\), which is less than 0. So for this reaction, \(K_p < K_c\).

Analyzing reaction (b)

For the reaction \( PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)} \), the number of moles of gaseous reactants is 1 (for \(PCl_5\)) and the number of moles of gaseous products are 1 (for \(PCl_3\)) + 1 (for \(Cl_2\)) = 2. So, \(\Delta n = 2 - 1 = 1\), which is more than 0. So for this reaction, \(K_p > K_c\).

Analyzing reaction (c)

For the reaction \(H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI(g)\), the number of moles of gaseous reactants is 1 (for \(H_2\)) + 1(for \(I_2\)) = 2 and the number of moles for gaseous products is 2 (for \(HI\)). So, \(\Delta n = 2 - 2 = 0\). So for this reaction, \(K_p = K_c\).

Analyzing reaction (d)

For the reaction \(2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}\), the number of moles of gaseous reactants is 2 (for \(SO_3\)) and the number of moles of gaseous products are 2 (for \(SO_2\)) + 1 (for \(O_2\)) = 3. So, \(\Delta n = 3 - 2 = 1\), which is more than 0. So for this reaction, \(K_p > K_c\).

Key concepts

Kp vs Kc

Understanding the difference between the equilibrium constants K_p and K_c is crucial for grasping the concepts of chemical equilibrium. K_p is used to express the equilibrium position of a gas-phase reaction in terms of partial pressures, while K_c is used for concentrations or molarity of the reactants and products. The relationship between these two constants is not always intuitive, but it can be described by the equation:
\[ K_p = K_c(RT)^{\Delta n} \]
where R is the gas constant, T is the temperature in Kelvin, and \Delta n is the change in the number of moles of gas (product moles minus reactant moles). When \Delta n is greater than 0, K_p is larger than K_c, and vice versa. If \Delta n is 0, then K_p equals K_c. To address the exercise, we must consider this relationship and analyze the stoichiometry of each reaction.

Reaction Quotient

The reaction quotient, Q, serves as a predictive aid in determining the direction in which a chemical reaction must proceed to reach equilibrium. Q is calculated in the same way as equilibrium constants, but with the initial concentrations or partial pressures instead of those at equilibrium. Subsequently, one can compare the reaction quotient to the equilibrium constant (K):

• If Q < K, the reaction will proceed in the forward direction.
• If Q > K, the reaction will shift in the reverse direction.
• If Q = K, the reaction is at equilibrium.

Understanding Q helps predict whether K_p or K_c would be greater or lesser in a given reaction scenario.

Chemical Equilibrium

Chemical equilibrium is the stage in a reversible chemical reaction where the rate of the forward reaction equals the rate of the reverse reaction, leading to no net change in the concentration of reactants and products. It’s important to note that equilibrium does not mean the reactants and products are equal in concentration. Instead, they are at a ratio defined by the equilibrium constant, K. In the context of our exercise, achieving chemical equilibrium allows us to use the constants K_p and K_c to describe the reaction's balance point.

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture would exert if it alone occupied the entire volume. It's vital when talking about gas-phase reactions, and K_p expressions are used in equilibrium calculations to represent how the gases contribute to the overall pressure. The partial pressure of a gas can be determined using the ideal gas law, as long as the volume, temperature, and moles of gas are known. In our step-by-step solutions, partial pressures would directly affect the value of K_p when relating to the change in moles of gas.

Molar Concentration

Molar concentration, or molarity, is the amount of substance (mol) per unit volume of solution (L), typically expressed as moles per liter (M). For chemical reactions in solution or for aqueous reactants and products, equilibrium calculations are based on these molar concentrations, leading to an equilibrium expression for K_c. The relationship between molar concentration and partial pressure is essential when we consider reactions involving gases, as we can compute one from the other, given the ideal gas law and the conditions of the system. In the exercise context, we use molarity when examining changes in the molecular count as it directly influences the equilibrium constant K_c.