Question

Determine \(K_{c}\) for each of the following equilibria from the value of \(K\) : (a) \(2 \mathrm{NOCl}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{~g}), K=1.8 \times 10^{-2}\) at 500 . K (b) \(\mathrm{CaCO}_{3}\) (s) \(\rightleftharpoons \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}), K=167\) at \(1073 \mathrm{~K}\)

Short answer

\( K_{c(a)} = 1.8 \times 10^{-2} \) and \( K_{c(b)} = 167 \)

Step-by-step solution

Identify the Equilibrium Expression for Reaction (a)

Write the equilibrium expression for the reaction of NOCl to NO and Cl2. For the reaction 2NOCl(g) ⇌ 2NO(g) + Cl2(g), the equilibrium constant expression is given by: \[ K_c = \frac{[NO]^2[Cl_2]}{[NOCl]^2} \]

Calculating Kc for Reaction (a)

Since the given K value of 1.8 x 10^-2 for reaction (a) represents the concentration-based equilibrium constant Kc, no conversion is necessary. Therefore, \( K_{c(a)} = 1.8 \times 10^{-2} \).

Identify the Equilibrium Expression for Reaction (b)

Write the equilibrium expression for the decomposition of CaCO3 into CaO and CO2. For the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), the equilibrium constant expression is: \[ K_c = [CO_2] \] Note that solids are not included in the expression. Solids (s) and pure liquids (l) do not appear in the equilibrium expression because their concentrations do not change during the reaction.

Calculating Kc for Reaction (b)

Given the K value for reaction (b) is 167, this implies that Kc only involves the concentration of CO2 because CaCO3 and CaO are solids and do not appear in the equilibrium expression. Therefore, \( K_{c(b)} = 167 \).

Key concepts

Chemical Equilibrium

In the realm of chemistry, the concept of chemical equilibrium plays a crucial role in understanding how reactions proceed. It occurs when a chemical reaction and its reverse reaction proceed at the same rate, leading to an unchanging ratio of reactants and products. Although both the forward and reverse reactions are ongoing, their speeds are equal, creating a dynamic balance rather than a static one. This state doesn't mean the reactants and products have vanished, but rather that their concentrations have stabilized and will remain steady, given that temperature and other external conditions are constant.

Equilibrium can be visualized on the molecular level as a busy interchange where two opposing lanes of traffic—the reactants and products—move at the same pace, allowing no overall change in traffic density. The point to grasp here is that equilibrium represents a balance in chemical processes, not the end point of a reaction. In the context of the exercise problem, for the reactions given, equilibrium will be when the rate of decomposition of NOCl is equal to the rate of formation of NO and Cl2 for reaction (a), and the rate of decomposition of CaCO3 is equal to the rate of formation of CaO and CO2 for reaction (b).

Concentration-Based Equilibrium Constant

Diving deeper into equilibrium, we encounter the concentration-based equilibrium constant, represented by the symbol Kc. It's a numerical value that represents the ratio of product concentrations to reactant concentrations at equilibrium, with each raised to the power of their respective coefficients in the balanced chemical equation. It is a key indicator of the extent to which a reaction proceeds before reaching equilibrium.

Understanding Kc helps predict the direction of the reaction—whether it favors the formation of products or the reactants. A larger Kc value suggests a reaction that predominantly results in products, while a smaller value indicates that the reactants are more favored at equilibrium. In our exercise, the given values of Kc for reactions (a) and (b) are 1.8 x 10^-2 and 167, respectively. In reaction (b), solids are excluded from the expression since their concentration remains constant and does not affect the equilibrium condition.

Reaction Quotient

When dealing with chemical reactions, it's not always the case that the system will be at equilibrium. This is where the reaction quotient, Qc, comes into play. It is calculated in the same way as the equilibrium constant, Kc, using the instantaneous concentrations of the reactants and products at any given point in time, not just at equilibrium.

The real power of Qc lies in its ability to predict the direction in which a reaction will shift to reach equilibrium. If Qc is less than Kc, the system will proceed forward, converting reactants to products, to achieve equilibrium. Conversely, if Qc is greater than Kc, the reaction will proceed in the reverse direction to reach equilibrium. When Qc equals Kc, the system is already at equilibrium. This comparative analysis is crucial for chemists to manipulate conditions in favor of desired product yield.