Question

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$ \mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g) $$ is \(3.8 \times 10^{-5}\) at \(727^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the equilibrium $$ 2 \mathrm{I}(g) \rightleftharpoons \mathrm{I}_{2}(g) $$ at the same temperature.

Short answer

The equilibrium constant \(K_{c}\) for the reaction \(2 \mathrm{I}(g) \rightarrow \mathrm{I}_{2}(g)\) is \(\sqrt{\frac{1}{3.8 \times 10^{-5}}}\) and the \(K_{P}\) for this reaction is \(K_{c}\times (RT)^{-1}\) where \(R\) is \(0.08206 \mathrm{L atm mol^{-1} K^{-1}}\) and \(T\) is \(1000\)K.

Step-by-step solution

Calculate the Equilibrium Constant for the Reversed Reaction

The equilibrium constant for the reversed reaction is the reciprocal of the equilibrium constant for the original reaction. So, \(K_{c1}\) for the reaction \(\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{I}(g)\) is \(3.8 \times 10^{-5}\). Therefore, the equilibrium constant \(K_{c2}\) for the reversed reaction, i.e., \(2 \mathrm{I}(g) \rightarrow \mathrm{I}_{2}(g)\), is the reciprocal of \(K_{c1}\), which equals \(\frac{1}{3.8 \times 10^{-5}}\).

Account for Reaction Coefficient

As the reaction \(\mathrm{I}_{2}(g) \rightarrow 2 \mathrm{I}(g)\) has been reversed and multiplied by a factor of \(1/2\), the equilibrium constant for the new reaction must be raised to the power of \(1/2\). So, the equilibrium constant \(K_{c}\) for the new reaction is \(\sqrt{\frac{1}{3.8 \times 10^{-5}}}\).

Calculate \(K_{P}\)

The relationship between \(K_{c}\) and \(K_{P}\) is given by the equation \(K_{P}=K_{c}(RT)^{\Delta n}\) (according to the ideal gas law), where \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin and \(\Delta n\) is the change in the number of moles of gas. The \(\Delta n\) for the reaction \(2 \mathrm{I}(g) \rightarrow \mathrm{I}_{2}(g)\) is \(-1\). Hence, the \(K_{P}\) for the reaction is \(K_{c}\times (RT)^{-1}\).

Convert Temperature to Kelvin

To use the above calculated \(K_{P}\), it is necessary that temperature should be in Kelvin. To convert, add 273.15 to the Celsius value. \(727^{\circ} \mathrm{C}\) = \(727+273.15\) K = \(1000\mathrm{K}\).

Substitute Values into the Equation

Substitute the values of \(K_{c}\), \(R\) (which is \(0.08206 \mathrm{L atm mol^{-1} K^{-1}}\), and \(T\) into the equation result from Step 3 to compute \(K_{P}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a fundamental concept in chemistry that occurs when a chemical reaction and its reverse reaction proceed at equal rates, leading to stable concentrations of the reactants and products. It is crucial to understand that equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their rates of formation are equal, resulting in no net change over time.

The dynamics of chemical processes are such that, initially, when reactants are present in high concentrations, the forward reaction is favored. As products accumulate, the reverse reaction starts to compete with the forward reaction until both reach the same rate. At this point, the system is said to be at equilibrium. Despite this dynamic balance, the system can be disturbed by changes in temperature, pressure, or concentration, as described by Le Chatelier's principle, but will re-establish a new equilibrium by adjusting the rate of the forward and reverse reactions.

Kc and Kp Relationship

The relationship between the equilibrium constants Kc and Kp is rooted in the different ways they express the concentration of reactants and products in a gas-phase equilibrium: Kc is based on molar concentrations, while Kp uses partial pressures. This distinction is particularly significant in reactions involving gases, where changes in pressure and volume can drastically affect the system.

To convert Kc to Kp and vice versa, you use the equation
\( K_{P} = K_{C}(RT)^{abla n} \).

Here, R represents the ideal gas constant, T stands for temperature in Kelvin, and \( abla n \) is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants. This equation underlines the intrinsic link between these two equilibrium constants and showcases the influence of temperature and molar changes on the position of equilibrium.

Reaction Quotient

The reaction quotient, Q, is a measure that tells us the direction in which a reaction will shift to reach equilibrium. It is calculated using the same formula as the equilibrium constant, but with the initial rather than the equilibrium concentrations or partial pressures of the reactants and products.

By comparing the reaction quotient Q to the equilibrium constant K, we can predict whether a reaction will proceed forward or in reverse to attain equilibrium:

• If \( Q < K \), the forward reaction is favored, and the concentration of products will increase.
• If \( Q > K \), the reverse reaction is favored, and the concentration of reactants will increase.
• If \( Q = K \), the system is already at equilibrium.

This comparative analysis between Q and K allows chemists to determine the direction of the dynamic shift required for the system to reach equilibrium.

Thermodynamics in Chemistry

Thermodynamics in chemistry deals with the energy changes associated with chemical reactions and physical transformations. It helps in providing a molecular-level understanding of enthalpy, entropy, and energy of a system. At the core, the laws of thermodynamics govern the direction of spontaneous processes and the concept of equilibrium.

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed in an isolated system. The second law introduces the concept of entropy, a measure of the randomness or disorder of a system, which generally increases in spontaneous processes. The third law is associated with absolute zero temperature and the entropy of a perfect crystal.

In the context of equilibrium, thermodynamics helps predict whether a reaction is exothermic or endothermic and how temperature changes affect the position of equilibrium. Furthermore, it lays the framework for understanding the Gibbs free energy, a quantity that combines enthalpy and entropy to predict the spontaneity of reactions at constant pressure and temperature.