Question

Calculate the \(\mathrm{pH}\) which would give the most rapid reaction rate for a carbonyl compound with \(\mathrm{K}_{\mathrm{B}}=10^{-14}\) and an \(\mathrm{RNH}_{2}\) derivative with \(\mathrm{K}_{\mathrm{B}}=10^{-11}\) assuming \(1 \mathrm{M}\) concentrations for the reactants, and the slow step being as in Eq.(1). How many times faster is the rate at this \(\mathrm{pH}\) than at \(\mathrm{pH} 0\) ? Than at \(\mathrm{ph} 7 ?\)

Short answer

The most rapid reaction rate occurs at a pH of \(12.5\). At this pH, the reaction is \(10^{-25}\) times faster than at pH \(0\) and \(10^{-11}\) times faster than at pH \(7\).

Step-by-step solution

Use the K_B values to find the base equilibrium constant

The equilibrium constant for a base is given by the formula: \[ K_B = \frac{ [\mathrm{OH}^-][\mathrm{BH^+}] }{ [\mathrm{B}] } \] Here, \(\mathrm{B}\) is the base, and \(\mathrm{BH^+}\) is its corresponding conjugate acid. The \(\mathrm{K}_{\mathrm{B}}\) values for the carbonyl compound and the amine derivative are given as \(10^{-14}\) and \(10^{-11}\). Keep in mind the assumption of 1 M concentrations for reactants.

Calculate the pH conditions for maximum reaction rate

The pH conditions for maximum reaction rate can be found by calculating the equilibrium constant for the reaction and setting this value equal to the product of the base dissociation constants for the carbonyl compound and the amine derivative. The equilibrium constant for this combined reaction can be obtained by multiplying the individual equilibrium constants, which are given as: \[ K = K_{B1} \times K_{B2} \] Substitute the values, \(K_{B1} = 10^{-14}\) and \(K_{B2} = 10^{-11}\), into the formula: \[ K = 10^{-14} \times 10^{-11} \] Which yields: \[ K = 10^{-25} \] Now, let's apply the formula to calculate the pH: \[ pH = \frac{-\log (K)}{2} \] Substitute the value of \(K = 10^{-25}\) into the formula: \[ pH = \frac{-\log (10^{-25})}{2} \] Simplifying the equation gives the pH value that results in the most rapid reaction rate: \[ pH = \frac{25}{2} = 12.5 \]

Calculate the rate acceleration factors

To calculate the rate acceleration factors at the optimal pH value compared to pH \(0\) and \(7\), we will need to find the ratios of the reaction rates at these pH values. The rates can be expressed in terms of the equilibrium constant: \[ \frac{Rate_{optimal}}{Rate_{0}} = \frac{K_{optimal}}{K_{0}} \] Since pH = 0 corresponds to \(K_{0} = 1\): \[ \frac{Rate_{optimal}}{Rate_{0}} = \frac{10^{-25}}{1} = 10^{-25} \] Then to compare the reaction rate at optimal pH to that at pH \(7\), use the same concept: \[ \frac{Rate_{optimal}}{Rate_{7}} = \frac{K_{optimal}}{K_{7}} \] Since pH = 7 corresponds to \(K_{7} = 10^{-14}\): \[ \frac{Rate_{optimal}}{Rate_{7}} = \frac{10^{-25}}{10^{-14}} = 10^{-11} \] So, at the optimal pH of \(12.5\), the reaction is \(10^{-25}\) times faster than at pH \(0\) and \(10^{-11}\) times faster than at pH \(7\).

Key concepts

Equilibrium Constant

The equilibrium constant, often denoted as K, is a crucial concept in chemical kinetics and equilibrium. It quantifies the ratio of the concentration of products to the concentration of reactants at equilibrium for a reversible reaction. In mathematical terms, for a reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant \( K_{eq} \) is expressed as \( K_{eq} = \frac{[C]^c \times [D]^d}{[A]^a \times [B]^b} \), where the bracketed terms represent molar concentrations of the respective species.

For a base dissociation reaction like \( B + H_2O \rightleftharpoons BH^{+} + OH^{-} \), the equilibrium constant is known as the base dissociation constant \( K_B \) and is defined as \( K_B = \frac{[BH^{+}][OH^{-}]}{[B]} \). The equilibrium constant is central in determining the position of equilibrium and gives insight into the favorability of the reaction. It remains constant at a given temperature, regardless of the changes in concentration of reactants or products, provided the temperature does not change.

Understanding how to utilize this constant can help students predict the extent of a reaction and the concentrations of different species in a mixture at equilibrium.

Base Dissociation Constant

The base dissociation constant, symbolized as \( K_B \), is a special type of equilibrium constant that applies to bases. It measures the strength of a base to dissociate into its respective ions in water. For the generic base \( B \), the dissociation in water is represented as \( B + H_2O \rightleftharpoons BH^{+} + OH^{-} \). The base dissociation constant is given by \( K_B = \frac{[BH^{+}][OH^{-}]}{[B]} \), where the brackets indicate the molar concentration of each species in equilibrium.

The larger the \( K_B \) value, the stronger the base, meaning it dissociates more completely in solution. Understanding \( K_B \) is essential when calculating the pH of solutions and for determining the optimal conditions for chemical reactions involving bases, as seen in the reaction rate problem where \( K_B \) values were used to find the most favorable pH for a reaction.

Rate Acceleration Factor

The rate acceleration factor is an indicator of how much faster a chemical reaction proceeds under certain conditions compared to a reference state. It is calculated by dividing the rate of the reaction under one set of conditions by the rate under another set. This value helps to understand the effect of different variables, such as temperature, pH, catalysts, or concentration, on the speed of a reaction.

In the context of pH and reaction rates, the rate acceleration factor can be expressed through the equilibrium constants at different pH levels. For example, if the equilibrium constant at the optimal pH is \( K_{optimal} \) and at pH 0 is \( K_{0} \), the rate acceleration factor would be \( \frac{K_{optimal}}{K_{0}} \). If the reaction is significantly faster at the optimal pH than at pH 0, the rate acceleration factor would be a large number, indicating a substantial increase in the reaction rate due to the change in pH. Knowing this factor is incredibly useful for optimizing reactions in both laboratory and industrial settings, ensuring that chemical processes are carried out efficiently and effectively.