Question

Consider the reaction A temperature-jump experiment is performed where the relaxation time constant is measured to be \(310 \mu\) s, resulting in an equilibrium where \(K_{e q}=0.70\) with \([\mathrm{P}]_{e q}=0.20 \mathrm{M}\) What are \(k\) and \(k^{\prime} ?\) (Watch the units!)

Short answer

The rate constants \(k\) and \(k'\) for the given reaction are approximately 1328.27 s⁻¹ and 1897.54 s⁻¹, respectively.

Step-by-step solution

Plug in the Given Values

We are given the time relaxation constant \(\tau = 310 \mu s\), which is equivalent to 310 × 10⁻⁶ s and the equilibrium constant \(K_{eq} = 0.70\). We are also given the concentration of the product, \([P]_{eq} = 0.20 M\).

Find Sum of Rate Constants Using Time Relaxation Constant Formula

We can find the sum of the rate constants \(k\) and \(k'\) using the formula \(\tau = \frac{1}{k + k'}\). Using the given relaxation time constant, we have: \(310 \cdot 10^{-6} s = \frac{1}{k + k'}\) Rearranging the equation to find the sum of rate constants: \(k + k' = \frac{1}{310 \cdot 10^{-6}}\)

Calculating the Sum of Rate Constants

Calculate the value of \(k + k'\) using the equation derived in Step 2: \(k + k' = \frac{1}{310 \cdot 10^{-6}} ≈ 3225.81 s^{-1}\)

Finding \(k\) and \(k'\) using Equilibrium Constant Formula

Now we can use the equilibrium constant formula to find the values of \(k\) and \(k'\), given that the ratio between them is already known (\(K_{eq} = 0.70\)). By using the formula \(K_{eq}=\frac{k}{k'}\), we have: \(0.70 = \frac{k}{k'}\)

Solve for the Rate Constants

To find the values of \(k\) and \(k'\), we can set up a system of equations using the information from Steps 3 and 4: \( \begin{cases} k + k' = 3225.81 s^{-1} \\ \frac{k}{k'} = 0.70 \end{cases} \) From the second equation, we have \(k = 0.70 k'\). Substituting this into the first equation, we get: \(0.70 k'+ k' = 3225.81 s^{-1}\) Dividing both sides by 1.70, we find the value of \(k'\): \(k' = \frac{3225.81 s^{-1}}{1.70} ≈ 1897.54 s^{-1}\) Now, we can find the value of \(k\) using the equation \(k = 0.70 k'\): \(k = 0.70 \cdot 1897.54 s^{-1} ≈ 1328.27 s^{-1}\)

Final Answer

The rate constants \(k\) and \(k'\) for the given reaction are approximately 1328.27 s⁻¹ and 1897.54 s⁻¹, respectively.

Key concepts

Relaxation Time Constant

The relaxation time constant, often denoted by \( \tau \), is a fundamental concept in studying the kinetics of chemical reactions. In a physical chemistry context, it represents the time it takes for a system to return to equilibrium after a disturbance - such as a temperature jump - has been applied.

In a temperature-jump experiment, the system's response to a sudden increase in temperature can be monitored by observing how quickly it recovers or 'relaxes' back to equilibrium. The relaxation time constant is mathematically related to the rate constants of the forward and reverse reactions by the equation \( \tau = \frac{1}{k + k'} \), where \( k \), and \( k' \) represent the rate constants of the forward and reverse reactions, respectively.

Understanding the relaxation time constant is instrumental in the study of reaction mechanisms and dynamics, as it provides insights into the speed of the reaction under study as well as the intermediate steps that it may involve.

Equilibrium Constant

The equilibrium constant, expressed as \( K_{eq} \), is a dimensionless quantity that reflects the ratio of the concentration of products to reactants at equilibrium state in a reversible chemical reaction. It is a crucial aspect when determining a reaction's spontaneity and extent.

For a general reaction \( aA + bB \rightleftharpoons cC + dD \), the equilibrium constant is given by the formula \( K_{eq} = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \) where \( [X] \) denotes the concentration of species \( X \) at equilibrium. In a temperature-jump experiment, knowing the equilibrium constant allows us to estimate the relationship between the rate constants for the forward and reverse reactions.

Furthermore, an understanding of the equilibrium constant can aid students in predicting how changes in conditions, such as temperature or pressure, will affect the position of equilibrium - allowing for the exploration of Le Chatelier's principle and its applications in real-world scenarios.

Rate Constants

Rate constants are fundamental parameters in physical chemistry kinetics that define the speed at which a chemical reaction proceeds. Represented by \( k \) for the forward reaction and \( k' \) for the reverse reaction, these constants are determined experimentally and are inherently related to the activation energy and the temperature of the reaction environment.

The relationship between rate constants and reaction rate is captured by the rate law, which is often written as \( rate = k[A]^{m}[B]^{n} \) for a reaction \( A + B \rightarrow products \), where \( m \) and \( n \) are the reaction orders and \( [A] \) and \( [B] \) are the reactant concentrations.

By investigating rate constants, chemists can deduce the overall rate of the reaction and gain a deeper understanding of the reaction mechanism - the step-by-step sequence of elementary reactions by which a chemical change occurs.

Physical Chemistry Kinetics

Physical chemistry kinetics encompasses the study of the rates of chemical processes and the factors affecting them. It provides insight into how quickly reactants turn into products and which variables - such as temperature, pressure, and concentration - can influence this rate.

One key aspect of kinetics is the study of rate laws and reaction orders, which help in formulating mathematical models for reaction rates. Reaction mechanisms are another focal area, showing the sequence of steps that lead to the formation of products, including potential intermediates and transition states.

A solid grasp of kinetics is essential for predicting and controlling the outcomes of chemical reactions in various industries, from pharmaceuticals to energy production. Moreover, kinetics plays a critical role in the development of catalytic processes that can speed up chemical reactions without being consumed by the reaction itself.

Reaction Mechanisms

Reaction mechanisms peel away the simple equation of a chemical reaction to reveal the intricate ballet of atoms and molecules as they rearrange to form new products. They depict a detailed, step-by-step portrayal of how reactants are transformed into products, including the formation of any short-lived intermediates or unstable transition states.

Understanding the reaction mechanism is vital because it explains not just the overall outcome of a reaction, but also how energy is transferred, how molecular geometry changes, and how various reaction pathways contribute to the speed and selectivity of the process.

Studying mechanisms helps chemists design new reactions and synthesize complex molecules with a high degree of specificity. It also has immense implications in areas such as drug development, where a profound understanding of a reaction can steer chemists toward more efficient and safer production methods.