Problem 4 - Basic Principles of Organic Chemistry

Question

Ethyl chloride \((0.1 \mathrm{M})\) reacts with potassium iodide \((0.1 \mathrm{M})\) in 2 -propanone (acetone) solution at \(60^{\circ}\) to form ethyl iodide and potassium chloride at a rate \((v)\) of \(5.44 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\). a. If the reaction proceeded by an \(S_{N} 2\) mechanism, what is the value of \(k\) (in proper units) and what would be the rate of the reaction in moles per liter per sec at \(0.01\) M concentrations of both reactants? Show your method of calculation. b. Suppose the rate were proportional to the square of the potassium iodide concentration and the first power of the ethyl chloride concentration. What would be the rate with \(0.01\) M reactants? c. It one starts with solutions initially \(0.1 \mathrm{M}\) in both reactants, the rate of formation of ethyl iodide initially is \(5.44 \times 10^{-7} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\), but falls as the reaction proceeds and the reactants are used up. Plot the rate of formation of ethyl iodide against the concentration of ethyl chloride as the reaction proceeds (remembering that one molecule of ethyl chloride consumes one molecule of potassium iodide). Assume that the rate of reaction is proportional to the first power of the ethyl chloride concentration; and to (1) the zeroth power, (2) the first power, and (3) the second power of the potassium iodide concentration. d. What kind of experimental data would one need to determine whether the rate of the reaction of ethyl chloride with potassium iodide is first order in each reactant or second order in ethyl chloride and zero order in potassium iodide?

Step-by-Step Solution

Answer

a. SN2 rate: 5.44×10^{-5} L/mol/s; rate at 0.01 M: 5.44×10^{-9} mol/L/s. b. Rate: 5.44×10^{-11} mol/L/s for alternative order. c. Plot showing dependencies. d. Vary reactant concentrations and measure rate changes.

Step 1: Define the reaction order and rate law for SN2

An SN2 mechanism involves a bimolecular concerted reaction where the rate depends on the concentration of both reactants. The rate law for an SN2 reaction is given by: \[ v = k [C_2H_5Cl][KI] \] Given that both concentrations are 0.1 M and the rate \( v \) is \( 5.44 \times 10^{-7} \text{ mol L}^{-1} ext{s}^{-1} \).

Step 2: Calculate the rate constant k

Using the rate law from Step 1, plug the values into the equation to find \( k \):\[ 5.44 \times 10^{-7} = k (0.1)(0.1) \]\[ k = \frac{5.44 \times 10^{-7}}{0.01} = 5.44 \times 10^{-5} \text{ L mol}^{-1} ext{s}^{-1} \]

Step 3: Calculate the rate at 0.01 M concentrations for SN2 mechanism

For a 0.01 M concentration of each reactant, use the rate constant from Step 2 in the rate law:\[ v = 5.44 \times 10^{-5} \times (0.01) \times (0.01) = 5.44 \times 10^{-9} ext{ mol L}^{-1} ext{s}^{-1} \]

Step 4: Consider an alternative reaction order and calculate rate

If the rate were dependent on the square of \([KI]\) and first power of \([C_2H_5Cl]\), the rate law would be: \[ v = k' [C_2H_5Cl][KI]^2 \]With 0.01 M concentrations for both:\[ v = 5.44 \times 10^{-7} \cdot \left(\frac{(0.01)(0.01)^2}{(0.1)(0.1)^2}\right) = 5.44 \times 10^{-11} ext{ mol L}^{-1} ext{s}^{-1} \]

Step 5: Plot the rate of ethyl iodide formation versus [C2H5Cl]

Assuming a rate proportional to the concentration of ethyl chloride, plot three scenarios: 1. Zero order with respect to KI: Rate is constant. 2. First order with respect to KI: Linear relationship (rate decreases as [KI] decreases linearly). 3. Second order with respect to KI: Rate decreases quadratically as [KI] decreases.

Step 6: Suggest experimental data to determine reaction order

To determine the reaction order, monitor how changes in the initial concentrations of each reactant affect the initial rate of reaction. Perform experiments varying the concentration of one reactant while keeping the other constant. Compare the patterns to zero, first, or second order dependence to establish the actual reaction order.

Key Concepts

Rate Law

The rate law is a mathematical relationship that connects the rate of a chemical reaction to the concentration of its reactants. It shows how the concentration of reactants affects the speed at which a reaction proceeds. In the context of an SN2 reaction, the rate law is often written as: \\[ v = k [ ext{C}_2 ext{H}_5 ext{Cl}][ ext{KI}] \] \This equation indicates that the rate of reaction (\(v\)) is directly proportional to the product of the concentrations of ethyl chloride (\([ ext{C}_2 ext{H}_5 ext{Cl}]\)) and potassium iodide (\([ ext{KI}]\)). \
\The rate constant (\(k\)) is a factor that integrates various conditions (such as temperature) under which the reaction occurs. This makes it a pivotal part of the rate law, as it helps predict how fast a reaction will proceed under specific concentrations of the reactants. \
\The initial rate of an SN2 mechanism suggests that if either reactant's concentration is doubled, the rate should also double, assuming constant temperature and pressure conditions. Understanding how the rate law is formulated allows chemists to predict reaction speeds and optimize conditions for industrial and laboratory processes.

Reaction Order

A reaction order indicates the power to which the concentration of a reactant is raised in the rate law. It defines the relationship between the concentration of reactants and the rate of reaction. For an SN2 reaction:\[ v = k [ ext{C}_2 ext{H}_5 ext{Cl}][ ext{KI}] \] we say it is a second-order reaction because the sum of the powers (exponents) in the equation is two (1 for each reactant).
Different reaction orders dictate different relationships:

• First-order reactions depend linearly on the concentration of one reactant.
• Second-order reactions can depend on either two first-order reactants or one reactant raised to the second power.
• Zero-order reactions have rates that do not depend on reactant concentrations.

Understanding reaction order is key to predicting how changes in concentration affect reaction velocity, a crucial factor when scaling reactions for industrial processes.

Rate Constant

The rate constant, denoted as \(k\), is a crucial parameter in the rate law that relates directly to the reaction speed. It varies with temperature and other specific conditions.
In our SN2 reaction example, the rate constant was calculated as:\[ k = rac{5.44 imes 10^{-7}}{0.01} = 5.44 imes 10^{-5} ext{ L mol}^{-1} ext{s}^{-1} \] This constant helps define the rate of reaction when the concentrations are known. It is affected by:

• Temperature: An increase in temperature typically increases \(k\), speeding up the reaction.
• Reactant nature: Different chemicals will naturally react at different rates.
• Solvent and pressure conditions: These can affect how molecules interact during the reaction.

Having a known rate constant enables chemists to control reaction conditions more precisely, optimizing yields in both small-scale laboratory settings and large-scale manufacturing.

Kinetic Experiments

Kinetic experiments are designed to understand how different variables affect the rate of chemical reactions. They involve monitoring and measuring how the concentration of reactants and products changes over time.
Experiments usually focus on these aspects:

• Initial rate method: Measuring how the rate of reaction changes with varying initial concentrations of reactants. This helps deduce reaction orders.
• Temperature variation: Experiments run at different temperatures to study effects on the rate constant.
• Concentration monitoring: Tracking how reactant concentration changes can help plot reaction curves and maintain consistency with the proposed reaction mechanism.

From these experiments, important data can be collected to elucidate reaction mechanisms and derive kinetic parameters, ensuring accurate and efficient chemical syntheses.