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Chapter 14: Problem 41

(a) The gas-phase decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}, \mathrm{SO}_{2} \mathrm{Cl}_{2}(g)\) \(\longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g),\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2} .\) At 600 \(\mathrm{K}\) the half-life for this process is \(2.3 \times 10^{5} \mathrm{s}\) . What is the rate constant at this temperature? (b) At 320 "C the rate constant is \(2.2 \times 10^{-5} \mathrm{s}^{-1} .\) What is the half-life at this temperature?

Short Answer

Expert verified
(a) The rate constant at 600K is \(3.01 脳 10^{-6} s^{-1}\). (b) The half-life at 320掳C is \(3.15 脳 10^4 s\).

Step by step solution

01

(a) Calculate the rate constant at 600K using the given half-life

As mentioned earlier, we will use the first-order reaction formula \(t_{1/2} = \frac{\ln{2}}{k}\). Rearranging it to solve for k: \(k = \frac{\ln{2}}{t_{1/2}}\) Given half-life, \(t_{1/2} = 2.3 脳 10^5 s\) Calculating k, \(k = \frac{\ln{2}}{2.3 脳 10^5 s}\) \(k = 3.01 脳 10^{-6} s^{-1}\) So, the rate constant at 600K is \(3.01 脳 10^{-6} s^{-1}\).
02

(b) Calculate the half-life at 320掳C using the given rate constant

The given rate constant, \(k = 2.2 脳 10^{-5} s^{-1}\). Now, we will use the first-order reaction formula \(t_{1/2} = \frac{\ln{2}}{k}\) to calculate half-life. \( t_{1/2} = \frac{\ln{2}}{2.2 脳 10^{-5} s^{-1}} \) \( t_{1/2} = 3.15 脳 10^4 s \) So, the half-life at 320掳C is \(3.15 脳 10^4 s\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding First-Order Reactions
A first-order reaction is a type of chemical reaction where the rate at which it occurs is directly proportional to the concentration of one of the reactants. In simpler terms, if you were to double the amount of this reactant, the reaction rate would also double.

In mathematical terms, a first-order reaction can be described by the equation: \[ \text{Rate} = k[\text{Reactant}] \]where \( k \) is the rate constant, and \( [\text{Reactant}] \) represents the concentration of the reactant. This kind of reaction has distinctive characteristics, such as a constant half-life regardless of the initial concentration, which makes them especially interesting in fields like pharmacokinetics and radioactive decay.
Rate Constant Calculation
The rate constant, symbolized by \( k \), is a crucial parameter in chemical kinetics as it dictates the speed of the chemical reaction. For first-order reactions, it is related to the half-life of the reaction, which is the time required for the concentration of a reactant to drop to half its initial value.

The formula connecting the half-life (\( t_{1/2} \)) and the rate constant for a first-order reaction is:\[ t_{1/2} = \frac{\ln{2}}{k} \]By rearranging this formula, we can solve for the rate constant if we know the half-life:\[ k = \frac{\ln{2}}{t_{1/2}} \]For instance, given a certain half-life, we can calculate the rate constant, which is essential to predict how fast the reaction proceeds under various conditions. Understanding this relationship is foundational for students studying reaction kinetics.
The Concept of Half-Life in Chemical Kinetics
The half-life of a chemical reaction is a term that describes the duration needed for half of the reactant to undergo the reaction. In the context of a first-order reaction, the half-life is independent of the initial concentration of the reactant, making it a valuable tool for scientists to understand the reaction's behavior over time.

The half-life is inversely proportional to the rate constant, as seen in the equation:\[ t_{1/2} = \frac{\ln{2}}{k} \]This means that a greater rate constant results in a shorter half-life, implying a faster reaction. Conversely, a smaller rate constant indicates a slower reaction with a longer half-life. By measuring the half-life, scientists can deduce the rate constant and vice versa, facilitating the prediction of how long it will take for a reaction to reach a specific stage or to complete.

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Most popular questions from this chapter

A flask is charged with 0.100 mol of A and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(A(g) \longrightarrow \mathrm{B}(g) .\) The following data are collected:(a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 s interval in units of mol/s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?

Consider the reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D} .\) Is each of the following statements true or false? (a) The rate law for the reaction must be Rate \(=k[\mathrm{A}][\mathrm{B}] .\) (b) If the reaction is an elementary reaction, the rate law is second order. (c) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be greater than that for the forward reaction.

Based on their activation energies and energy changes and assuming that all collision factors are the same, rank the following reactions from slowest to fastest.\( \begin{aligned} \text { (a) } E_{a} &=45 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-25 \mathrm{kJ} / \mathrm{mol} \\ \text { (b) } E_{a} &=35 \mathrm{kJ} / \mathrm{mol} ; \Delta E=-10 \mathrm{kJ} / \mathrm{mol} \\ \text { (c) } E_{a} &=55 \mathrm{kJ} / \mathrm{mol} ; \Delta E=10 \mathrm{kJ} / \mathrm{mol} \end{aligned}\)

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{B}]^{2}\) . (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B} ?\) What is the overall reaction order? (c) What are the units of the rate constant?

The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a two-step mechanism: $$ \begin{array}{c}{\mathrm{H}_{2} \mathrm{O}_{2}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{IO}^{-}(a q) \text { (slow) }} \\ {\mathrm{IO}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{g})+\mathrm{I}^{-}(a q) \text { (fast) }}\end{array} $$ \(\begin{array}{l}{\text { (a) Write the chemical equation for the overall process. }} \\ {\text { (b) Identify the intermediate, if any, in the mechanism. }} \\ {\text { (c) Assuming that the first step of the mechanism is rate }} \\ {\text { determining, predict the rate law for the overall process. }}\end{array}\)
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