91Ó°ÊÓ

The reaction of thioacetamide with water is shown by the equation below: $$\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{CH}_{3} \mathrm{C}(\mathrm{O}) \mathrm{NH}_{2}(a q)$$ The rate of reaction is given by the rate law: $$\text { Rate }=k\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\right]$$ Consider one liter of solution that is \(0.20 \mathrm{M}\) in \(\mathrm{CH}_{3} \mathrm{C}(\mathrm{S}) \mathrm{NH}_{2}\) and \(0.15 \mathrm{M}\) in \(\mathrm{HCl}\) at \(25^{\circ} \mathrm{C}\). a. For each of the changes listed below, state whether the rate of reaction increases, decreases, or remains the same. Why? i. A \(4-g\) sample of \(\mathrm{NaOH}\) is added to the solution. ii. \(500 \mathrm{~mL}\) of water is added to the solution. b. For each of the changes listed below, state whether the value of \(k\) will increase, decrease, or remain the same. Why? i. A catalyst is added to the solution. ii. The reaction is carried out at \(15^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C}\).

Step by step solution

01

Understand the Rate Law

The given rate law is Rate \( = k[\mathrm{H}_3\mathrm{O}^+][\mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2] \). This means the rate of reaction is directly proportional to the concentrations of \( \mathrm{H}_3\mathrm{O}^+ \) and \( \mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2 \).

02

Analyze the Addition of NaOH

Adding \( \mathrm{NaOH} \) will neutralize \( \mathrm{HCl} \), reducing \( [\mathrm{H}_3\mathrm{O}^+] \). Since the rate of the reaction depends on \( [\mathrm{H}_3\mathrm{O}^+] \), the rate will decrease.

03

Analyze the Addition of Water

Adding 500 mL of water will dilute the solution, decreasing the concentrations of \( \mathrm{HCl} \) and \( \mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2 \). Both are in the rate law; thus, the rate will decrease.

04

Evaluate Catalyst Addition Effect on k

Adding a catalyst does not change the rate constant \( k \); it lowers the activation energy, increasing the rate, but \( k \) remains the same.

05

Evaluate Temperature Change Effect on k

Temperature affects the rate constant \( k \); lowering the temperature from \( 25^{\circ} \mathrm{C} \) to \( 15^{\circ} \mathrm{C} \) will decrease \( k \), as reactions generally slow down at lower temperatures.

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Key Concepts

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

Rate Law

A rate law is an expression that describes the reaction rate in terms of the concentration of reactants. For the reaction involving thioacetamide and water, the rate law is given as \( \text{Rate} = k[\mathrm{H}_3\mathrm{O}^+][\mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2] \). This means that the rate of the reaction depends directly on the concentration of both \([\mathrm{H}_3\mathrm{O}^+]\) and \([\mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2]\). The rate law is crucial because it helps us to predict how the speed of a chemical reaction relates to the concentrations of its reactants.
Changes in these concentrations will directly affect the reaction rate. For instance, increasing the concentration of either reactant will increase the rate of reaction, while a decrease will do the opposite. Understanding this relationship is key in controlling reactions efficiently and safely.

Reaction Rate

The reaction rate tells us how quickly a reaction proceeds. It is usually expressed in terms of the change in concentration of a reactant or product over time. In our given reaction, the speed can be altered by adjusting the concentrations of \([\mathrm{H}_3\mathrm{O}^+]\) and \([\mathrm{CH}_3\mathrm{C}(\mathrm{S})\mathrm{NH}_2]\). When you add sodium hydroxide (NaOH) to the solution, it neutralizes the hydronium ions \([\mathrm{H}_3\mathrm{O}^+]\), effectively reducing their concentration. This leads to a decrease in the reaction rate because the rate law shows that the reaction is proportional to \([\mathrm{H}_3\mathrm{O}^+]\).
Similarly, adding water dilutes the solution, which lowers the concentration of both reactants present in the rate law. This dilution also reduces the reaction rate. Thus, any change in the concentration of reactants, whether by neutralization or dilution, directly impacts the reaction rate in predictable ways.

Rate Constant

The rate constant \(k\) in a rate law expression is a crucial factor in chemical kinetics. It provides the constant of proportionality between the reaction rate and the concentrations of reactants. Hence, different reactions have different rate constants due to the varied nature of interactions among reactant molecules. It is important to note that while changes in concentration affect the reaction rate, they do not affect the rate constant \(k\).
However, \(k\) is temperature-dependent and catalytic influence-independent. This means that adding a catalyst will increase the rate of reaction by lowering the activation energy needed, but it does not alter \(k\). Catalysts work by providing an alternative pathway with a lower activation energy, enhancing the reaction speed without affecting the intrinsic rate constant. Changes in temperature, on the other hand, can influence \(k\) as explained further.

Effect of Temperature on Reaction Rate

Temperature has a significant impact on the reaction rate and the rate constant \(k\). Generally, increasing the temperature results in an increase in \(k\), thereby accelerating the reaction. This is due to the fact that molecules at higher temperatures have more kinetic energy; thus, they collide more frequently and with greater energy.
When the reaction temperature is lowered from \(25^{\circ} \mathrm{C} \) to \(15^{\circ} \mathrm{C}\), the decreased kinetic energy leads to fewer successful collisions and hence a decrease in \(k\). This results in a slower reaction rate as it becomes less frequent and less energetic. In most cases, a rule of thumb in chemical kinetics is that the reaction rate doubles with every 10-degree Celsius increase in temperature. However, in scenarios where the temperature is reduced, we observe an inverse effect on the reaction rate and rate constant.