Question

The ketone acid \(\left(\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)_{2} \mathrm{CO}\) undergoes a first-order decomposition in aqueous solution to yield acetone and carbon dioxide: \(\left(\mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)_{2} \mathrm{CO} \rightarrow\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}+2 \mathrm{CO}_{2}\) (b) The rate (a) Write the expression for the reaction rate, constant \(k\) has been determined experimentally as \(5.48 \times\) \(10^{-2} / \mathrm{sec}\) at \(60^{\circ} \mathrm{C}\). Calculate \(t_{1 / 2}\) at \(60^{\circ} \mathrm{C}\). (c) The rate constant at \(0^{\circ} \mathrm{C}\) has been determined as \(2.46 \times 10^{-5} / \mathrm{sec}\). Calculate \(t_{1 / 2}\) at \(0^{\circ} \mathrm{C}\). (d) Are the calculated half-lives in accord with the stated influence of temperature on reaction rate?

Short answer

The half-life at 60°C is approximately 12.7 seconds, while the half-life at 0°C is approximately 28247 seconds (about 7 hours and 51 minutes). This difference in half-lives is in accordance with the general concept that reaction rates increase with increasing temperatures, meaning that the ketone acid decomposes more rapidly at higher temperatures.

Step-by-step solution

Write the expression for the reaction rate

The rate equation for this first-order decomposition is: \[ \textit{rate} = k [\text{(CH\(_2\)CO\(_2\)H)\(_2\)CO}] \]

Calculate half-life for a first-order reaction

The half-life of a first-order reaction(series) is given by the formula: \[ t_{1/2} = \frac{\ln 2}{k} \]

Calculate the half-life at 60°C

Use the given rate constant at 60°C and plug it into the half-life formula \[ t_{1/2} = \frac{\ln 2}{5.48 \times 10^{-2}} \] Calculate the result: \[ t_{1/2} \approx 12.7 \, \text{sec} \] So the half-life at 60°C is 12.7 seconds.

Calculate the half-life at 0°C

Use the given rate constant at 0°C and plug it into the half-life formula \[ t_{1/2} = \frac{\ln 2}{2.46 \times 10^{-5}} \] Calculate the result: \[ t_{1/2} \approx 282\,47 \, \text{sec} \] So the half-life at 0°C is approximately 28247 seconds, which is about 7 hours and 51 minutes.

Analyze the influence of temperature on the reaction rate

As we can see, the half-life at 0°C is significantly longer than at 60°C. This is in accordance with the general concept that reaction rates increase with increasing temperatures. In other words, increasing the temperature causes the reaction to take place faster, and the molecule decomposes more rapidly.

Key concepts

Reaction Rate

In a chemical reaction, the reaction rate determines how quickly a reactant is converted into a product. For first-order reactions, such as the decomposition of ketone acid, the rate of the reaction is directly proportional to the concentration of the reactant.
This can be expressed by the equation \[ \text{rate} = k [\text{(CH}_2\text{CO}_2\text{H})_2\text{CO}] \] where \( k \) is the rate constant, and \([\text{Reactant}]\) is the concentration of the ketone acid.
This equation highlights that as the concentration of the reactant decreases, the reaction rate decreases as well. Understanding this relationship is crucial in predicting how long a reaction will take under given conditions.

Half-Life Calculation

The half-life of a first-order reaction is the time required for half of the reactant to decompose. It is independent of the initial concentration of the reactant and relies solely on the rate constant \( k \).
The formula used for calculating the half-life \( t_{1/2} \) is: \[ t_{1/2} = \frac{\ln 2}{k} \] This formula shows that the half-life is inversely proportional to the rate constant.
Hence, a larger \( k \) results in a shorter half-life, meaning the reaction proceeds faster. By plugging the rate constants at specific temperatures into this formula, we can calculate the half-lives at those temperatures, providing insight into the reaction's dynamics.

Temperature Effect on Reaction Rate

Temperature is a significant factor affecting reaction rates. An increase in temperature typically leads to faster reaction rates. This is because higher temperatures provide reactant molecules with more energy, increasing the frequency and effectiveness of their collisions.
In the decomposition reaction of ketone acid, as the temperature increases from 0°C to 60°C, the rate constant \( k \) also increases, which in turn shortens the reaction's half-life from approximately 28247 seconds to 12.7 seconds.
This observation aligns with the broader concept that chemical reactions generally proceed more quickly at higher temperatures, due to the increased kinetic energy and collision frequency of the molecules involved.

Rate Constant

The rate constant \( k \) is a crucial parameter in the rate equation of a reaction. It signifies the speed of the reaction under specific conditions and varies with changes in temperature.
For first-order reactions, the rate constant remains unaffected by the concentration of the reactants. Instead, it is determined by factors such as temperature and the nature of the reactants.

• At 60°C, the rate constant is \(5.48 \times 10^{-2} \, \text{sec}^{-1}\),
• At 0°C, it is \(2.46 \times 10^{-5} \, \text{sec}^{-1}\).

These values indicate that the reaction proceeds faster at the higher temperature, reflecting the kinetic aspect of reaction dynamics. Understanding the rate constant's role helps in predicting the behavior of reactions under different conditions.