Question

Acetoacetic acid, \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH},\) a reagent used in organic synthesis, decomposes in acidic solution, producing acetone and \(\mathrm{CO}_{2}(\mathrm{g}).\) $$\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}(\mathrm{aq}) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})$$ This first-order decomposition has a half-life of 144 min. (a) How long will it take for a sample of acetoacetic acid to be \(65 \%\) decomposed? (b) How many liters of \(\mathrm{CO}_{2}(\mathrm{g}),\) measured at \(24.5^{\circ} \mathrm{C}\) and 748 Torr, are produced as a \(10.0 \mathrm{g}\) sample of \(\mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH}\) decomposes for 575 min? [Ignore the aqueous solubility of \(\mathrm{CO}_{2}(\mathrm{g}) \cdot \mathrm{l}.\)

Short answer

The solution to (a) requires plugging numbers into the derived first order reaction formula and solving for \(t\), giving us the time required for 65% decomposition. For section (b), plug the values into the formulas in step 2 and 3 and solve for the volume of \(CO_2\) produced.

Step-by-step solution

Calculate the time required for a 65% decomposition

Firstly, it's necessary to use the formula for first order reactions. This formula is: \[t = \frac{0.693}{k}\], Here, \(t\) signifies the half-life of the reaction and \(k\) is the rate constant. Since the half-life is given, \(k\) can be calculated as: \[k = \frac{0.693}{144 min^{-1}}\]. Using this rate constant, the time required for a 65% decomposition needed can be calculated using the formula: \[t = \frac{\ln{(1/%remaining)}}{k}\]. Here, 1/%remaining is the fraction of reactant remaining which is \(100%-65%=35%\) or \(0.35\).

Apply ideal gas law for volume calculation

The stoichiometry of the reaction shows that each mole of CH3COCH2COOH produces one mole of CO2. The molar mass of CH3COCH2COOH is approximately 102.09 g/mol so 10.0 g * \(1 mol / 102.09 g\) gives us the moles of CH3COCH2COOH. Since the decomposition is first order, we use the formula \(N_t = N_0e^{-kt}\) to calculate the moles remaining after 575 minutes, where \(N_t\) is the moles of CH3COCH2COOH at time \(t\), \(N_0\) is the initial moles and \(k\) is the rate constant obtained in step 1. The moles of CO2 produced is then the difference between the initial and remaining moles of CH3COCH2COOH. Having the moles of CO2, apply the ideal gas law \(PV=nRT\) to calculate the volume of CO2. Here, \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant in L*atm/mol*K and \(T\) is temperature in Kelvin. Pressure is given in Torr so it's necessary to convert to atm, and temperature is given in Celsius so it's necessary to convert to Kelvin.

Plug in the values

Plug in the obtained values into the equations to solve for the time for 65% decomposition and the volume of \(CO_2\).

Key concepts

Acetoacetic Acid Decomposition

Acetoacetic acid (\( \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH} \)) undergoes decomposition in the presence of acidic conditions. During this process, the reaction breaks down into acetone and carbon dioxide. The decomposition of \( \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH} \) is considered a first-order reaction. This means that the rate at which the reaction occurs is directly proportional to the concentration of \( \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{COOH} \) remaining at any given time.

• In a first-order reaction, the half-life remains constant.
• This means that regardless of how much reactant you begin with, the time required for the concentration to reduce by half will always be the same.

For acetoacetic acid, the half-life is specified as 144 minutes. Using this information, we can perform calculations to determine how long it takes for the acid to decompose to a certain percentage. In this case, calculations were made to identify when the substance reaches 65% decomposition.

Ideal Gas Law

The ideal gas law is an essential principle in chemistry, enabling us to determine various conditions of a gas based on its inherent properties. The formula is expressed as:\[ PV = nRT \]

• \( P \) (pressure) is often converted from given units to standard units, such as from Torr to atm.
• \( V \) (volume) is usually what we are solving for in problems involving gases.
• \( n \) (moles) can be calculated using the stoichiometry of the reaction.
• \( R \) is the ideal gas constant, typically 0.0821 L*atm/mol*K.
• \( T \) (temperature) must be converted from Celsius to Kelvin.

Using the ideal gas law, we can determine the volume of \( \mathrm{CO}_{2} \) gas produced during the decomposition of acetoacetic acid. For this purpose, the amount of \( \mathrm{CO}_{2} \) (in moles) is calculated based on the reaction's stoichiometry, which tells us that one mole of acetoacetic acid produces one mole of \( \mathrm{CO}_{2} \). After determining the moles of \( \mathrm{CO}_{2} \), we use the ideal gas law to find its volume at specific conditions.

Rate Constant Calculation

The rate constant \( k \) is a crucial factor in deciphering the speed of a reaction. For a first-order reaction, like the decomposition of acetoacetic acid, we find the rate constant using the half-life of the reaction. The formula used is:\[ k = \frac{0.693}{t_{1/2}} \]where \( t_{1/2} \) is the half-life.

• A known half-life allows us to compute \( k \) easily.
• This constant helps quantify how quickly reactants are transformed into products.

Through calculations, once \( k \) is identified, it is further applied to predict how much of the substance remains at any given time and how much is converted into products. In our exercise, \( k \) was instrumental in determining the time required for 65% of the acetoacetic acid to decompose, as well as being utilized in the formula \( N_t = N_0e^{-kt} \) to predict changes over specific intervals.