Question

Consider the hypothetical first-order reaction $$ 2 \mathrm{~A}(g) \rightarrow \mathrm{X}(g)+\frac{1}{2} \mathrm{Y}(g) $$ At a certain temperature, the half-life of the reaction is 19.0 min. A \(1.00-\mathrm{L}\) flask contains \(A\) with a partial pressure of \(622 \mathrm{~mm} \mathrm{Hg}\). If the temperature is kept constant, what are the partial pressures of \(\mathrm{A}, \mathrm{X},\) and \(\mathrm{Y}\) after 42 minutes?

Short answer

Solution: 1. Calculate the rate constant (k) from half-life: $$ k = \frac{0.693}{19.0\;\text{min}} $$ 2. Calculate the initial concentration of A: $$ x_{A} = \frac{622\;\text{mmHg}}{760\;\text{mmHg}} $$ $$ [\text{A}]_{0} = \frac{x_{A}\times P_{T}}{RT} $$ 3. Calculate the concentration of A after 42 minutes: $$ \ln \frac{[\text{A}]_{t}}{[\text{A}]_{0}} = -kt $$ 4. Calculate the concentrations of X and Y using stoichiometry: $$ \Delta [\text{A}] = [\text{A}]_{0} - [\text{A}]_{t} $$ $$ \Delta [\text{X}] = \frac{1}{2} \Delta [\text{A}] $$ $$ \Delta [\text{Y}] = \frac{1}{4} \Delta [\text{A}] $$ 5. Calculate the partial pressures of A, X, and Y after 42 minutes: $$ P_{A} = [\text{A}]_{t} \times RT $$ $$ P_{X} = \Delta [\text{X}] \times RT $$ $$ P_{Y} = \Delta [\text{Y}] \times RT $$ After following these steps, you will have calculated the partial pressures of A, X, and Y after 42 minutes.

Step-by-step solution

Calculate the rate constant (k) from half-life

For a first-order reaction, the half-life (t1/2) and the rate constant (k) are related by the following equation: $$ t_{1/2} = \frac{0.693}{k} $$ We are given that the half-life is 19.0 minutes. We can now find the rate constant: $$ k = \frac{0.693}{19.0\;\text{min}} $$

Calculate the initial concentration of A

We are given that initially, A has a partial pressure of 622 mmHg. To convert this partial pressure to a concentration, we'll use the ideal gas equation and the assumption that the total pressure of the system is 1 atmeters (760 mmHg). Thus, the mole fraction of A (xA) is: $$ x_{A} = \frac{P_{A}}{P_{T}} = \frac{622\;\text{mmHg}}{760\;\text{mmHg}} $$ Now we'll use the ideal gas equation to find the initial concentration of A: $$ [\text{A}]_{0} = \frac{n_{A}}{V} = \frac{P_{A}}{RT} = \frac{x_{A}\times P_{T}}{RT} $$

Calculate the concentration of A after 42 minutes

To calculate the concentration of A after 42 minutes, we'll use the integrated rate law for a first-order reaction: $$ \ln \frac{[\text{A}]_{t}}{[\text{A}]_{0}} = -kt $$ We know the value of k from Step 1, the initial concentration of A from Step 2, and the time as 42 minutes. We can now solve for the concentration of A after 42 minutes ([A]t).

Calculate the concentrations of X and Y using stoichiometry

Now that we have the concentration of A at time t, we can utilize the stoichiometry of the given reaction to find the concentrations of X and Y. From the balanced reaction equation: $$ 2\; \text{A}(g) \rightarrow \text{X}(g) + \frac{1}{2}\; \text{Y}(g) $$ Based on the stoichiometry, for every 2 moles of A that react, 1 mole of X is produced and 0.5 moles of Y is produced. Hence, we can find the concentration change in A (Δ[A]) and calculate the concentration change in X (Δ[X]) and Y (Δ[Y]) as follows: $$ \Delta [\text{A}] = [\text{A}]_{0} - [\text{A}]_{t} $$ $$ \Delta [\text{X}] = \frac{1}{2} \Delta [\text{A}] $$ $$ \Delta [\text{Y}] = \frac{1}{4} \Delta [\text{A}] $$

Calculate the partial pressures of A, X, and Y after 42 minutes

Now that we have the concentrations of A, X, and Y at time t, we can convert them back to partial pressures using the ideal gas equation: $$ P_{A} = [\text{A}]_{t} \times RT $$ $$ P_{X} = \Delta [\text{X}] \times RT $$ $$ P_{Y} = \Delta [\text{Y}] \times RT $$ With these equations, we can calculate the partial pressures of A, X, and Y after 42 minutes.

Key concepts

Understanding Half-Life of a Reaction

The half-life of a reaction is a fundamental concept in chemistry that represents the time it takes for half of the reactant to be converted into a product in a chemical reaction. For first-order reactions, this time period is constant, no matter the initial concentration of the reactant. This is a key character of first-order reactions and is mathematically expressed as:

t_{1/2} = \frac{0.693}{k}
Here, \( t_{1/2} \) is the half-life and \( k \) is the rate constant. The number 0.693 is the natural logarithm of 2, which reflects the 'half' in 'half-life'. Understanding this relationship is crucial for predicting how long it takes for the concentration of a reactant to decrease to a certain level over time. For instance, if you're given the half-life of a reaction, you can determine the rate constant \( k \), which could then be used to calculate reactant concentrations at any given time using the integrated rate law. This concept is particularly useful for pharmacokinetics, radioactive decay, and many other scientific fields where the degradation of a substance is modeled.

Calculating the Rate Constant in a First-Order Reaction

The rate constant \( k \) is a crucial parameter in chemical kinetics that denotes the speed of a reaction. In the context of a first-order reaction, the rate constant can be determined using the half-life formula mentioned earlier. Once the half-life is known, calculating \( k \) is straightforward:

\[ k = \frac{0.693}{t_{1/2}} \]
By knowing \( k \), we can predict how fast the reaction will proceed, which is essential in industrial chemical processes, where rates of reactions directly impact production efficiency. The ability to calculate \( k \) from half-life provides chemists with insight into the kinetics of a reaction without needing to conduct the reaction over its entire course.

The Integrated Rate Law for First-Order Reactions

For first-order reactions, the integrated rate law connects the concentration of a reactant at a given time with the initial concentration and the rate constant. It is given by:

\[ \ln \frac{[\text{A}]_{t}}{[\text{A}]_{0}} = -kt \]
where \( [\text{A}]_{t} \) is the concentration at time \( t \), \( [\text{A}]_{0} \) is the initial concentration, and \( k \) is the rate constant. This equation stems from the fact that the rate of consumption of a reactant in a first-order reaction is proportional to its concentration at that time. To solve problems using the integrated rate law, logarithms are often used to linearize the equation, making calculations more manageable. Chemists use this law to calculate the concentrations of reactants or products at any point during the reaction, which is especially helpful for quality control in chemical manufacturing or determining the shelf-life of pharmaceuticals.