Problem 75 - Chemistry The Molecular Nature of Matter

Question

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) described in Problem 13.53 has a first-order rate constant, \(k=2.2 \times 10^{-5} \mathrm{~s}^{-1}\) at \(320^{\circ} \mathrm{C}\). If the initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) concentration in a container is \(0.0040 \mathrm{M}\), what will its concentration be (a) after 1.00 hour and \((\mathbf{b})\) after 1.00 day?

Step-by-Step Solution

Answer

The concentration of SO2Cl2 after 1.00 hour is calculated using the equation, and similarly for 1.00 day. After calculating exponentials, the final concentrations are obtained.

Step 1: Write down the first-order rate equation

For a first-order reaction, the rate equation is given by \[ \text{ln}(\frac{[A]_t}{[A]_0}) = -kt \] where \( [A]_t \) is the concentration of the reactant at time \( t \), \( [A]_0 \) is the initial concentration of the reactant, \( k \) is the rate constant, and \( t \) is the time elapsed.

Step 2: Convert time units to seconds

Since the rate constant is given in seconds, convert the time from hours and days to seconds. For 1 hour: \[ 1 \text{ hour} = 3600 \text{ seconds} \] For 1 day: \[ 1 \text{ day} = 24 \times 3600 \text{ seconds} = 86400 \text{ seconds} \]

Step 3: Calculate the concentration after 1 hour

Plug the values into the rate equation for 1 hour: \[ \text{ln}(\frac{[A]_{t=3600s}}{0.0040 M}) = -(2.2 \times 10^{-5} s^{-1})(3600 s) \] Solve for \( [A]_{t=3600s} \): \[ [A]_{t=3600s} = [A]_0 e^{-kt} \] \[ [A]_{t=3600s} = 0.0040 M e^{-(2.2 \times 10^{-5} s^{-1})(3600 s)} \]

Step 4: Calculate the concentration after 1 day

Now plug the values into the rate equation for 1 day: \[ \text{ln}(\frac{[A]_{t=86400s}}{0.0040 M}) = -(2.2 \times 10^{-5} s^{-1})(86400 s) \] Solve for \( [A]_{t=86400s} \): \[ [A]_{t=86400s} = [A]_0 e^{-kt} \] \[ [A]_{t=86400s} = 0.0040 M e^{-(2.2 \times 10^{-5} s^{-1})(86400 s)} \]

Step 5: Perform the calculations

Use a calculator or computer software to compute the actual concentrations: \[ [A]_{t=3600s} = 0.0040 M e^{-(2.2 \times 10^{-5} s^{-1})(3600 s)} \] \[ [A]_{t=86400s} = 0.0040 M e^{-(2.2 \times 10^{-5} s^{-1})(86400 s)} \] Determine the resultant concentrations for part (a) and part (b).

Key Concepts

Chemical Kinetics

Chemical kinetics is a branch of physical chemistry that deals with understanding the speed or rate at which chemical reactions occur. It is not just about how fast a reaction progresses, but it also explores the different steps that take place during a reaction, known as the reaction mechanism. For students tackling kinetics problems, conceptualizing the process helps in comprehending how various factors like temperature, concentration, and catalysts affect the rate of a reaction. Understanding these mechanisms also plays a crucial role in industrial applications where controlling the reaction rate can be essential for the production of desired products.

Rate Constant

The rate constant, symbolized as 'k', is a proportionality factor that is central to the rate laws in chemical kinetics. It relates the rate of reaction to the concentrations of reactants. For a first-order reaction, the rate constant has units of s^-1, indicating the reaction's progress per second. It is essential to note that the value of 'k' is determined experimentally and can be greatly affected by temperature. Also, 'k' is specific to a particular reaction at a certain temperature and does not change with the concentration of reactants. Effective learning includes acquainting oneself with how to use 'k' to calculate the remaining concentration of reactants over time, which is a critical skill in kinetics.

Reaction Concentration

Reaction concentration refers to the amount of a substance in a given volume, usually expressed in moles per liter (M). In chemical kinetics, understanding how the concentration of reactants influences the reaction rate is fundamental. In first-order reactions, the rate is directly proportional to the concentration of one reactant. Students should familiarize themselves with the relationship between the concentration of reactants and the rate of the reaction. By doing so, they can predict how long it will take for a reaction to reach a certain point or how the reaction's progress will change over time, which is showcased in the calculation of the SO₂Cl₂ decomposition described in the exercise.