Question

A flask is charged with \(0.100 \mathrm{~mol}\) of \(\mathrm{A}\) and allowed to react to form \(\mathrm{B}\) according to the hypothetical gas-phase reaction \(\mathrm{A}(g) \longrightarrow \mathrm{B}(g)\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (s) } & 0 & 40 & 80 & 120 & 160 \\ \hline \text { Moles of A } & 0.100 & 0.067 & 0.045 & 0.030 & 0.020 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 -s interval in units of \(\mathrm{mol} / \mathrm{s}\). (c) What additional information would be needed to calculate the rate in units of concentration per time?

Short answer

a) The number of moles of B at each time in the table: at 0 s: 0 mol, at 40 s: 0.033 mol, at 80 s: 0.055 mol, at 120 s: 0.070 mol, and at 160 s: 0.080 mol. b) The average rate of disappearance of A for each 40-s interval: 0-40 s: -0.000825 mol/s, 40-80 s: -0.00055 mol/s, 80-120 s: -0.000375 mol/s, and 120-160 s: -0.00025 mol/s. c) Additional information needed to calculate the rate in units of concentration per time: The volume of the system is needed to calculate the molar concentration at each time point and subsequently the rate in units of concentration per time.

Step-by-step solution

a) Calculate the number of moles of B at each time in the table

Since the reaction is given by A(g) → B(g), and there are no intermediates, every mole of A is cleanly converted to B. So, we can calculate the moles of B formed at each time by subtracting the moles of A remaining at that time from the initial moles of A (0.100 mol). \(Moles~of~B = Initial~moles~of~A - Moles~of~A~remaining\) Now let's calculate the moles of B at each given time. At 0 s: \(Moles~of~B = 0.100 - 0.100 = 0\) At 40 s: \(Moles~of~B = 0.100 - 0.067 = 0.033\) At 80 s: \(Moles~of~B = 0.100 - 0.045 = 0.055\) At 120 s: \(Moles~of~B = 0.100 - 0.030 = 0.070\) At 160 s: \(Moles~of~B = 0.100 - 0.020 = 0.080\)

b) Calculate the average rate of disappearance of A for each 40-s interval

The average rate of disappearance of A can be calculated as the change in the number of moles of A divided by the time interval. We have to calculate the average rate for each 40-s interval. For interval 0-40 s: \(\mathrm{Average~rate = \frac{moles~of~A_{final} - moles~of~A_{initial}}{time~interval} = \frac{0.067 - 0.100}{40} = -0.000825~mol/s}\) For interval 40-80 s: \(\mathrm{Average~rate = \frac{0.045 - 0.067}{40} = -0.00055~mol/s}\) For interval 80-120 s: \(\mathrm{Average~rate = \frac{0.030 - 0.045}{40} = -0.000375~mol/s}\) For interval 120-160 s: \(\mathrm{Average~rate = \frac{0.020 - 0.030}{40} = -0.00025~mol/s}\) #c) What additional information would be needed to calculate the rate in units of concentration per time?

Additional information for rate in units of concentration per time

To calculate the rate in units of concentration per time, we need information about the volume of the system, as concentration is measured in moles per unit volume (e.g., mol/L). Knowing the volume, the molar concentration can be calculated as: \(Concentration (M) = \frac{moles}{volume}\) Once we have the molar concentration at each time point, we can calculate the rate of disappearance of A in units of concentration per time.

Key concepts

Understanding Reaction Rate Calculation

Understanding the rate at which chemical reactions occur is crucial in the field of chemical kinetics. The reaction rate indicates how quickly reactants are converted into products over time. To calculate this rate, we look at changes in the concentration or amount of a reactant or product in a given time interval. As in our exercise, the average rate of disappearance of a gas-phase reactant A can be calculated by taking the difference in the moles of A at two times and dividing by the time elapsed.

For example, if we have 0.100 mol of A initially and after 40 seconds have 0.067 mol, we determine the average rate over that interval by subtracting the final amount from the initial amount and dividing by 40 seconds. Our exercise demonstrated this with:
\(\text{Average rate} = \frac{\text{moles of A}_{final} - \text{moles of A}_{initial}}{\text{time interval}}\), resulting in a negative value indicating the disappearance of A. Negative signs typically denote consumption, while positive values signify formation rates of substances.

The Mole Concept in Gas-Phase Reactions

The mole is a fundamental concept in chemistry, serving as a bridge between the microscopic world of atoms and molecules and the macroscopic world we measure in the lab. When dealing with gases, understanding the mole concept is essential to account for substances in reactions without needing to calculate vast numbers of individual molecules. A mole corresponds to Avogadro's number (6.022 × 10²³) of particles, which could be atoms, ions, or, in the case of our exercise, molecules of a gas.

In gas-phase reactions, we often work with volumes where the moles of a gas are related to volume through the ideal gas law under conditions of constant temperature and pressure. This relation is critical when translating the stoichiometry of the reactions into measurable quantities and is integral in calculating the conversion of reactants to products, as we did by subtracting the moles of A remaining from the initial moles to find the moles of B formed.

Conversion of Reactants to Products

The conversion of reactants to products is at the heart of a chemical reaction. It involves the transformation of substances, wherein reactants lose their identity and form new products. In our exercise, for each mole of reactant A that disappears, one mole of product B is formed. This 1:1 stoichiometry means we can calculate the amount of product formed at any time simply by knowing how much reactant has been used up.

To make sense of this conversion, we use the concept of moles and the conservation of matter, which states that atoms are neither created nor destroyed in a chemical reaction. Thus, in a closed system with no side reactions or intermediates, the initial amount of reactant will equal the sum of the remaining reactant and the amount of product formed at any point in time.

The Concentration-Time Relationship

The concentration-time relationship in chemical kinetics provides insights into how the concentrations of reactants and products change over the course of a reaction. This relationship can often be represented by a mathematical equation or model that indicates whether the reaction rates are constant or change over time. In our problem, the concentrations of reactants decrease over time, which could suggest a first-order relationship where the rate of the reaction is directly proportional to the concentration of the reactant.

To precisely analyze this relationship, we need to know the volume in which the reaction occurs. This allows us to convert moles into concentrations (molarity, M), by dividing moles by volume (in liters). With the concentrations at different times, we could then plot a concentration-time graph and potentially determine the order of the reaction and the rate constant. Such concepts are foundational in kinetics, enabling chemists to predict and control reaction outcomes effectively.