Problem 33 - Chemical Principles

Question

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

Step-by-Step Solution

Answer

Substitute the values into the second-order reaction time formula and solve for \( t \) for each concentration level (one-sixteenth, one-fourth, and one-fifth) using the calculated rate constant \( k \).

Step 1: Understanding the Given Information

The problem states that we have a second-order reaction with a half-life of 50.5 seconds when the initial concentration \( [A]_0 \) is 0.84 mol/L. For a second-order reaction, the half-life is inversely proportional to the initial concentration. The half-life formula for a second-order reaction can be written as \( t_{1/2} = \frac{1}{k[A]_0} \) where \( k \) is the rate constant.

Step 2: Calculate the Rate Constant

First, we need to calculate the rate constant \( k \) using the given half-life and initial concentration. We rearrange the half-life formula to solve for \( k \) with \( t_{1/2} = 50.5 s \) and \( [A]_0 = 0.84 mol/L \) which gives us \( k = \frac{1}{t_{1/2}[A]_0} \) or \( k = \frac{1}{50.5 \times 0.84} \) s\textsuperscript{-1}.

Step 3: Determine the Final Concentration Values

We need to calculate the times for different final concentrations: (a) \( \frac{[A]_0}{16} \) (b) \( \frac{[A]_0}{4} \) (c) \( \frac{[A]_0}{5} \) for substance A.

Step 4: Apply the Second-Order Reaction Formula

For a second-order reaction, the formula relating time, initial concentration, and final concentration is \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) where \( t \) is the time required for the concentration to change from \( [A]_0 \) to \( [A] \). Now, we will use this formula to calculate the times for each final concentration.

Step 5: Calculate Time for One-Sixteenth Concentration

Using the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) with \( [A] = \frac{[A]_0}{16} \) and the rate constant \( k \) already calculated, we solve for time \( t \).

Step 6: Calculate Time for One-Fourth Concentration

Similarly, use the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) with \( [A] = \frac{[A]_0}{4} \) and the known value of \( k \) to find the time \( t \) for this concentration level.

Step 7: Calculate Time for One-Fifth Concentration

Lastly, apply the formula \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \) using \( [A] = \frac{[A]_0}{5} \) to calculate the time \( t \) for the substance A to reach one-fifth of its original concentration.

Key Concepts

Chemical Kinetics

Chemical kinetics is a branch of physical chemistry that deals with understanding the rates of chemical reactions. It’s not just about identifying how long a reaction will take to complete, but also about grasping the various factors that influence this rate. A fundamental aspect of chemical kinetics is the identification of the order of a reaction, which indicates how the rate depends on the concentration of the reactants.

For instance, a second-order reaction involves the rate of a reaction being proportional to the square of the concentration of one reactant or the product of the concentrations of two reactants. In the exercise provided, we delved into a second-order reaction, where the half-life depends on the initial concentration and provides essential insights into how quickly reactant concentrations decrease over time.

Reaction Rate Constant

The reaction rate constant, denoted as k, is a crucial term in the equation that describes the speed of a chemical reaction. In the context of a second-order reaction, the rate constant bridges the relationship between the concentration of reactants and the rate at which they are converted to products.

The value of k is determined experimentally and can vary with temperature, pressure, and the presence of a catalyst. Our step-by-step example began with the computation of the rate constant using the provided half-life and initial concentration. Knowing the rate constant is essential for predicting the concentration of reactants at any point in time during a reaction.

Half-Life of Reaction

Half-life, commonly symbolized as t_1/2, is a term that describes the time required for the concentration of a reactant to decrease to half of its original value. This is a particularly handy concept when discussing radioactive decay, but it also plays a significant role in chemical kinetics.

In second-order reactions, the half-life is uniquely dependent on the initial concentration of the reactant, inversely so. As we observed in our exercise example, a given half-life and initial concentration allowed for the calculation of the reaction rate constant. Understanding half-life can provide insights into the durability and effectiveness of substances over time, which is valuable in various fields such as pharmacology and environmental science.

Concentration-Time Relationship

The concentration-time relationship is a key concept in chemical kinetics that describes how the concentration of reactants change as a reaction progresses. In a second-order reaction, this relationship can be expressed by the equation \( \frac{1}{[A]} - \frac{1}{[A]_0} = kt \), where [A] represents the concentration of the reactant at time t, [A]_0 is the initial concentration, and k is the rate constant.

The equation showcases that, unlike in zero or first-order reactions, the change in the inverse of concentration with time is linear. In practical terms, this means that as time goes on, the concentration decreases more slowly than in a first-order reaction, which is a logarithmic relationship. This relationship allows us to calculate not only the time it takes for the concentration to fall to a certain level, as demonstrated in our textbook example, but also to predict future concentrations at any given time during the reaction.