Problem 2 - Chemistry in 30 days

Question

For a first order reaction, \(A \rightarrow B, t_{1 / 2}=1\) h, what fraction of the initial conc. of A reacts in \(4 \mathrm{~h}\).? (a) \(15 / 16\) (b) \(1 / 16\) (c) \(7 / 8\) (d) \(1 / 8\)

Step-by-Step Solution

Answer

(a) \(15 / 16\)

Step 1: Identify the half-life formula

The half-life of a first-order reaction is given by the formula \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant. Given \( t_{1/2} = 1 \) hour, we can use this to calculate \( k \).

Step 2: Calculate the rate constant

Using the half-life formula \( t_{1/2} = \frac{0.693}{k} \) and the given half-life of 1 hour, we find \( k = 0.693 \) per hour.

Step 3: Use the first order reaction formula

The formula for the concentration of a reactant in a first order reaction after time \( t \) is \( [A] = [A]_0 e^{-kt} \).

Step 4: Plug in known values

Given that the time \( t \) is 4 hours and \( k = 0.693 \), the concentration \( [A] \text{ in 4 hours} = [A]_0 e^{-0.693 \times 4} \).

Step 5: Calculate remaining fraction of A

Calculate \( e^{-0.693 \times 4} \) which approximates \( e^{-2.772} \approx 0.0625 \). Thus, \( [A] / [A]_0 = 0.0625 \).

Step 6: Determine the fraction reacted

The fraction of A reacted can be found as \( 1 - [A]/[A]_0 = 1 - 0.0625 = 0.9375 \), which equals \( \frac{15}{16} \).

Key Concepts

Understanding Half-life in Reactions

In chemistry, the concept of half-life is crucial, especially for reactions. The term half-life refers to the time it takes for half of the initial concentration of a reactant to be consumed in a reaction. For a first-order reaction, half-life is independent of the initial concentration and remains constant throughout the reaction.
It can be calculated using the formula: \[ t_{1/2} = \frac{0.693}{k} \]
Here, \( k \) is the rate constant. In the context of our exercise, the half-life is given as 1 hour. This constant duration for the concentration to halve simplifies calculations, enabling more straightforward prediction of concentration changes over time.

Exploring the Rate Constant

The rate constant \( k \) is a pivotal factor in determining the speed of reactions. For a first-order reaction, it indicates how quickly the reactant is depleted over time.
Its unit is typically per time (e.g., \( \text{h}^{-1} \) for hours). As seen in our half-life equation, reaching a rate constant of \( 0.693 \text{ h}^{-1} \) indicates the reaction's speed given the half-life of 1 hour.
Understanding \( k \) allows chemists to predict how long it takes for specific proportions of reactants to react, which is essential in industrial and laboratory settings.

The Dynamics of Reaction Kinetics

Reaction kinetics is the study of the speed or rate at which chemical reactions occur. It's integral in understanding how quickly a reaction proceeds and reaching equilibrium.
In first-order reactions, the rate is directly proportional to the concentration of a single reactant. Reaction kinetics involves both experimental data and the mathematical formulas used to deduce reaction rates.
Analyses often involve plotting concentration versus time, resulting in a negative exponential graph for first-order reactions, reflecting the gradual decrease in concentration over time.

Exponential Decay in Reactions

Exponential decay is a mathematical concept used to describe the reduction in quantity at a rate proportional to its current value. In first-order reactions, the concentration of a reactant decreases exponentially over time according to the formula:\[ [A] = [A]_0 e^{-kt} \]
This equation shows how the initial concentration \([A]_0\) lessens by a factor of \(e^{-kt}\). Over time, as we calculate with a rate constant \(k = 0.693\) and time \( t = 4 \) hours, the concentration vastly diminishes.
The calculation of \( e^{-2.772} \approx 0.0625 \) showcases the power of exponential decay, leaving only a small fraction of the initial reactant, illustrating the reaction's progress.