Problem 69
Question
The rate constant of first-order reaction is \(3 \times 10^{-6}\) per second. The initial concentration is \(0.10 \mathrm{M}\). The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)
Step-by-Step Solution
Verified Answer
The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\).
1Step 1: Understand the Reaction Order and Formula
We are dealing with a first-order reaction, which means the rate depends linearly on the concentration of the reactant. The general rate equation for a first-order reaction is given by: \[ \text{Rate} = k \times [A] \]where \( k \) is the rate constant, and \( [A] \) is the concentration of the reactant.
2Step 2: Identify Given Values
From the problem statement, we know:- The rate constant \( k = 3 \times 10^{-6} \mathrm{s}^{-1} \).- The initial concentration \( [A] = 0.10 \mathrm{M} \).
3Step 3: Substitute Values into the Rate Equation
Now, we substitute the known values into the first-order rate equation:\[ \text{Rate} = (3 \times 10^{-6} \mathrm{s}^{-1}) \times (0.10 \mathrm{M}) \]
4Step 4: Perform the Calculation
Multiply the rate constant by the concentration:\[ \text{Rate} = 3 \times 10^{-6} \times 0.10 = 3 \times 10^{-7} \mathrm{M} \mathrm{s}^{-1} \]
5Step 5: Match Your Answer with Given Options
Compare the calculated rate with the options provided:(a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\) (b) \(3 \times 10^{-4} \mathrm{Ms}^{-1}\) (c) \(3 \times 10^{-5} \mathrm{Ms}^{-1}\) (d) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)The correct answer is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\).
Key Concepts
First-order reactionRate constantInitial concentrationReaction rate calculation
First-order reaction
In chemical kinetics, a first-order reaction is quite straightforward. It is a type of reaction where the rate depends on the concentration of a single reactant. This means that if you double the concentration of the reactant, the rate of the reaction will also double. First-order reactions are common in many natural processes. For example, radioactive decay and some biological reactions often follow first-order kinetics. To mathematically express the rate of a first-order reaction, we use the equation:
- Rate = k × [A]
- k is the rate constant.
- [A] is the concentration of the reactant.
Rate constant
The rate constant, often represented as k, is a critical factor in the rate law equation. It provides a direct measure of how fast a reaction takes place. The units of the rate constant differ based on the order of the reaction. For a first-order reaction, the unit is typically per second
i.e.
- s-1
Initial concentration
Initial concentration refers to the concentration of a reactant at the beginning of a reaction. In the context of a first-order reaction, this initial concentration plays a significant role in determining the rate of the reaction, since the rate is directly proportional to the concentration of the reactant.
In calculations, the concentration is often expressed in molarity (M), which is moles per liter. Understanding the starting concentration helps predict how much reactant will remain after a certain period and how long it will take for the reaction to reach completion. When working through problems involving reaction rates, knowing the initial concentration allows us to confidently use the rate law formula to find how fast the reaction proceeds right from the start.
Reaction rate calculation
To calculate the reaction rate, especially in a first-order reaction, you should substitute the values you have into the rate formula. Here’s a simple step-by-step guide:
- Start with the equation: Rate = k × [A]
- Identify your values. In this case:
- k is the rate constant: 3 × 10-6 s-1
- [A] (initial concentration) is 0.10 M
- Substitute these values into the equation and calculate: Rate = (3 × 10-6 s-1) × (0.10 M)
- Do the math, and you get: Rate = 3 × 10-7 M s-1
Other exercises in this chapter
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