Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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) \(3 \times 10^{-6} \mathrm{Ms}^{-1}\)

Short Answer

Expert verified
The initial rate is (a) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\).

Step by step solution

01

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.
02

Identify Given Values

From the problem statement, we know:- The rate constant \( k = 3 \times 10^{-6} \mathrm{s}^{-1} \).- The initial concentration \( [A] = 0.10 \mathrm{M} \).
03

Substitute Values into the Rate Equation

Now, we substitute the known values into the first-order rate equation:\[ \text{Rate} = (3 \times 10^{-6} \mathrm{s}^{-1}) \times (0.10 \mathrm{M}) \]
04

Perform the Calculation

Multiply the rate constant by the concentration:\[ \text{Rate} = 3 \times 10^{-6} \times 0.10 = 3 \times 10^{-7} \mathrm{M} \mathrm{s}^{-1} \]
05

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) \(3 \times 10^{-7} \mathrm{Ms}^{-1}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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]
where:
  • k is the rate constant.
  • [A] is the concentration of the reactant.
Understanding the order of a reaction helps predict how a reaction will behave under various conditions, which is crucial for both practical applications and theoretical understandings of chemical kinetics.
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
The magnitude of the rate constant can indicate the speed of the reaction. A larger rate constant means the reaction occurs faster, while a smaller rate constant means it is slower. Factors such as temperature can affect the rate constant. Generally, as temperature increases, the rate constant and hence the reaction rate also increase. This highlights the importance of the rate constant in understanding and controlling reaction conditions.
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
This straightforward calculation shows how fast a first-order reaction occurs given the rate constant and initial concentration. The calculated rate tells us the speed at which the reactant is converting into products at the start of the reaction.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A catalyst increases rate of reaction by (a) decreasing enthalpy (b) decreasing activation energy (c) decreasing internal energy (d) increasing activation energy

The rate constant of a reaction at temperature 200 is 10 times less than the rate constant at \(400 \mathrm{~K}\). What is the activation energy \(\left(\mathrm{E}_{\alpha}\right)\) of the reaction? \((\mathrm{R}=\) gas constant) (a) \(1842.4 \mathrm{R}\) (b) \(921.2 \mathrm{R}\) (c) \(460.0 \mathrm{R}\) (d) \(230.3 \mathrm{R}\)

A redox reaction is carried out at \(127^{\circ} \mathrm{C}\). If the same reaction is carried out in presence of a catalyst at the same temperature, the rate of reaction is doubled. To what extent is the energy barrier lowered by the catalyst? [Use \(\mathrm{R}=2\) cal \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\) and \(\left.\log 2=0.301\right]\) (a) \(455 \mathrm{cal}\) (b) \(231 \mathrm{cal}\) (c) \(693 \mathrm{cal}\) (d) \(554 \mathrm{cal}\)

Hydrogenation of vegetable ghee at \(27^{\circ} \mathrm{C}\) reduces the pressure of \(\mathrm{H}_{2}\) from \(3 \mathrm{~atm}\) to \(2.18 \mathrm{~atm}\) in 40 minutes. The rate of reaction in terms of molarity per second is \(\left(\mathrm{R}=0.082 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-t}\right)\) (a) \(1.357 \times 10^{-6}\) (b) \(1.537 \times 10^{-5}\) (c) \(1.375 \times 10^{-5}\) (d) \(6.250 \times 10^{-4}\)

For the reaction \(\mathrm{A} \longrightarrow\) Products, it is found that the rate of reaction increases by a factor of \(6.25\), when the concentration of \(\mathrm{A}\) is increased by a factor of \(2.5 .\) The order of reaction with respect to \(\mathrm{A}\) is (a) \(0.5\) (b) 1 (c) 2 (d) 3

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free