The rate of a first order reaction is 0.04 at 10 sec and 0.03 at 20 sec after initiation of reaction. The half life period of the reaction is ?

Question

Answer

Answer

Answer

See Ananya we know its 1st order therefore via rate law
rate=k (conc.)1
now substitute
values of rate into two given cases then
divide thos two equations then u will get the ratio of conc at ten sec to conc at 20 sec then put this ratio in the first order equation ie
k=2.303/t log c at 10 sec / c at 20 sec
here u will get the value of ratewqq constant which according to me is 0.029 sec-1
then just apply t half =0.693 / k
i got approx 24min 24.1 to be exact

nice question hope u understand

Answer

Introduction:
The rate of a chemical reaction is determined by the rate law, which expresses the relationship between the rate of reaction and the concentrations of the reactants. First-order reactions are those where the rate of reaction is directly proportional to the concentration of a single reactant raised to the power of 1. In this case, we are given the rates of a first-order reaction at two different time intervals.

Given Information:
Rate at 10 sec: 0.04
Rate at 20 sec: 0.03

Rate Law for a First-Order Reaction:
The rate law equation for a first-order reaction is given by the following equation:
rate = k[A]

Where 'rate' is the rate of reaction, 'k' is the rate constant, and '[A]' is the concentration of the reactant.

Determining the Rate Constant:
To determine the rate constant, we can use the given information to set up two equations:

0.04 = k[A] at 10 sec
0.03 = k[A] at 20 sec

Since the concentration of the reactant, [A], is the same in both equations, we can set up a ratio:

0.04/0.03 = (k[A] at 10 sec)/(k[A] at 20 sec)

Simplifying the equation:

4/3 = (k[A] at 10 sec)/(k[A] at 20 sec)

We can cancel out the [A] term:

4/3 = (k at 10 sec)/(k at 20 sec)

Determining the Half-Life:
The half-life of a first-order reaction is the time it takes for the concentration of the reactant to decrease by half. In this case, we need to find the time it takes for the reaction rate to decrease from 0.04 to 0.02.

Using the rate equation for a first-order reaction, we can set up an equation:

0.02 = k[A] at t (where t is the time)

Since the concentration of the reactant, [A], is the same in both equations, we can set up a ratio:

0.04/0.02 = (k[A] at 10 sec)/(k[A] at t)

Simplifying the equation:

2 = (k at 10 sec)/(k at t)

We know that the rate constant, k, is the same for both equations, so we can set up a ratio:

2 = (k at 10 sec)/(k at 20 sec)

From the previous equation, we determined that:

4/3 = (k at 10 sec)/(k at 20 sec)

Setting the two equations equal to each other:

4/3 = 2

This equation is not true, so there is no specific time at which the rate decreases to 0.02. Therefore, we cannot determine the half-life of the reaction based on the given information.

Similar NEET Doubts

The rate of a first-order reaction is 0.04 mol-1 s-1 at 10 seconds and 0.03 mol -1 s-1 at 20 secondsafter initiation of the reaction. The half-life period of the reaction is

The correct difference between first- andsecond-order reaction is that

The correct difference between first and second order reactions is that

The half life of a first order reaction is 2000 years if the concentration after 8000 years is 0.02 then the initial concentration was?

In the following gaseous phase first order reaction A -> 2B+C Initial pressure was observed to be 200 mmHg and after 20 minutes, it changed to 400mmHg. The rate constant for the reaction is?

Top Courses for NEETView all

Biology Class 11

Daily Test for NEET Preparation

NEET Mock Test Series 2025

Topic-wise MCQ Tests for NEET

Physics Class 11

Top Courses for NEET

Top Courses for NEET

Related Content

About NEET

NEET Preparation Material

How to Prepare for NEET Exam?

NCERT Based Study Material

NEET Mock Tests

Other Popular Exams

Company