Chemistry, 26.07.2019 05:20 reycaden

Afirst-order reaction has a half-life of 27.4 s . how long does it take for the concentration of the reactant in the reaction to fall to one-fourth of its initial value?

Answers: 3

Show answers

Answers

Answer from: Chad5978

54.8 seconds

Explanation:

Given

Reaction is first order

half life is 27.4 seconds

Half life is defined as the time required to complete the reaction 50% of its initial value or the time in which the concentration of reactant reaches to 50% of its initial value.

It will take one half life to reach concentration half of the initial value.

It will take another half life to reach concentration half of the half of the initial concentration which is one fourth of the initial concentration.

So the total time taken to reach one fourth of the initial value = two half lives = 27.4 + 27.4 = 54.8 seconds.

Another approach to solve the same problem is :

Let us calculate the rate constant of the reaction from its half life:

rate constant = k = 0.693 / half life = 0.693 /27.4 = 0.0253

The rate expression for first order reaction is:

$time=\frac{1}{k}[ln\frac{A_{0}}{A_{t}}] =\frac{1}{0.0253}ln[\frac{100}{25}] =54.8 seconds$

Answer from: Quest

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: 1 - α α α = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

$k_{c}$ = $\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)v }$

or, $k_{c}$ = $\frac{\alpha ^{2} }{(1-\alpha) }$ (at v = 1l)

the reaction: so₂cl₂ (g) = so₂ (g) + cl₂ (g) total

the moles at equilibrium: 1 - α α α = 1+α

concentrations: (1 - α)/0.5 α/0.5 α/0.5 = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

$\frac{[so2cl2] x [cl2]}{[so2cl2]}$

or, $\frac{\alpha ^{2} }{(1-\alpha)0.5 }$

$\frac{\alpha ^{2} }{(1-\alpha) }$ = $\frac{\alpha ^{2} }{(1-\alpha) 0.5}$

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

Answer from: Quest

A. 17 hope this dudet

Another question on Chemistry

Chemistry, 21.06.2019 13:00

Note the ph and poh values labeled with letters on the ph scale below. based on log rules and the way ph is calculated, what is the difference in [oh– ] concentration between point a and point b. a) 10^1 b) 10^5 c) 10^6 d) 10^7

Answers: 1

Answer

Chemistry, 21.06.2019 16:30

How does decreasing the gas volume affect the pressure of a gas?

Answers: 1

Answer

Chemistry, 21.06.2019 19:00

Asyringe contains 56.05 ml of gas at 315.1 k. what volume will that gas occupy if the temperature is increased to 380.5 k? a) 12.41 b) 46.42 c) 67.68 d) 81.74

Answers: 1

Answer

Chemistry, 22.06.2019 11:30

Compare and contrast refraction of light and sound will give brainliest

Answers: 1

Answer

You know the right answer?

Afirst-order reaction has a half-life of 27.4 s . how long does it take for the concentration of the...

Questions