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# write the integrated rate law for the first order reaction.

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### Mohammed

Guys, does anyone know the answer?

get write the integrated rate law for the first order reaction. from screen.

## Using the First

Learn how to use the first-order integrated rate law, and see examples that walk through sample problems step-by-step for you to improve your chemistry knowledge and skills.

## Using the First-order Integrated Rate Law

High School Chemistry Skills Practice

Cesar Prugue, Melissa Bush

Example Solutions Practice Questions

## Steps for Using the First-order Integrated Rate Law

Step 1: Read the question carefully and determine what is being asked.Step 2: Use the first-order integrated rate law to solve for the variable in question.

## Vocabulary and Equations for Using the First-order Integrated Rate Law

First-Order Integrated Rate Law: This equation can be used for any first-order reactions of the form

r a t e = k [ A ] rate=k[A] where k k

is the rate constant in units of

s − 1 s−1 , [ A ] [A]

is the concentration of reactant A, and the rate is in units of concentration per time.

The integrated rate law for a first-order reaction is then:

ln [ A ] = − k t + ln [ A ] 0 ln⁡[A]=−kt+ln⁡[A]0 Here, t t is time, and [ A ] 0 [A]0

is the initial concentration of reactant A. All other variables are similar to the first-order rate law.

Notice that the integrated rate law is similar to the equation of a line (

y = m x + b y=mx+b

). The integrated rate law allows us to plot

ln [ A ] ln⁡[A]

vs time. Then, the slope of the line is

− k −k

, and the y-intercept is

ln [ A ] 0 ln⁡[A]0 .

The following three sample problems will test your understanding of the first-order integrated rate law, and allow you to use the equation to solve for missing variables.

## Example Problem 1 - Identifying a First-order Reaction From a Plot

The following plots were obtained from different types of reactions. Which plot follows a first-order rate law?

Step 1: Read the question carefully and determine what is being asked. We need to find the plot that follows a first-order rate law.Step 2: The first-order rate law will always have a negative slope when

ln [ A ] ln⁡[A]

vs time is plotted. Therefore, this is the correct plot:

## Example Problem 2 - Using the First-order Integrated Rate Law

In a certain chemical reaction, ethyl chloride creates hydrochloric acid and ethene. This reaction follows a first-order rate law. If the rate constant is

6.4 ⋅ 10 − 2 s − 1 6.4⋅10−2s−1

, how long will it take for 75% of the initial reactant concentration to react?

Step 1: Read the question carefully and determine what is being asked. We need to solve for time.Step 2: Use the first-order integrated rate law to solve for time, given the rate constant and the reactant concentration information. We can rearrange the first-order integrated rate law to solve for time:

ln [ A ] = − k t + ln [ A ] 0 ln⁡[A]=−kt+ln⁡[A]0 ln [ A ] − ln [ A ] 0 = − k t ln⁡[A]−ln⁡[A]0=−kt ln [ A ] [ A ] 0 = − k t ln⁡[A][A]0=−kt t = ln ( [ A ] [ A ] 0 ) ⋅ 1 − k = ln ( [ A ] 0 [ A ] ) ⋅ 1 k

t=ln⁡([A][A]0)⋅1−k=ln⁡([A]0[A])⋅1k

We know that 25% of the initial concentration will be left, so we can set

[ A ] [A] equal to 0.25 [ A ] 0 0.25[A]0

Now we have everything necessary to plug into the equation:

t = ln ( [ A ] 0 [ A ] ) ⋅ 1 k = ln ( [ A ] 0 0.25 [ A ] 0 ) ⋅ 1 6.4 ⋅ 10 − 2 s − 1

t=ln⁡([A]0[A])⋅1k=ln⁡([A]00.25[A]0)⋅16.4⋅10−2s−1

t = ln ( 1 0.25 ) ⋅ 1 6.4 ⋅ 10 − 2 s − 1 = 22 s

t=ln⁡(10.25)⋅16.4⋅10−2s−1=22s

It will take 22s for 75% of the initial concentration to react.

## Example Problem 3 - Using the First-order Integrated Rate Law

स्रोत : study.com

## Derive an integrated rate law expression for first order reaction: A → B + C

Derive an integrated rate law expression for first order reaction: A → B + C

Derive an integrated rate law expression for first order reaction: A → B + C

### SOLUTION

Consider first order reaction, A → B + C

The differential rate law is given by

rate = − dA dt d[A]dt

= k[A]      .....(1)

where, [A] is the concentration of reactant at time t.

Rearranging Eq. (1) dA A d[A][A]

= - k dt     ....(2)

Let [A]0 be the initial concentration of the reactant A at time t = 0.

Suppose [A]t is the concentration of A at time = t

The equation (2) is integrated between limits [A] = [A]0 at t = 0 and [A] = [A]t at t = t

At A dA A kt

∫[A]0[A]td[A][A]=-k∫0t

dt On integration, lnAAt Aktt

[ln[A]][A]0[A]t=-k(t)0t

Substitution of limits gives

ln[A]t − ln[A]0 = –kt

or ln At A [A]t[A]0 = - kt     ....(3) or k = t 1t ln A At [A]0[A]t

Converting ln to log10, we write

k = t A At

2.303tlog10 [A]0[A]t

....(4)

Eq. (4) gives the integrated rate law for the first order reactions.

Concept: Integrated Rate Law

Is there an error in this question or solution?

Chapter 6: Chemical Kinetics - Short answer questions (Type- II)

Q 1 Q 7 Q 2.1

### APPEARS IN

SCERT Maharashtra Question Bank 12th Standard HSC Chemistry Maharashtra State Board 2022

Chapter 6 Chemical Kinetics

Short answer questions (Type- II) | Q 1

स्रोत : www.shaalaa.com

## First

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## Want to join the conversation?

Posted 2 years ago. Direct link to Minsu Park's post “why does the natural log ...”

why does the natural log of pressure have no unit?

• Richard 2 years ago

Posted 2 years ago. Direct link to Richard's post “It still has the original...”

It still has the original unit of pressure because all we've done to it by taking the ln(P) is a mathematical operation. We haven't added a new unit or eliminated the original unit. So technically the unit of ln(P) should still be Torr and they should have indicated the unit on graph.

Hope that helps. Michael YANG 2 years ago

Posted 2 years ago. Direct link to Michael YANG's post “Could anyone please show ...”

Could anyone please show the derivation of the function for the first-order reactions?

• BilalKawsara a year ago

Posted a year ago. Direct link to BilalKawsara's post “This is grade-12/college-...”

This is grade-12/college-level but if you're curious I will show you below.

So for a first order reaction -- we have the reaction equals the rate constant times the concentration of the (only) reactant --> R = k[A]

1. Then we choose to re-write R as -Δ[A]/Δt

and we get -Δ[A]/Δt = k[A]

2. Then we bring -Δt to the right side

Δ[A] = -k[A]Δt

3. Then we bring [A] to the left side

Δ[A]/[A] = -kΔt

4. Then we integrate (the left side with respect to A and the right side with respect to t)

∫Δ[A] 1/[A] = -k∫Δt Ln[A] = -kt

5. Then we evaluate both integrals from 0 to t

Ln[A]ₜ - Ln[A]₀ = -kt-(-k0)

6. Then we bring Ln[A]₀ to the right

Ln[A]ₜ = -kt -0 + Ln[A]₀

7. Finally, we have our answer:

Ln[A]ₜ = -kt + Ln[A]₀

8.Notes:

i. For the connection to y=mx+b

The natural log of the concentration of A at a given time t --> Ln[A]ₜ is basically Y, and is equal to the natural log of the initial concentration of A --> Ln[A]₀ which is basically b, minus the rate constant -->k (basically m, aka the slope of the line) multiplied by time (basically x). So we get a linear graph of the form Y=mx+b

ii. The reason the it is negative at the beginning -Δ[A]/Δt and at -kt is because the rate is positive, but the change in reactant is negative because it is decreasing, so we build in a negative sign to cancel it and make the rate positive.

Kokes, Joshua 2 years ago

Posted 2 years ago. Direct link to Kokes, Joshua's post “What does he mean by the ...”

What does he mean by the "natural log" at 0:43?

• Davin V Jones 2 years ago

Posted 2 years ago. Direct link to Davin V Jones's post “https://www.khanacademy.o...”

bob ross 4 months ago

Posted 4 months ago. Direct link to bob ross's post “hi, at 2:25, Jay said if...”

hi,

at 2:25, Jay said if the coefficient of A is 2, then -kt will become -2kt, so why don't we generalize it as -akt, where a is the coefficient of reactant A?

then according to this new convention, won't the half life equation be [At] = [Ao]e^(-akt) and t1/2 = ln2÷ak?

thank you! • Richard 4 months ago

Posted 4 months ago. Direct link to Richard's post “Yeah, the calculus works ...”

Yeah, the calculus works out to that.

Ria Rose 2 months ago

Posted 2 months ago. Direct link to Ria Rose's post “I’m having trouble findin...”

I’m having trouble finding how you got -2.08 x 10 ^-4. When I add up the y and x numbers to do m= change in y divided by change in x, I get -16.189 divided by - 39800 = 4.067x10^-4

• Richard a month ago

Posted a month ago. Direct link to Richard's post “The slope of a line is de...”

The slope of a line is defined as the change in the y-direction divided by the change in the x-direction (rise over run). As a formula looks like m = Δy/Δx; where m is the slope, Δy is the change in y, and Δx is the change in x. Change here being a difference (subtraction) between two point's x and y coordinates. So we can also write the slope formula as m = (y2-y1)/(x2-x1); where x1 and y1 are the coordinates for the first point and x2 and y2 are the coordinates for the second point.

Now assuming it is a perfectly straight line, the slope should be constant at all points on the line and so we can pick any two points to calculate the slope. I'll choose the first and last points; (0,6.219) and (15000,3.109). So x1 is 0, y1 is 6.219, x2 is 15000, and y2 is 3.109. Substituting these into the previous formula yields: m = (3.109 - 6.219)/(15000 - 0) = -2.07 x 10^(-4). Which is reasonably close to what Jay got in the video. The discrepancy between my answer and Jay's is due to him using a graphing computer of some sort which takes into account all the coordinate points and also shows that the line is not perfectly straight. At any case even doing it by hand our answers should agree for the most part.

With your calculation I'm not sure why or what you added together, but having a change in y of -16.189 and a change in x of -39800 is wildly wrong. Additionally if you divide those two numbers you get a positive slope (dividing a negative by a negative) and judging solely off the graph of the line it should have a negative slope.