Integrated Rate Law - Online Tutor, Practice Problems & Exam Prep

The integrated rate law is crucial for understanding the relationship between reactant concentrations and time, revealing how long it takes for a certain amount of reactant to be consumed. It varies according to the reaction order: zero, first, or second. For zero order reactions, the rate constant (K) has units of molarity per time, and the concentration decreases linearly over time. First order reactions, often seen in radioactive processes, have a rate constant with units of time inverse, and a plot of the natural logarithm of concentration versus time is linear. Second order reactions feature a rate constant with units of molarity inverse per time, and a plot of the inverse of concentration versus time shows an increasing linear relationship. Recognizing these patterns is key to identifying reaction orders and understanding the kinetics of chemical processes.

Integrated Rate Law Concept 1

Now the integrated rate law describes the relationship between reactants and their concentrations as well as time. Now this helps to determine how long it takes for X amount of moles per liter of reactant to become consumed or used up. And we're going to say here that the integrated rate law depends on the order of the reaction.

Now the first one we talk about is our zero order integrated rate law. Now here this is for reactions following zero order rate laws. And with it we're going to use the following equation. Here it is aT=-KT+a0. Here aT equals the final reactant concentration. a0 equals the initial reacting concentration because zero time has passed. K equals your rate constant in.

Now remember to figure out the units for K We say K=m-n+1×t-1, and we're going to say that n equals the order of the reaction. So if we're dealing with 0 TH order, that means n is equal to 0. So zero -0 + 1 just means plus one. This would mean that for 0 TH order reaction the units for K are in molarities to the one times time inverse. Now time here could be seconds, days, years, whatever. Here T would be our time.

So these are the components that make up our zero order integrated weight law equation. Now this equation is also connected to the equation of a straight line. Here AT would be connected to Y. Your rate constant K would be connected to M which is your slope. And remember, it's not just K, it's -KT here will be connected to X and then your initial concentration connected to B.

If we look at this graphically, remember a graph is a plot of Y versus X. Your Y again would be your AT, so the concentration of your reactant and then T is your X. And then remember here this is our initial concentration where we first start out our reaction. Notice that the slope is decreasing because remember slope is equal to M which is equal to -K negative K meaning that it's decreasing over time.

Also remember that slope is equal to change in Y over change in X, which in this case we can say is really the change in concentration over the change in time. So keep these in mind when discussing any question dealing with a zero order reaction.

Integrated Rate Law Example 1

A plot of the concentration of nitrogen trioxide versus time with a slope of 0.260 gives a straight line. What was the initial concentration of nitrogen trioxide if after 35 seconds its concentration dropped to 2.75×10-2 molar? All right, so here they're telling us it's a plot of concentration versus time. Remember, if our plot is of your reacting concentration versus time, that would mean that it is a zeroth order reaction or zero order reaction, so we know it's zero order.

So that means that our final reacting concentration equals negative KT initial reacting concentration. Now here they're asking us to figure out the initial concentration. So we don't know what this portion is. We know that it drops to this final number here. So that's our final concentration. So that's 2.75×10-2 equals. Now remember this equation for zeroth order reaction is also equal to the equation for a straight line, and here is our slope which is equal to our negative K.

So when they tell me my slope is 0.260, they're really telling me what K is. So we're going to plug that in for K. So negative 0.260 T here is time, which we're told is 35 seconds. All right, so then we're going to have 2.75×10-2 = -9.1 plus the initial concentration of your reactant. Here you're going to add 9.1 to both sides, and when we do that, we're going to get as our initial concentration 9.12 75 molar.

So this would represent our initial concentration here in our question. This has three sig figs, 3 sig figs, 2 sig figs. So here if we had it in terms of two sig figs, it would come out to be 9.1 molar as our initial concentration, right? So that would be our final answer.

Integrated Rate Law Concept 2

For reactions with first order mechanics we're going to say we use the following equation which is ln⁡(AT)=−KT+ln⁡(AO). Now here again AT is our final concentration of your reactant. AO is the initial concentration of your reactant. K here is your rate constant. And remember units for K is M to the negative NN being the order of the reaction, which in this case would be 1 + 1 times time. Inverse.

Here let's just do seconds, because a lot of the times time inverses in seconds, so -1 + 1 comes out to 0. Anything to the zero power is equal to just one, so it drops out. So that would mean that K here is in units of time inverse. So here T again is time. And we're going to say here that our equation for first order processes follows the equation for straight line, which means that ln⁡(AT) is equal to Y. Your K negative K is equal to M which is your slope, T is equal to X and then ln⁡(AO) is equal to B.

Now anytime we see a plot of ln⁡(reactant) concentration versus time, that's a dead giveaway that it's first order because remember a plot is always AY versus X. So here are Y is ln⁡(A) and our X is RT. So here is our Y axis with the ln⁡(reactant) concentration. Here's our X which is time ln⁡(AO) is just our initial starting amount. And then remember slope is equal to change in Y over change in X, which is the same thing as change in reactant concentration over change in time.

Now lastly, what's important to remember is that all first order process all radioactive processes follow a first order mechanism of first order rate law. OK, so not all first order processes are radioactive, It's just that the ones that are radioactive happen to be first order, so that's a key giveaway. So if a question is talking about a radioactive isotope, if a question says the word radioactive, that's a dead giveaway that you're dealing with their first order process. So keep in mind these little tips in terms of identifying the order of any type of word problem, we have to determine if it's first order or not.

Integrated Rate Law Example 2

A certain reaction has a rate constant of 0.289 seconds^-1. How long in seconds would it take for the concentration of reaction reactant A to decrease from 1.43 molar to 0.850 molar? All right, so they tell us our rate constant is 0.289 seconds^-1. Remember, a dead giveaway for first order reaction is that K is in units of time inverse, so that's unique to 1st order. The fact that it's second inverse tells us that it's first order.

So since its first order, we know that we're dealing with lnAt=-kt+lnA0. Now here our initial is 1.43, so ln1.43 and then our final is 0.850, so ln0.850 over here equals. All right, so here are rate constant is 0.289. And then here we're looking for time T. What we're going to do here is we're going to subtract ln1.43 from both sides.

When we do that, we're going to get 0.520193 equals -0.289T. Divide both sides by -0.289. So here, since K is in seconds, that means time would also be in seconds. Remember that their units must always match. If K was in minutes^-1, then time would be in minutes. OK, they're going to match in terms of the units.

So here when we work all this out, we get 1.80 seconds. Here our answer has three sig figs because 0.289 has three sig figs as well as 1.43 and 0.850. So again, 1.80 seconds will be the time for this first order process.

Integrated Rate Law Concept 3

For reactions with second order processes we use the following equation and it is one over at equals positive Kt plus one over AO. So at here is a final reacting concentration. AO is the initial reacting concentration. Now K is our rate constant in units of. Remember K = m-n + 1 × time^-1. n equals the order of R process. Here since its second order it's -2 + 1 × time^-1.

This would mean that when it comes to 2nd order processes, the units for K would be in molarity's inverse times time inverse. So always be on the lookout for that when looking to see if a reaction is second order or not. t is of course time. Now the 2nd order integrated rate law equation is related to the equation of a straight line. Here 1at is connected to YK. Here is equal to our slope, mt is X and one over AO is B.

Now if we took a plot of 1a versus time, that's a dead giveaway. That's a second order process. Remember plots are of Y versus X, so one over reacting concentration matches with this for Y. That's why it's on the Y axis and then the X axis has time because t is equal to X here. Now here our initial starting is down here and notice that the line is increasing overtime. That's because for 2nd order processes, K is positive.

Here it's a positive Kt. We're cussing negative K when it deals with 0 TH order and 1st order processes. Second order is unique, it's the only one with the increasing slope over time, right? So just keep these little things in mind that are giving you clues on 2nd order processes.

Integrated Rate Law Example 3

Here we're told that the reactant concentration for a second order reaction was 0.7670 molar after 300 seconds, and 7.3×10-2 molar after 750 seconds. What is the rate constant K for this reaction? All right. So they tell us it's second order. So that means 1AT=KT+1AL.

Here we're told that we initially have 0.670 molar after 300 seconds. That's going to be our initial. OK, no one says that initial has to start at 0 seconds. It's just in this case we're starting at 300 seconds and then one over. The final concentration is 7.3×10-2. Here we're looking for K, but we need to figure out what our time is, how much time has elapsed.

Well, we're starting at 300 seconds, and this is the initial concentration, and we go to 750 seconds. If we subtract those two numbers, that tells us how much time has elapsed, and that's 450 seconds have elapsed. All right, so we're going to subtract 10.670 from both sides. Alright, so then when we do that, we're going to get 1+2.20609=K450.

So then we're going to divide both sides now by 450 and when we divide both sides by 450, that's going to give us our K value. K here will equal 2.71×10-2. Now here's the thing. What are the units for K? Remember for K it's M-n+1 times time inverse. Since it's second order, N is 2 so -2+1 times time inverse.

Here the units for time where we see are in seconds, so this would be seconds. Inverse, so -2+1 is -1 so M-1 times seconds to the -1. So this will represent the value for our rate constant K as well as its units.

For the reaction A →  B, the rate constant is 0.0837 M^–1•sec^–1. How long would it take for [A] to decrease by 85%?

• A. 0.0211 s
• B. 1.94 s
• C. 179 s
• D. 0.677 s

The following reaction has a rate constant of 3.7 × 10^–3 M•s^–1 at 25°C:

A → B + C

Calculate the concentration of C after 2.7 × 10^–3 sec where [A]₀ was 0.750 M at 25°C; assume [C]₀ = 0 M.

• A. 0.75 M
• B. 9.99×10^–6 M
• C. 3.7×10^–3 M
• D. 7.5×10^–6 M

For the decomposition of urea, NH₂CONH₂ (aq) + H⁺(aq) + 2 H₂O (l) → 2 NH₄⁺ (aq) + HCO₃^– (aq), the rate constant is 3.24 × 10^–4 s^–1 at 35°C. The initial concentration of urea is 2.89 mol/L. What fraction of urea has decomposed after 3.5 minutes?

• A. 0.934
• B. 0.0658
• C. 1.13×10^–3
• D. 0.0235

Iodine-123 is used to study thyroid gland function. As this radioactive isotope breaks down, after 5.7 hrs the concentration of iodine-123 is 56.3% complete. Find the rate constant of this reaction.

• A. 0.359 hr^–1
• B. 0.15 hr^–1
• C. 1.36×10^–3 M^–1•hr^–1
• D. 2.26×10^–3 M^–1•hr^–1

Here’s what students ask on this topic:

What is the integrated rate law?

The integrated rate law is an equation that describes the concentration of a reactant in a chemical reaction as a function of time. It's derived from the rate law, which expresses the rate of a reaction in terms of the concentration of reactants. For a reaction of order n, the rate law is:

$\text{rate =}k{\left[A\right]}^n$

where k is the rate constant, [A] is the concentration of the reactant, and n is the order of the reaction.

There are different forms of integrated rate laws for different reaction orders. For example:

1. Zero-order reactions:

$\left[A\right]=\left[A\right]0_{}-\mathrm{kt}$

where [A]₀ is the initial concentration and [A] is the concentration at time t. The plot of [A] vs. time is linear for a zero-order reaction.

2. First-order reactions:

Which of the following statements correctly describe an integrated rate law? Can you select all that apply?

An integrated rate law expresses the concentration of a reactant in a chemical reaction as a function of time. Unlike the differential rate law, which shows the rate of reaction at a particular moment (instantaneous rate), the integrated rate law provides a means to calculate the amount of reactant remaining after a certain period. It is derived by integrating the differential rate law.

The integrated rate law can be used to determine the order of a reaction, which indicates the power to which the concentration of the reactant is raised in the rate law. For example, a first-order reaction will have a linear relationship between the natural logarithm of the reactant concentration and time, while a second-order reaction will exhibit a linear relationship between the inverse of the reactant concentration and time.

Furthermore, the integrated rate law allows for the calculation of the half-life of a reaction, which is the time required for half of the reactant to be consumed. In first-order reactions, the half-life is constant and does not depend on the initial concentration, while in second-order reactions, the half-life is dependent on the initial concentration.

Which of the following represents the integrated rate law for a first-order reaction?

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

$\ln \left[A\right]=-kt+\ln \left[A\right]0_{}$

In this equation, $\left[A\right]$ represents the concentration of the reactant at time $t$, $k$ is the first-order rate constant, and $\left[A\right]0_{}$ is the initial concentration of the reactant when $t$ = 0. The negative sign indicates that the concentration of the reactant decreases over time. This equation can also be rearranged to solve for $t$ or $k$ depending on the given data. When plotted on a graph with

Which of the following represents the integrated rate law for a second-order reaction?

The integrated rate law for a second-order reaction, where the concentration of the reactant decreases over time as the reaction proceeds, is represented by the following equation:

$\frac{1}{A_t}=kt+\frac{1}{A_0}$

In this equation, $A_t$ is the concentration of the reactant A at time t, $A_0$ is the initial concentration of the reactant, k is the second-order rate constant, and t is the time. The units of the rate constant k in a second-order reaction are typically M^-1s^-1 (inverse molarity per second).

This form of the integrated rate law shows that the inverse of the concentration of the reactant at any given time is directly proportional to the time elapsed. A plot of $\frac{1}{A_t}$ versus time (t) for a second-order reaction will yield a straight line, with the slope equal to the rate constant k.