Integrated Rate Law - Video Tutorials & Practice Problems
When we include the variable of time to our Rate Law then we obtain the Integrated Rate Laws.
Understanding the Integrated Rate Laws
Integrated Rate Law Concept 1
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
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%?
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.
0.75 M
9.99×10–6 M
3.7×10–3 M
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?
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.
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