Any universal tricks for all Rate Law question? Thanks!!!

[fifi1314]

1. given data below, find the rate law for the following rxn at 300k. (From kaplan blue book, their method is using the formula and it takes too much time).
A+B --> C+D
Trial [A]initial(M) initial(M) r initial(M/sec)
1 1 1 2
2 1 2 8.1
3 2 2 15.9
Answer: order of [A] is 1, is 2, overall rxn order is 1+2=3

2. For the rxn A+B --> C, determine the order of the rxn with respect to B from the infor given below.
[A]initial(M) initial(M) r initial rate of formation of C(M/sec)
2.5x10^-6 3x10^-4 5x10^-3
5x10^-6 3x10^-4 1x10^-2
1x10^-5 9x10^-4 1.8x10^-1

I am struggling and confused, please help!
Thanks!!!!!

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[nze82]

1. given data below, find the rate law for the following rxn at 300k. (From kaplan blue book, their method is using the formula and it takes too much time).
A+B --> C+D
Trial [A]initial(M) initial(M) r initial(M/sec)
1 1 1 2
2 1 2 8.1
3 2 2 15.9
Answer: order of [A] is 1, is 2, overall rxn order is 1+2=3
This one should be easy...
When you hold the concentration of A constant (Trials 1&2), doubling the concentration of B causes the rate to jump from 2 to 8 (a 4x increase). In other words, doubling the concentration quadruples the rate. This means the reaction order with respect to B is 2. What if you can't figure it out like this? Let's look at the math:

We know that: Rate = k [A]^x**y
So:
Trial#1: 2 = k [1]^x*[1]^y
Trial#2: 8 = k [1]^x*[2]^y

Now, divide trial#2 by trial#1:

8/2 = (k [1]^x*[2]^y)/(k [1]^x*[1]^y)

Cancel out what's necessary, and you'll get:

4 = [2]^y --> y = 2 which is the order of reaction with respect to B.

Now, when we hold the concentration of B constant (Trials 2&3), doubling the concentration of A causes the rate to jump from 8 to ~16 (a 2x increase). This means the reaction order with respect to A is 1. Let's see if our math confirms this result:

Trial#2: 8 = k [1]^x*[2]^2
Trial#3: 16 = k [2]^x*[2]^2

16/8 = (k [2]^x*[2]^2)/(k [1]^x*[2]^2)

2 = [2]^x--> x = 1 which is the order of reaction with respect to A.


2. For the rxn A+B --> C, determine the order of the rxn with respect to B from the infor given below.
[A]initial(M) initial(M) r initial rate of formation of C(M/sec)
2.5x10^-6 3x10^-4 5x10^-3
5x10^-6 3x10^-4 1x10^-2
1x10^-5 9x10^-4 1.8x10^-1

Let's look at trials 1&2. When concentration of B is held constant, doubling the concentration of A causes the rate to jump from 5x10^-3 to 1x10-2 (a 2x increase). So, the order of reaction with respect to A is 1 (You can check this by doing the math). Now, let's write the rate law for trials 3&4:

Trial#3: 1x10^-2 = k [5x10^-6]^1*[3x10-4]^y
Trial#4: 1.8x10^-1 = k [1x10^-5]^1*[9x10^-4]^y

Now divide trial#4 by trial#3:

1.8x10^-1/1^10-2 = (1x10^-5/5x10^-6)^1*(9/3)^y

18 = 2*(3)^y

9 = 3^y --> y = 2 which is the order of the reaction with respect to B.

I am struggling and confused, please help!
Thanks!!!!!

These problems are quite time consuming, and you can only get good at them by doing lots of problems, so you'll get comfortable with setting up the equations and doing all the math.

 

[ajlst11]

Just look at the changes from one trial to the next.
Example, in your first example, when [A] doubles and stays the same, the Rate (r) ALSO doubles. This AUTOMATICALLY means that the order of [A] is 1, because when it doubles, so does the rate. Now, when [A] is constant, and doubles, the rate is multiplied by 4. This means that is 2nd order because it is (x)^2, so when x, which is in this, doubles, the rate is multiplied by: (2)^2= 4. So [A] is 1st order, and is second.

NOW, the next one is a little harder. You didn't give me the answer to it, but I will try, tell me if it is right.

So, when [A] doubles, and is constant, the Rate ALSO doubles. Thats easy, 1st order. NOW, to get you have to already account for [A] in this because there aren't two trials where [A] stays the same. So, from trials 2 to 3, [A] doubles and triples. The rate is multiplied by 18. Automatically take in account the change in [A] and its effect on rate (it is 1st order so when it doubles so does the rate). This leaves us with the Rate being multiplied by 9 (the new rate from the effect of changing [A]0, now since triples, and the Rate is multiplied by 9, this is 2nd order ((x)^2), just like before bc (3)^2=9. Let me know if this is right, I tried.

 

[BiomajorPreDent]

I havent read the other posts, but let me just say, I was worried about having these problems after reading kaplans wierd and long explanations

But I had these questions on my chem section, and it was reallly basic

very simple, and similar to the OP first example

Forget the 2nd example its not representative of the real thing

 

[americanpierg]

Forget the 2nd example its not representative of the real thing

are you sure? the first seems like a 8th grade problem. the second problem actually involves a thinking step.

 

[nze82]

are you sure? the first seems like a 8th grade problem. the second problem actually involves a thinking step.

It does involve some thinking. But he/she is right! They rarely give you something that involves this much thinking and math on the real exam. I think the first one is more representative of what one may see on the exam. That being said, I'm not encouraging anyone to ignore the second example and not learn it.

 

[jdpaul14]

so, are you always going to divide trial 3 by trial 2? or could you have divided trial 3 by trial 1 and received the same answer? how do you know which trials to divide by?

 

[americanpierg]

so, are you always going to divide trial 3 by trial 2? or could you have divided trial 3 by trial 1 and received the same answer? how do you know which trials to divide by?

you can divide any trial by any other trial, you just want to find one that gives you clean numbers where A is to the 0th power (1). the answers will always be the same (or roughy the same if you're given numbers that are rounded like going from 8.9 to 18.1

 

[nze82]

so, are you always going to divide trial 3 by trial 2? or could you have divided trial 3 by trial 1 and received the same answer? how do you know which trials to divide by?

That's where you need to start thinking. I always try to find two trials, between which the concentration of one of the reactions is held constant, while the concentration of the other reactant changes.

 

[fifi1314]

Let me know if this is right, I tried.

YES! The correct answer is: order of B is 2.
Thanks alot for the speedy solution!!!

 

[fifi1314]

These problems are quite time consuming, and you can only get good at them by doing lots of problems, so you'll get comfortable with setting up the equations and doing all the math.

Thanks very much for the explanation, it helps alot!

 

[fifi1314]

Ok, ajlst11 and nez 82.
I am having trouble again on this question.... ajlst11's speedy solution give me 1 as correct answer, nez82 dividing rule give me 16.... Please help! Thnx!!!
Here is the question (from kaplan subject test 2)
Trial [A]o (M) o (M) rate (M/min)
1 1 x 10–2 3 x 10–3 2 x 10–3
2 1 x 10–2 6 x 10–3 8 x 10–3
3 2 x 10–2 1.2 x 10–2 3.2 x 10–2
17. What is the reaction order with respect to A?
A. 1
B. 1.5
C. 2
D. 3
E. 0 correct answer

 

[nze82]

Ok, ajlst11 and nez 82.
I am having trouble again on this question.... ajlst11's speedy solution give me 1 as correct answer, nez82 dividing rule give me 16.... Please help! Thnx!!!
Here is the question (from kaplan subject test 2)
Trial [A]o (M) o (M) rate (M/min)
1 1 x 10–2 3 x 10–3 2 x 10–3
2 1 x 10–2 6 x 10–3 8 x 10–3
3 2 x 10–2 1.2 x 10–2 3.2 x 10–2
17. What is the reaction order with respect to A?
A. 1
B. 1.5
C. 2
D. 3
E. 0 correct answer

You don't even need to do much math here:

Compare the change in concentration of B between trials 1 & 2. The concentration is doubled. This changes causes the rate to quadruple. So:

2^x = 4

x = 2 (This is the order of reaction with respect to B)

Now, let's move on to trials 2 &3:
Notice how both the concentrations of A and B has changed. The concentration of B has been doubled. Look at the rate. What has happened? The rate has quadrupled! What does this tell you about A?
Essentially change in concentration of A had no effect on the reaction rate. The reaction rate quadrupled, when B's concentration was double. This is just like the first two reactions, where concentration of A was held constant.
Since, change in concentration of A has no affect on the reaction rate, then we can conclude that the order of reaction with respect to A is ZERO.

You really don't need to do math for all of these questions.

Hope this helps!

 

[fifi1314]

You don't even need to do much math here:

Compare the change in concentration of B between trials 1 & 2. The concentration is doubled. This changes causes the rate to quadruple. So:

2^x = 4

x = 2 (This is the order of reaction with respect to B)

Now, let's move on to trials 2 &3:
Notice how both the concentrations of A and B has changed. The concentration of B has been doubled. Look at the rate. What has happened? The rate has quadrupled! What does this tell you about A?
Essentially change in concentration of A had no effect on the reaction rate. The reaction rate quadrupled, when B's concentration was double. This is just like the first two reactions, where concentration of A was held constant.
Since, change in concentration of A has no affect on the reaction rate, then we can conclude that the order of reaction with respect to A is ZERO.

You really don't need to do math for all of these questions.

Hope this helps!

Thanks alot!!!! Err... Yes, I did think too much...