were doing an experiment.
and i did the calculation for r3/r2 and r2/r1
each the answer is 2. but the teacher
did this one the board in the example
answer
r2/r1= 2n
log 1.23/log 2.0 = n
where does the 1.23 come from? if the answer is a whole number do i have to do the log? if so how do i calculate a log for a single number? please help!
It may have helped if you had provided numbers for r2 and r1 (unless those are r^2 and r^1). As it is I don't have a clue what is going on.
r2: o.532
r1: 0.266
OK, I see where the 2 in your statement comes from. What experiment are you doing. What is the relationship of r2/r1 = 2n. What is n.
study of kinetics these are the numbers:
run 1
MnO4^-1 = 0.0125m
h2c2o4= 0.266M
average time: 371 sec
rate of reaction: 1.68 x 10^-5
run 2
MnO4^-1 = 0.0125m
h2c2o4= 0.532M
ave time: 239 sec
rate of reaction: 2.61 x 10^-5
run 3
MnO4^-1 = 0.0250m
h2c2o4= 0.266M
ave time 312 sec
rate of reaction: 4.01 x10^-5
r2/r1 =2
r3/r1=2
in class during this step he converted the answer to log using
1.23=2n
m is MN04 and n is h2c2o4
last i find k
1.68x10^-5/ [0.0125M][0.266]= 5.07 X 10^-3
im only confused about the log thing...
Here is the way it is to be done but it doesn't connect all of the steps you have. Apparently some of yours have been omitted. You may know all of this so don't get insulted if I put in too much detail.
You take run 1 and run 2 because the MnO4 is the same and oxalic acid is different.
Same for run 1 and run 3 because the H2C2O4 is the same (but MnO4 is different). Also, from your k calculation, I am assuming you were supposed to obtain 1 as the order for both MnO4 and H2C2O4.
rate 3 = 4.01 x 10^-5
rate 1 = 1.68 x 10^-5
rate 3 = k(KMnO4)x(H2C2O4)y
rate 1 = k(KMnO4)x(H2C2O4)y
Substitute the values and divide the first equation I've written by the second.
4.01E-5....k(0.025)^x(0.266)^y
------- = ----------------------
1.68E-5....k(0.0125)^x(0.266)^y
2.39 = the ks cancel, the 0.266^y cancel to leave
2.39 = 2x
This becomes
log 2.39 = x*log 2
0.3778/0.301 = x = 1.25 which apparently is rounded to 1 and the order of the reaction with respect to permanganate is 1
I don't know where the 1.23 came from. Perhaps what I've done here will fit somewhere with what you have done and you will see that number pop out. One thing I need to point out is that r3/r1 and r2/r1 don't appear to be 2. That are 2.38 and 1.55 respectively.
In the example given by your teacher, the expression "log 1.23" is used to calculate the value of n, which is then used in the equation "r2/r1 = 2n". Let me explain how this works.
To calculate the value of n, your teacher is using logarithms. In this case, they are using the logarithm to the base 2.0, denoted as "log 2.0". The value of "log 2.0" is a constant and can be easily looked up in a logarithm table or calculated using a scientific calculator.
Now let's talk about the number 1.23. In the example, this number is given to you as a specific value. However, in a real experiment, the value of "r2/r1" may not always be a whole number. In such cases, you can use logarithms to find the approximate value of n.
To calculate the logarithm of a single number, like 1.23 in this case, you divide the logarithm of that number by the logarithm of the base. In other words, you divide "log 1.23" by "log 2.0" to get the value of n.
If you don't have a logarithm table or a scientific calculator, you can also calculate logarithms using the following steps:
1. Divide the number for which you want to find the logarithm (in this case, 1.23) by the base of the logarithm (in this case, 2.0).
1.23 / 2.0 = 0.615
2. Take the logarithm of the result using any calculator that has a logarithm function (e.g., a scientific calculator). The result will be the value of n.
log 0.615 = -0.212
Therefore, in this case, the value of n is approximately -0.212.
Just remember, logarithms are used when the ratio is not a whole number, and the 1.23 is a specific value given for illustration purposes.