[earthnation112]

I just want to confirm if I have got the answers right to some revision questions for chemistry. Would be very grateful if someone can give productive input to what I post. Thank You

The question is:
Q5. In a particular reaction, the order of reaction with respect to compound C is known to be fractional. The rates of reaction were measured for different concentrations of C (all other reagents were present in large excess). What is the order of reaction with respect to C?

[C]/M           0.842  0.739   0.644   0.557   0.476  0.401
rate/M s^-1   1.40     1.36   1.33      1.29   1.25     1.21

The answer is ONE of the following: 1/5, ¼, 1/3, ½, 2/3, ¾, 5/4, 4/3, 3/2.

To work out the fractional order of reaction I used the “Initial rate method” by applying the following formula:
ln (rate) = aln[A] + ln(k)
y            =  m x   +  c

I read in a book that “So if we know the rate at different [A] we can plot ln (rate) vs ln [A], the gradient will be a, the order of the reaction”.

So I produced the following table from the table above:

ln(A)          -0.17198   -0.30246   -0.44006   -0.58519   -0.74234   -0.91379
ln(rate)   0.336472    0.307485    0.285179    0.254642    0.223144    0.19062

So based of this I made a graph (image attached) in excel and the equation produced was:
y = 0.1957x + 0.3692

The gradient is m so I used 0.1957, but I know excel isn’t accurate when predicating the gradient as its better to do it physically using graph paper. So I just rounded 0.1957 to 0.2 and choose the fractional order 1/5. 

Is this right? Have I made a mistake anywhere? Thanks

[mjc123]

I know excel isn’t accurate when predicating the gradient as its better to do it physically using graph paper

What makes you think that?
0.1975 is pretty close to 0.2, I'd be very happy with that if I was doing the experiment!
Do a regression (not a trendline) in Excel. What are the confidence limits for the slope?

[earthnation112]

The reason is because I'm always told by my professors that excel isn't as accurate as doing it using graph paper. I think they are referring to the line of best fit when you have anomalies.

Is the answer I got not right?

I've never done a regression using excel or been thought how to produce confidence limits for the slope, without these two would I still be able to get the order of reaction using the method I used? Thanks

[earthnation112]

0.1975 is pretty close to 0.2, I'd be very happy with that if I was doing the experiment!
Do a regression (not a trendline) in Excel. What are the confidence limits for the slope?

Just performed regression in excel using the following data:

x                 y                  xy       x2                        y2
-0.17198   0.336472   -0.057866   0.02957712   0.113213407
-0.30246   0.307485   -0.093002   0.091482052   0.094547025
-0.44006   0.285179   -0.125496   0.193652804   0.081327062
-0.58519   0.254642   -0.149014   0.342447336   0.064842548
-0.74234   0.223144   -0.165649   0.551068676   0.049793245
-0.91379   0.19062   -0.174187   0.835012164   0.036335984
            
sum(x)    sum(y)      sum(xy)        sum(x2)      sum(y2)
-3.15582   1.597542   -0.765214     2.043240151   0.440059271
            
(sum x)^2     (sum y)^2      
9.959199872   2.552140442      

After having produced the above I used an equation which once solved equals m, the value for m which is the gradient was:
m=0.197            

I got the same value approximately when using the trend line in excel which was:
y = 0.1957x + 0.3692

After using both a trend line and performing rergression on my data and having got the same answers can I be confident now that the gradient is 0.197 and say the fractional order is 1/5? Thanks

[mjc123]

Yes.
If you use the Regression tool in Excel (you need to have installed the Data Analysis add-in) you can get statistics of the coefficients. For the slope, m = 0.195749 with a standard error of 0.002833, and the lower and upper 95% confidence limits are 0.187883 and 0.203614. So yes, your result is consistent wih an order of 0.2.

After having produced the above I used an equation which once solved equals m, the value for m which is the gradient was: m=0.197           

Do you mean m = 0.1957?

[earthnation112]

Yes 0.1957, just used the regression tool now using excel. The method I used before to calculate m was regression also but i didn't use the built in excel add on rather I calculated the data manually, basically followed the tutorial on the following website:
https://www.clemson.edu/ces/phoenix/tutorials/excel/regression.html

Thanks for the reply. 

[Babcock_Hall]

I know excel isn’t accurate when predicating the gradient as its better to do it physically using graph paper

What makes you think that?
0.1975 is pretty close to 0.2, I'd be very happy with that if I was doing the experiment!
Do a regression (not a trendline) in Excel. What are the confidence limits for the slope?

mjc123,

Would you care to say a bit more about the difference between a regression versus a trendline in Excel?  A little bit of Googling did not turn up much of value.

[mjc123]

Basically it's the same calculation - the slope and intercept values should be the same. It differs in the information it presents to you. Trendline draws a regression line on your graph (non-linear fit options are also available) and gives you the equation. Regression gives you (if you want it) the statistics of the coefficients - standard error, confidence limits, t and P values etc. Thus you can not only get a value for the slope, but also (e.g.) see whether, at the 95% confidence level, your data are consistent with a slope of 0.2.