The study of kinetics
group 1:
Temp: 12.6 C
T(K): 12.6+273= 285.6
1/T(K): 3.50X10^-3 this one is the x value
Time: 816 sec
Rate ®: 7.66 x10^-6
LnR: -11.8 this is the y value
group 2:
Temp: 45C
T(K): 45+273= 318
1/T(K): 3.14X10^-3 this one is the x value
Time: 495 sec
Rate ®: 1.26 x10^-6
LnR: -11.3
group 3:
Temp: 12.6 C
T(K): 34.8+273= 285.6
1/T(K): 3.28X10^-3 this one is the x value
Time: 107 sec
Rate ®: 5.84 x10^-5
LnR: -9.75
now I need to find the
group 1:
Temp: 12.6 C
T(K): 12.6+273= 285.6
1/T(K): 3.50X10^-3 this one is the x value
Time: 816 sec
Rate ®: 7.66 x10^-6
LnR: -11.8
1. im supposed to find the
slope= -EA/R = triangle y/triangle x = (y2-y1) ________/(x2-x1)____________
how do I solve this? Also the teacher said our k from last week was strong, and to use LnR instead.
2. also how do I find the Ea?
3. can someone check to see if my graph looks right?
img207.imageshack.us/img207/6375/2ndgraph.jpg
We are supposed to find the y inter, and the R2.
any help would be appreciated.
To solve the first part of your question, let's calculate the slope using the given data points for group 1:
(x1, y1) = (3.50x10^-3, -11.8)
(x2, y2) = (3.14x10^-3, -11.3)
The formula for the slope is:
slope = (y2 - y1) / (x2 - x1)
Plugging in the values:
slope = (-11.3 - (-11.8)) / (3.14x10^-3 - 3.50x10^-3)
= 0.5 / (-0.36x10^-3)
≈ -1.389 × 10^3
This value represents the negative of the activation energy divided by the gas constant (EA/R). The negative sign indicates that the activation energy is positive.
To find the activation energy (Ea) from the slope, you can rearrange the equation:
Ea = -slope × R
Given that the gas constant (R) is approximately 8.314 J/(mol·K), you can substitute the value and solve for Ea.
For the second part of your question, you mentioned that the teacher suggested using LnR instead of Rate (®). To find the activation energy using this approach, you would typically plot LnR as a function of 1/T(K) for all the data points. Then, the slope of the resulting line would represent the activation energy (Ea).
Regarding your third question about checking your graph, unfortunately, as an AI text-based bot, I'm unable to view the image you mentioned. However, I can guide you on how to find the y-intercept and R2 value of a graph.
To find the y-intercept (the point where the line crosses the y-axis), you can set 1/T(K) equal to zero and solve for LnR or Rate (®), depending on what you are plotting. This value will give you the y-intercept.
The R2 value, also known as the coefficient of determination, indicates how well the data fits the line. It ranges from 0 to 1, with 1 being a perfect fit. You can calculate it by using statistical software or spreadsheet programs that have the capability to calculate regression values. The higher the R2 value, the better the fit of the data to the line.
I hope this explanation helps you solve the questions. If you have any further queries, feel free to ask!