Is calculating the sloope of the tangent line the same as calculating the slope? (y2-y1) / (x2/x1)
for instance if i took the points 60, 0.023 for X1Y1 and points 100, 0.034 for X2Y2, the slope of the line would be :
(0.034-0.023) / (100-60) = 0.011/40 = 0.000275 (the slope) - m
y = 0.000275x + b?
Using the following data, determine the initial rate by calculating the slope of the tangent line at time zero.
Time (s). [I2] (mM)
0. 0
20. 0.009
40. 0.017
60. 0.023
80. 0.029
100. 0.034
120. 0.039
140. 0.043
160. 0.046
180. 0.049
I can't figure out how to make a tangent line at time zero. Rate of the reaction = 8.59 x10^-7 M/s.
Is it just finding the slope? I made a chart in excel, it gave the y=mx+b: y = 0.0003x + 0.0048, R^2 = 0.9719.
Is that the slope ( y = 0.0003x + 0.0048)?
Please any help would be greatly appreciated.
2 answers
Nope, it is not linear, it is a curve whose slope decreases.
AT zero, that means the initial slope.
your first slope is .009/20=.00045mmoles/s
or 4.5E-7moles/sec.
At the end, between 160-180 sec
rate=.003/20=.00015E-3=1.5E-7Mol/sec
The other way, to get initial slope, is to plot this on a graph, then connect the points (with a french curve, then draw a tangent line at t=0, you should get about what I calculated above.
AT zero, that means the initial slope.
your first slope is .009/20=.00045mmoles/s
or 4.5E-7moles/sec.
At the end, between 160-180 sec
rate=.003/20=.00015E-3=1.5E-7Mol/sec
The other way, to get initial slope, is to plot this on a graph, then connect the points (with a french curve, then draw a tangent line at t=0, you should get about what I calculated above.