[At] = -kt + [Ao]
likewise.. for 1st order
-d[A]/dt = k x [A]^1
1/[A] d[A] = -kt
ln[At] - ln[Ao] = -kt
ln[At] = -kt + ln[Ao]
likewise for 2nd order
-d[A]/dt = k x [A]^2
1/[A]^2 d[A] = -k dt
-1/[At] + 1/[Ao] = -kt
+1/[At] - 1/[Ao] = +kt
1/[At] = +kt + 1/[Ao]
***********
got all that?
zero order.. [At] = -kt + [Ao]
1st order... ln[At] = -kt + ln[Ao]
2nd order.. 1/[At] = +kt + 1/[Ao]
now.. [Ao] is a constant.. we start with a fixed concentration and measure [A] as a function of time..
so.. if we have a zero order rxn. and we let y = [A] and x = t and m = -k and b = [Ao] we get this equation
y = mx + b
ie... a plot of [A] vs t will give a straight line with slope = -k and intercept = [Ao]
for a 1st order rxn.. a plot of ln[A] vs t will give a line with slope = -k and intercept = ln[Ao]
for a 2nd order rxn.. a plot of 1/[A] vs t will give a line with slope = -k and intercept = 1/[Ao]
*** part a ***
ln[At] = -kt + ln[Ao]
we want the slope of the line ln[At] vs t
so..
k = - (Δln([At])) / (Δt) = -(ln([At1] - ln([At2]) / (t1 - t2) = ln([At2] / [At1]) / (t1-t2)
k = ln(5.30e-3/7.00e-2) / (30.0s - 90.0s) = __ /s
i'll let you calculate that out. notice the units of "M" have cancelled out?
*** part b ***
some deal.. but watch the signs and the units!
1/[At] = +kt + 1/[Ao]
so.. k = + ( 1/[At1] - 1/[At2] ) / (t1 - t2) = ( 1 / (0.710M) - 1 / (3.30e-2M)) / (225s - 770s) = __ /(Mxsec)