Answer:
Zero order
Explanation:
Looking at the data we can note a linear dependence between concentration and time.
Time Conc.
0 2
15 1.82
30 1.64
48 1.42
75 1.10
In the first 15 min it was consumed 2-1.82=0.18. So the rate is [tex]r=\frac{\Delta C}{\Delta t} = \frac{0.18}{15}=0.012[/tex]
From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is [tex]r=0.012[/tex].
From 30 to 48 (it has passed 18 min)the rate is [tex]r= \frac{0.22}{18}\approx 0.012[/tex]
From 48 to 75 (it has passed 27 min) the rate is [tex]r= \frac{0.32}{27}\approx 0.012[/tex]
So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as [tex]r= k^{0} = 0.012 mol L^{-1} min^{-1}[/tex] that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.