_______________
Answer:
The given reaction is
2R Products
The rate law is -dR/dt = k [R]2
The integrated form of the above rate law is
[ 1/Rt - 1/R0] = kt
Where R0 is the initial concentration of R
Rt is the concentration of R at time t
Given R0 = 0.265 mol L-1
t = 3.5 min = 3.5x 60 = 210 s
k = 0.0520 L mol-1s-1
Therefore 1/Rt = 1/R0 + kt
= [1 / 0.265 mol L-1 ]+ 0.0520 L mol-1s-1 x 210 s
= 3.774 L mol-1 + 10.92 L mol-1
= 14.694 L mol-1
Rt = [1 / 14.694] mol L-1
= 0.068 mol L-1
Thus the concentration of R after 3.5 minutes is 0.068 mol L-1
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