The integrated rate laws for zero-, first-, and second-order reaction may be arr

Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line,y=mx+b.

The reactant concentration in a zero-order reaction was 0.100 M after 195 s and 2.00×102M after 365 s . What is the rate constant for this reaction?

What was the initial reactant concentration for the reaction described in Part A?

Order Integrated Rate Law Graph Slope 0 [A]=kt+[A]0 [A] vs. t k 1 ln[A]=kt+ln[A]0 ln[A] vs. t k 2 1[A]= kt+1[A]0 1[A] vs. t k

Explanation / Answer

For zero order reaction

CAO- CA =kt

where CAO =initial concentration, CA =concentration at time t

k= rate constant

at 195 seconds

CAO= K*195+0.1 (1)

at 365 seconds

CAO= K*365+ 0.02 (2)

equation (1) and (2) gives

195*K+0.1= K*365+0.02

K(365-195)= (0.1-0.02)

K= 0.08/170 =0.000471 Msec-1

Subssiting this value in equation(1)

gives

CAO= 0.000471*195+0.1=0.191756M

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