please help me. thanks The integrated rate laws for zero-, first-, and second-or

Question

please help me. thanks

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.

Part A

The reactant concentration in a zero-order reaction was 0.100 M after 170 s and 3.00×102M after 380 s . What is the rate constant for this reaction?

Part B

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

Part C

The reactant concentration in a first-order reaction was 6.10×102M after 40.0 s and 4.10×103M after 70.0 s . What is the rate constant for this reaction?

Part D

The reactant concentration in a second-order reaction was 0.260 M after 295 s and 7.70×102M after 860 s . What is the rate constant for this reaction?

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

Explanation / Answer

Part A

k = - m = - ([A2] - [A1]) / (t2 - t1) = - (3.00x10-2M - 0.100M) / (380s - 170s) = 3.33 x10-4 M/s

Part B

[At] = -k x t + [Ao]

[Ao] = (0.100M) + (3.33 x10-4 M/s) x (170s) = 0.157M

Part C

k = - (ln[A2] - ln[A1]) / (t2 - t1) = - ln([A2] / [A1]) / (t2 - t1)
k = - ln(4.10×103 / 6.10×102 ) / (70s - 40.0s)=3.26

Part d

k = + (1 / [A2] - 1 / [A1]) / (t2 - t1): k = + (1/7.7x10^-2M - 1/0.26M) / (860s - 295s) = 0.0162

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