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.

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Part A

The reactant concentration in a zero-order reaction was 5.00×102M after 110 s and 4.00×102M after 375 s . What is the rate constant for this reaction?

3.77×105 Ms

Correct

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Part B

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

I only need B

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

Zero orden then

A = A0 - kt

from A; you got 3.77*10^-5

then

if we need A0

solve for A0

A = A0 - kt

choose any point, such as A = 5*10^-2 at t = 110 s nad k = 3.77*10^-5

substitute in equation

A = A0 - kt

5*10^-2 = A0 - (3.77*10^-5)(119)

solve for A0

A0 =5*10^-2+ (3.77*10^-5)(119)

A0 = 0.0544863 M

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