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 may be arranged such that they resemble the equation for a straight line, y = mx + b. The reactant concentration in a first-order reaction was 5.60 times 10^-2 M after 50.0 s and 1.00 times 10^-3 M after 70.0 s. What is the rate constant for this reaction? Express your answer with the appropriate units. Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash. The reactant concentration in a second-order reaction was 0.150 M after 160 s and 4.20 times 10^-2 M after 720 s. What is the rate constant for this reaction? Express your answer with the appropriate units Indicate the multiplication of units, as necessary, explicitly either with a multiplication dot or a dash.

Explanation / Answer

PART C:

Given that the reaction is of 1st order, the rate equation is:

ln[A] = -kt + ln[A]o

ln([Ao]/[A]) = kt

[A] = 5.6*10-2 M at t = 50s

[A] = 1*10-3 M at t = 70s

Substitute to get 2 equations as:

eqn 1: -50k + ln[A]0 = ln[5.6*10-2]

-50k + ln[A]0 = -2.882

Eqn 2:

-70k + ln[A]0 = -6.9077

solving equation 1 and 2, we get:

ln[A]0 = 7.18225

k = 0.201285 s-1

PART D:

Given that the reaction is of 2nd order, the rate law equation is:

1/[A] = kt + 1/[A]0

[A] = 0.15 M at t = 160 s

[A] = 4.2*10-2 M at t = 720 s

Similarly as above get 2 equations as:

1/0.15 = 160t + 1/[A]0

1/4.2*10-2 = 720t + 1/[A]0

solve to get,

k = 0.0306 M-1s-1

1/[A]0 = 1.7687 M

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