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

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 reaction concentration in a zero-order reaction was 7.00 times 10^-2 M after 155 s and 4.00 times 10^-2 M after 385 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. What was the initial reactant concentration for the reaction described in Part A? 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 first-order reaction was 6.30 times 10^-2 M after 45.0 s and 6.00 times 10^-3 M after 95.0 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.

Explanation / Answer

A)
for 0 order reaction,

[A]2 = -k*(t2-t1) +[A]1
4.00*10^-2 = -k*(385-155)+(7.00*10^-2)
3.00*10^-2 = k*(385-155)
3.00*10^-2 = k*(230)
k = 1.30*10^-4 M/s
Answer: 1.30*10^-4 M/s

B)
for 0 order reaction,
[A]1 = -k*t + [A]o
7.00*10^-2 = - (1.30*10^-4)*155 + [A]o
7.00*10^-2 = -0.02015 + [A]o
[A]o = 9.02*10^-2 M
Answer: 9.02*10^-2 M

C)
for 1st order reaction,
ln[A]2 = -k*(t2-t1) + ln [A]1
ln (6.00*10^-3) = -k*(95 - 45) + ln (6.30*10^-2)
-5.116 = -k*(50) - 2.765
k=4.70*10^-2 s-1

Answer: 4.70*10^-2 s-1

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