Question

The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...

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.

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

Part A

The reactant concentration in a zero-order reaction was 7.00×10−2M after 135 s and 2.50×10−2M after 315 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part B

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

Express your answer with the appropriate units.

Part C

The reactant concentration in a first-order reaction was 9.40×10−2M after 50.0 s and 1.50×10−3M after 75.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

Part D

The reactant concentration in a second-order reaction was 0.250 M after 170 s and 2.60×10−2M after 835 s . What is the rate constant for this reaction?

Express your answer with the appropriate units. Include an asterisk to indicate a compound unit with mulitplication, for example write a Newton-meter as N*m.t B

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

Express your answer with the appropriate units.

Homework Answers

Answer #1

Part A.

For a zero order reaction,

[A] = [A]o - kt

with,

[A] at 135 s = 7 x 10^-2 M

[A] at 315 s = 2.50 x 10^-2 M

So the rate constant k for the reaction becomes,

k = -[2.5 x 10^-2 - 7 x 10^-2)/(315 - 135)] = 2.5 x 10^-4 M/s

Part B.

Initial concentration [A]o

[A]o = 7 x 10^-2 + 2.5 x 10^-4 x 135 = 0.104 M

Part C

rate constant for a first order reaction,

k = -[(ln(1.5 x 10^-3) - ln(9.4 x 10^-2)/(75 - 50)] = 0.165 s-1

Part D

Rate constant for a second order reatcion,

k = (1/2.6 x 10^-2 - 1/0.25)/(835 - 170) = 0.052 M-1*s-1

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...
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. 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 Part A The reactant concentration in a zero-order reaction was 8.00×10−2M after 200 s and 2.50×10−2Mafter 390 s . What is the rate constant for this reaction? Express your answer with the...
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...
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. 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 ------------ Part A The reactant concentration in a zero-order reaction was 5.00×10−2M after 110 s and 4.00×10−2M after 375 s . What is the rate constant for this reaction? ---------- Part B...
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...
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. 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 Part A The reactant concentration in a zero-order reaction was 5.00×10−2M after 200 s and 2.50×10−2M after 310 s . What is the rate constant for this reaction? Express your answer with...
The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they...
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. 1.) The reactant concentration in a zero-order reaction was 6.00×10−2M after 165 s and 3.50×10−2Mafter 385 s . What is the rate constant for this reaction? 2.)What was the initial reactant concentration for the reaction described in Part A? 3.)The reactant concentration in a first-order reaction was 6.70×10−2 M after 40.0 s and 2.50×10−3Mafter 95.0 s ....
Item 4 The integrated rate laws for zero-, first-, and second-order reaction may be arranged such...
Item 4 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. Order Integrated Rate Law Graph Slope 0 [A]t=−kt+[A]0 [A]t vs. t −k 1 ln[A]t=−kt+ln[A]0 ln[A]t vs. t −k 2 1[A]t= kt+1[A]0 1[A]t vs. t k Part A The reactant concentration in a zero-order reaction was 6.00×10−2 mol L−1 after 140 s and 3.50×10−2 mol L−1 after 400 s . What is the rate constant for...
Learning Goal: To understand how to use integrated rate laws to solve for concentration. A car...
Learning Goal: To understand how to use integrated rate laws to solve for concentration. A car starts at mile marker 145 on a highway and drives at 55 mi/hr in the direction of decreasing marker numbers. What mile marker will the car reach after 2 hours? This problem can easily be solved by calculating how far the car travels and subtracting that distance from the starting marker of 145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110 miles=milemarker 35 If we...
Part A A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45...
Part A A certain first-order reaction (A→products) has a rate constant of 7.20×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration? Express your answer with the appropriate units. Answer: 6.42 min Part B A certain second-order reaction (B→products) has a rate constant of 1.35×10−3M−1⋅s−1 at 27 ∘Cand an initial half-life of 236 s . What is the concentration of the reactant B after...
For a first-order reaction, the half-life is constant. It depends only on the rate constant k...
For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t1/2=0.693k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t1/2=1k[A]0 Part A A certain first-order reaction (A→products) has a rate constant of 4.20×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A],...
For a first-order reaction, the half-life is constant. It depends only on the rate constant k...
For a first-order reaction, the half-life is constant. It depends only on the rate constant k and not on the reactant concentration. It is expressed as t1/2=0.693k For a second-order reaction, the half-life depends on the rate constant and the concentration of the reactant and so is expressed as t1/2=1k[A]0 Part A A certain first-order reaction (A→products ) has a rate constant of 5.10×10−3 s−1 at 45 ∘C . How many minutes does it take for the concentration of the...
Part A The rate constant for a certain reaction is k = 3.70×10−3 s−1 . If...
Part A The rate constant for a certain reaction is k = 3.70×10−3 s−1 . If the initial reactant concentration was 0.200 M, what will the concentration be after 11.0 minutes? Express your answer with the appropriate units. Part B A zero-order reaction has a constant rate of 2.00×10−4M/s. If after 65.0 seconds the concentration has dropped to 3.50×10−2M, what was the initial concentration? Express your answer with the appropriate units.
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT