Question

Part 11: Integrated Rate Law 8 pts each If a second-order reaction with a rate constant of 0.056 Mshas a reactant concentration of 1.2 M after 8.0 seconds have elapsed, what was the initial reactant c
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Fox Second corder でea el an 1191 I = 0,385 TA10 1S .595 M

Add a comment
Know the answer?
Add Answer to:
Part 11: Integrated Rate Law 8 pts each If a second-order reaction with a rate constant...
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. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • ± Using Integrated Rate Laws Part A The reactant concentration in a zero-order reaction The integrated...

    ± Using Integrated Rate Laws Part A The reactant concentration in a zero-order reaction 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 was 9.00x102 M after 155 s and 3.50x102 M after 320 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...

  • The integrated rate law allows chemists to predict the reactant concentration after a certain amount of...

    The integrated rate law allows chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A]=[A]0e−kt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute [A]02 for [A] and rearrange the equation to: t1/2=0.693k This equation calculates the time...

  • 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+by=mx+b. Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0[A]=−kt+[A]0 [A] vs. t[A] vs. t −k−k 1 ln[A]=−kt+ln[A]0ln⁡[A]=−kt+ln⁡[A]0 ln[A] vs. tln[A] vs. t −k−k 2 1[A]= kt+1[A]01[A]= kt+1[A]0 1[A] vs. t1[A] vs. t kk A.) The reactant concentration in a zero-order reaction was 0.100 MM after 165 ss and 4.00×10−2 MM after 305 ss . What is 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+by=mx+b. Order Integrated Rate Law Graph Slope 0 [A]=−kt+[A]0[A]=−kt+[A]0 [A] vs. t[A] vs. t −k 1 ln[A]=−kt+ln[A]0ln⁡[A]=−kt+ln⁡[A]0 ln[A] vs. tln[A] vs. t −k 2 1[A]= kt+1[A]01[A]= kt+1[A]0 1[A] vs. t1[A] vs. t k Part A The reactant concentration in a zero-order reaction was 8.00×10−2 MM after 130 ss and 4.00×10−2 MM after 380 ss . What is...

  • The integrated rate law allow chemists to predict the reactant concentration after a certain amount of...

    The integrated rate law allow chemists to predict the reactant concentration after a certain amount of time, or the time it would take for a certain concentration to be reached. The integrated rate law for a first-order reaction is: [A] = [A]oe -Rt Now say we are particularly interested in the time it would take for the concentration to become one-half of its initial value. Then we could substitute Z" for [A] and rearrange the equation to: A) 1/2= 0093...

  • HW 4 ± Using Integrated Rate Laws Resources previous | 1 of 11 | next» ±...

    HW 4 ± Using Integrated Rate Laws Resources previous | 1 of 11 | next» ± Using Integrated Rate Laws Part A 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 Mafter 125 s and 3.00x10 M The reactant concentration in a zero-order reaction was 6.00x10 after 305 s. What is the rate constant for this reaction? Express your answer with the...

  • + Using Integrated Rate Laws The integrated rate laws for zero-, first-, and second-order reaction may...

    + Using Integrated Rate Laws 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. Slope Order O 1 2 Integrated Rate Law Graph [A] = - kt + [A] [A] vs. t In[A] = -kt + In[A], In[A] vs. t LÀ=kt + TA LÀ vs. t -k Review Constants Periodic Table Part A The reactant concentration in a zero-order reaction was...

  • + Using Integrated Rate Laws The integrated rate laws for zero-, first-, and second-order reaction may be arrang...

    + Using Integrated Rate Laws 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. Slope Order O 1 2 Integrated Rate Law Graph [A] = - kt + [A] [A] vs. t In[A] = -kt + In[A], In[A] vs. t LÀ=kt + TA LÀ vs. t -k Review Constants Periodic Table Part A The reactant concentration in a zero-order reaction was...

  • ing Integrated Rate Laws < 10 of 11 > Review Constants Periodic Table ne integrated rate...

    ing Integrated Rate Laws < 10 of 11 > Review Constants Periodic Table ne integrated rate laws for zero, first- and second-order eaction may be arranged such that they resemble the equation or a straight line, ymr + b Part A Order 0 Integrated Rate Law Graph Slope [A] = - kt +(Alo (A) vs. t -k In A-kt+In Al In A vs. t -k Avst k 1 The reactant concentration in a zero order reaction was 8.00-10-2 Mafter 160...

  • 14.1 Question 3 Learning Goal: To understand how to use integrated rate laws to solve for...

    14.1 Question 3 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...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT