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 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

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

-------------

Part C

The reactant concentration in a first-order reaction was 9.60×10−2M after 30.0 s and 5.20×10−3M after 60.0 s . What is the rate constant for this reaction?

-------------

Part D

The reactant concentration in a second-order reaction was 0.650 M after 175 s and 5.60×10−2M after 815 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 multiplication, for example write a Newton-meter as N*m.

A. rate constant, k = - slope

= - d[A]/dt

= - [0.04-0.05] M /[375-110]s

= 3.77*10-5 M.s-1

B. Initial reactant concentration = [A]0

= intercept in y axis

look at the graph(the graph is qualitative). Draw your own graph in a graph paper to determine the initial concentration.

C. rate constant, k = - slope

= - dln[A]/dt

= - [ln 0.0052 M - ln 0.096 M]/(60-30)s

= - ln (0.0052/0.096)/30 s

= 0.097 s-1

D. rate constant, k = slope

= d(1/[A])/dt

= (1/0.056 - 1/0.65) M-1 /(815-175)s

= 0.0255 M-­1*s-1

Earn Coins

Coins can be redeemed for fabulous gifts.