Part A The reactant concentration in a zero-order reaction was 5.00×10 ?2 M afte

Question

Part A

The reactant concentration in a zero-order reaction was 5.00×10?2M after 200 s and 3.00×10?2Mafter 380 s . What is the rate constant for this reaction?

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 8.00×10?2M after 15.0 s and 3.40×10?3Mafter 85.0 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 0.180 M after 175 s and 6.10×10?2M after 750 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

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 La Graph Slope 2 vs. tk

Explanation / Answer

a) For zero order reaction

[A]=-kt+[A0] (1)

[A]0= initial concentration, [A] =concentration at any time, t

At two different time intervals t1 and t2

Given t1= 200 seconds [A]1 =5*10-2 and t2= 380 seconds [A]2= 3*10-2M

Substituting these values in (1) gives

5*10-2 =-k*200 +[A0] (2)

3*10-2 =-k*380 + [A0]   (3)

(2)-(3) gives

2*10-2 =k*(380-200) K= 2*10-2/180 M/s =0.00011 M/sec

b)

From (2)

5*10-2 +k*200 =[A0]

[A0]= 5*10-2+ 0.00011*200=0.072 M

c)

The integrated equation for first order is given by

ln [A] =ln[AO]-kt (1)

at time, t=15 sec [A]= 8/100=0.08M

ln [8*10-2] = ln [AO]-15k   (2)

at time t=80 sec, [A]=3.4*10-5M

ln [3.4*10-5] = ln [AO]-80k   (3)

subtracting (2) and (3)

ln (8*10-2/ 3.4*10-5)= 65k   7.763=65 k

hence rate constant k= 7.763/65=0.119431 sec-1

d)

For a second order reaction

1/[A]= kt+1/[A0] (1)

At t1=175 second [A]1= 0.180 M and t2= 750 seconds [A]2= 6.1*10-2 M

Therefore from (1)

1/0.180 = 175k+1/[A0] (2)

1/(6.1*10-2= 750K +1/[A0] (3)

Eq.3 –Eq.2 gives

16.4-5.5 6= (750-175)k , k =0.0188 / M..s

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