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

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. 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

The reactant concentration in a zero-order reaction was 7.00×102M after 150 s and 4.00×102M after 360 s . What is the rate constant for this reaction?

Express your answer with the appropriate units.

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

Express your answer with the appropriate units.

The reactant concentration in a first-order reaction was 0.100 M after 50.0 s and 4.20×103M after 85.0 s . What is the rate constant for this reaction?

Express your answer with the appropriate units

The reactant concentration in a second-order reaction was 0.710 M after 175 s and 7.90×102M after 745 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.

Explanation / Answer

Given [A]= -kt+ [A]0 where [A]0 is initial concentration and [A] is concentration at time t

At t= 150 second [A] =7*10-2M

7*10-2= -k*150+ [A]0 (1) at t= 360 seconds [A]= 4*10-2M  

4*10-2 = k*360 +[A]0 (2)

Eq.2 –Eq.1 give 3*10-2= k*210 k= 3*10-2/210 M/s=0.000143 M/s

From (1) 7*10-2 =-0.000143*150+ [A]0 , [A]0= 7*10-2+0.021429= 0.091429 M

2.

b) ln[A]=kt+ln[A]0    from t= 50 s [A] = 0.1M and t=85 s [A] =4.2*10-3

ln [0.1)= -k*50 +ln [A]0 (3)

ln (4.2*10-3)= -k*85+ ln [A]0 (4)

Eq.3- Eq.4 ln { 0.1/4.2*10-3} = k* 35, k= 0.090574/sec

From (3) ln [A]0= ln (0.1)+0.090574*50, [A]0= 9.263748 M

3.

/A= [1/A]0 +kt

at t= 175sec, A= 0.71   1/0.71 = [1/A]0 + k*175 (5)

1.408 =1/[A]0 +175K (5A)

At t=745 sec

12.66= 1/[A]0 +k*745 (6)

at Eq.6-Eq.5A   gives k= 0.01974/M.s

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