Not good at these. Please if possible show steps as well as answer so I can unde

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Not good at these. Please if possible show steps as well as answer so I can understand. Thank you.

) 7of12 (2) ± Using Integrated Rate Laws Part A Constants | Periodic Table The integrated rate laws for zero-, first-, and second-order reaction may be arranged such that they resemble the equation for a straight line, The reactant concentration in a zero-order reaction was 7.00x10 M after 170 s and 4.00102 M after 305 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 dot or a dash. Integrated Rate Law [A] = _ kt + IAlo Graph Slope | [A] vs. t | -k Order View Available Hint(s) xxbLb k0th= 1 Value Units Submit Previous Answers X Incorrect; Try Again; 5 attempts remaining Your answer does not have the correct dimensions.

Explanation / Answer

For any order reacction

The rate (-dA/dt)= K[A]n    (1), n is ther order and K ia the rate constant.

For zero order, n=0, A is the reacting species

And the equation becomes –dA/dt= K

When integrated noting that at t=0 , [A] =[A]o and at t=t , [A] =[A]

[A]= [A]0- Kt (2)

When t=170 sec, [A] = 7*10-2M

When t= 305 sec, [A]= 4*10-2M

Substituting these values in Eq. 2 twice, we get

7*10-2= [A]o-K*170   (3)

4*10-2= [A]o-K*305   (4)

Eq.3-Eq.4 gives 7*10-2- 4*10-2 = K(305-170)

K=0.00022M/sec

Substituting this value in eq. 3

7*10-2 =[A]0-0.00022*170

[A]0= 0.1077M=initial concentration of A.

2. for 1st order reaction, n=1 and -dA/dt= K[A]

when integrated, the equation becomes

lnA= ln[A]0-Kt (5)

where K is the rate constant

whem t= 15 sec, [A]= 9.8*10-2 Eq.5 becomes,   ln (9.8*10-2)= ln A0- K* 15 (6)

and when t= 90 sec, [A] =8.5*10-3M, Eq.5 becomes ln (8.5*10-3)= lnAo-K*90 (7)

Eq.5-Eq.6 gives ln(9.8*10-2/(8.5*10-3)= K*(90-15), K=0.033/sec

for the second order n=2 and Eq.1 becomes -dA/dt= K[A]2,

when integrated noting that at t=0, [A]= [A]0 and at t= t, [A] =[A]0

1/A= 1/[A]o+ Kt (7), given when t=220 sec, [A]= 0.15M, Eq. 7 becomes 1/0.15= 1/[A]o+ 220*t (8)

when t= 775 sec, [A]= 1.9*10-2M, Eq. 7 can be written as 1/1.9*10-2 = 1/[A]o+ 775*t (9)

Eq.9-Eq.8   1/(1.9*10-2)- 1/0.15= K*(775-220)

K=0.083/M.sec

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