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« previous| 9 of 21 | next » Question 9 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 + The reactant concentration in a zero-order reaction was 9.00x10-2 M after 195 s and 3.50x102 M after 300 s. What is the rate constant for this reaction? Express your answer with the appropriate units Order Integrated Rate Law Graph Slope Vs. t 1 In[A] =-kt InAlo In[A] vs. t -k VS Vs.t k kothValue Units Submit Hints My Answers Give Up Review Part Part B What was the initial reactant concentration for the reaction described in Part A? Express your answer with the appropriate units AloValue Units Submit Hints My Answers Give Up Review PartExplanation / Answer
For a zero order reaction
CA= CAO- kt (1)
Where CA= Concentration at time t, CAO= initial concentration k= rate constant and t is the time
At 195 seconds , CA= 9*10-2
9*10-2 =CAO- k*195 (1)
At 300 seconds 3.5*10-2 = CAO- k*300 (2)
Eq.1- Eq.2 gives (9-3.5)*10-2 =t(300-195)
5.5*10-2 =k*105, k= 0.00052M/s
b)
From Eq.1
9*10-2= CAO- 0.00052*195
CAO= 9*10-2+0.00052*195 =0.1914M
c)
For a first order reaction , CA/CAO= e(-kt)
Where CA= Concentration at time t, CAO= initial concentration k= rate constant and t is the time
given CA=8.7/100 at t1= 45seconds and CA= 4.5*10-3 M after t2= 100 seconds
at t1 and t2 , 8.7*10-2 /CAO= e(-k*45) (1)
4.5*10-3/CAO= e(-k*100) (2)
Eq.2/ Eq.1 ,8.7*10-2/ 4.5*10-3 = ek*(100-45)
19,33 == e k*55
taking ln
ln (19.33)= 55k, k=0.053848 /s
d)
For a second order reaction 1/CA= 1/CAO+ kt (1)
Where CA= Concentration at time t, CAO= initial concentration k= rate constant and t is the time
Given CA =0.15M after 185 seconds and 5.1*10-2M after 875 seconds
Writing Eq.1 twice for given data
1/0.15= 1/CAO+ k*185 (2) 6.7 =1/CAO+ k*185 (2A)
1/5.1*10-2= 1/CAO+ K*875 (3) 196.1= 1/CAO+ k*875 (3A)
Eq.3A- Eq.2 give 196.1-6.7= k*(875-185)
k=189.4/690=0.2744/M.s
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