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a first order parallel reaction given as"a-----b and c .given that k1 is rate constant for formation of b and k2 is rate constant for c.k1=1.3 * 10 to power -5 ,sec to power -1.also k1\k2=1\9.find conc.ratio of c to a if an experiment is allowed to start with only a for 1 hr. I HAVE ALREADY ASKED ANS.FOR THIS Q.THE ANS GIVEN IS 0.537.THE ANSWER U GAVE IS 9/11.PLZ GIVE ME A RIGHT SOLUTION.
Hi Aditya,
Hi, A---->B k1 A---->C k2 d[B]/dt = k1*[A] d[B] = k1*[A] dt d[C]/dt = k2*[A] d[C] = k2*[A] dt therefore [B]/[C] = k1/k2 = 1/9 [A]0 = [A]+[B]+[C] [A]=[A]0e-(k1+k2)t [A]=[A]0e-0.468 [A]0 = [A]e0.468 [A]0 = [A]+[B]+[C] [A]e0.468= [A]+[B]+[C] [A]*1.656= [A]+[B]+[C] [A]*0.656=[B]+[C] 0.656*[A]/[C] = [B]/[C] + 1 = 1/9 +1 = 11/9 [C]/[A]=0.656*9/11=0.537
Hi,
A---->B k1
A---->C k2
d[B]/dt = k1*[A] d[B] = k1*[A] dt
d[C]/dt = k2*[A] d[C] = k2*[A] dt
therefore [B]/[C] = k1/k2 = 1/9
[A]0 = [A]+[B]+[C]
[A]=[A]0e-(k1+k2)t
[A]=[A]0e-0.468 [A]0 = [A]e0.468 [A]0 = [A]+[B]+[C] [A]e0.468= [A]+[B]+[C] [A]*1.656= [A]+[B]+[C] [A]*0.656=[B]+[C] 0.656*[A]/[C] = [B]/[C] + 1 = 1/9 +1 = 11/9 [C]/[A]=0.656*9/11=0.537
[A]=[A]0e-0.468
[A]0 = [A]e0.468
[A]e0.468= [A]+[B]+[C]
[A]*1.656= [A]+[B]+[C]
[A]*0.656=[B]+[C]
0.656*[A]/[C] = [B]/[C] + 1 = 1/9 +1 = 11/9
[C]/[A]=0.656*9/11=0.537
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