Determine the values of rate constants k1 and k2 in the following reactions: (a) A k1 B K2 C (b) A k1 B k2 C (c) A k1 k2 B

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ANSWER:1)(a) Alongrightarrow^(K_(1))C^(K_(2))This is Parallel reaction if Ne assume that both of them are first order, we get-(d[A])/(dt)=(K_(1)+K_(2))[A](b) Ararr^(K_(1))Brarr^(K_(2))CThis is consecutive reaction{:[-(d[A])/(dt)=k_(1)[A]quadk_(1)[A]-k_(2)[B]rarr(ii)],[(d[B])/(dt)=k_(2)[B]],[(d[C])/(dt)=(iii)]:}inte frating e aration (i), we getNow we shall inte frate eauation (ii) and find the concentration of B related to time t.{:[(d[B])/(dt)=K_(1)[A]-K_(2)[B]],[(d[B])/(dt)+K_(2)[B]=K_(1)[A]]:}Substituting [A] as [A]_(0)e^(-k,t)(d[B])/(d)!=k_(2)[B]=k_(1)[A]_(0)e^(-k_(1)t)multiplying both side by e^(-k_(2)t){:[((d[B])/(dt)+k_(2)[B])e^(k_(2)t)=k_(1)[A]_(0)e^((k_(2)-k_(1))t)],[(d)/(dt)([B]e^(k_(2)t))=K_(1)[A]_(0)e^((K_(2)-K_(1))t)],[d([B]e^(k_(2)t))=k_(1)[A]_(0)e^((k_(2)-k_(1))t)dt],[((d[B])/(tt)+k_(2)[B])e^(k_(2)t)=k_(1)^(')[A]_(0)(e^(k)-k_(1))z],[:.(d[c])/(dt)=k_(2)[B]],[=(k_(1)k_(2)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)t)-e^(-k_(2)t)]],[d[c]=(k_(1)k_(2)[A]_(0))/(k_(2)-k_(1))[e^(-k_(1)t)-e^(-k_(2)t)]dt]:}on integrating{:[int d[C]=(k_(1)k_(2)[A]_(0)^(t)t)/((k_(2)-k_(1)))int_(0)^(t)[e^(-k_(1)t)-e^(-k_(2)t)]dt],[[C]=(k_(1)k_(2)[A]_(0))/((k_(2)-k_(1)))[((e^(-k_(1)t))/(-k_(1)))_(0)^(t)-((e^(-k_(2)t))/(-k_(2)))_(0)^(t)]],[[C]=(k_(1)k_(2)[ ... See the full answer