Home
Class 12
CHEMISTRY
An organic compound (A) decomposes accor...

An organic compound (A) decomposes according to two parallel first order mechanism.
`(k_(1))/(k_(2))=1.303 and k_(2) =2hr^(-1)`
Calculate the ratio of concentration of C to A, if an experiment is allowed to start with only A for one hour?

Text Solution

Verified by Experts


`(k_(1))/(k_(2)) =1.303 k_(2) =2hr^(-1)`
sol. `([B])/([C])=(k_(1))/(k_(2))=1.303" where "A_(t)=` concentration after 1hr.
`(-d[A])/(dt) -[k_(1)+k_(2)][A]`
`(k_(1)+k_(2))=(2.303)/(t)log_(10)""[([A_(t)]+[B]+[C])/([A_(t)])]`
`2.303 k_(2) =(2.303)/(1)log_(10)""[1+(2.303 [C])/([A_(t)])]`
`k_(2) =log_(10) [1+(2.303[C])/(A_(t))]`
`10^(2) =1+(2.303 [C])/(A_(t))`
So `(2.303 [C])/([A_(t)])=99," "([C])/([A_(t)])=(99)/(2.303)`
So `([C])/([A_(t)])=42.987 +43`.
Promotional Banner

Similar Questions

Explore conceptually related problems

For the first-order reaction (C=C_(0)e^(-k_(1)^(t))) and T_(av)=k_(1)^(-1) . After two average lives concentration of the reactant is reduced to :

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as , k_(1)=1.26xx10^(-4)" sec"^(-1) and k_(2)=3.6xx10^(-5)" sec"^(-1) Calculate the % distribution of B & C.

A substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as : k_(1) =1.26xx10^(-4)s^(-1), k_(2) =3.8xx10^(-5) s^(-1) The percentage distribution of B and C are :

For the two step reaction in which both the steps are first order R overset(k_(1))rarr I I overset(k_(2))rarr P the rate of change of concentration of I is given by the plot

For the consecutive unimolecular first order reaction A overset(k_(1))rarr R overset(k_(2))rarr S , the concentration of component A , C_(A) at any time t is given by

The substance undergoes first order decomposition.The decomposition follows two parallel first order reactons as : K_1=10^(-2) sec^(-1) and K_2=10^(-2)sec^(-1) If the corresponding activation energies of parallel reaction are 100 and 120 kJ "mol"^(-1) then the net activation energy of A is/are:

The substance undergoes first order decomposition. The decomposition follows two parallel first order reactions as : K_(1)=1.26xx10^(-4) sec^(-1) and K_(2) =3.8xx10^(-5) sec^(-1) The percentage distribution of B and C