Very high concentration, where C 1 ≪ k 2

Given: k 1 t + ln C + k 2 C = k 3 At very high concentration: C 1 ≪ k 2 Differentiate the equation with respect to time: k 1 + ( C 1 + k 2 ) d t d C = 0 At high concentration, C 1 is negligible. k 1 + k 2 d t d C = 0 − d t d C = k 2 k 1 = co n s t an t The rate is independent of concentration; therefore: O r d ero f re a c t i o n = 0

Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsExperimental determination of order of reactionMedium2 minQB

ChemistryMediumnumerical

Passage / Comprehension

For a certain reaction, concentration of reactant (C) & time (t) are related as

$k_{1} t + ln C + k_{2} C = k_{3}$

where $k_{1}, k_{2} & k_{3}$ are constants, then what is order of reaction at

Very high concentration, where $\frac{1}{C} ≪ k_{2}$

Answer:

0.00

Solution:

Given:

$k_{1} t + ln C + k_{2} C = k_{3}$

At very high concentration:

$\frac{1}{C} ≪ k_{2}$

Differentiate the equation with respect to time:

$k_{1} + (\frac{1}{C} + k_{2}) \frac{d C}{d t} = 0$

At high concentration, $\frac{1}{C}$ is negligible.

$k_{1} + k_{2} \frac{d C}{d t} = 0$

$- \frac{d C}{d t} = \frac{k _{1}}{k _{2}} = co n s t an t$

The rate is independent of concentration; therefore:

$O r d ero f re a c t i o n = 0$

Stream:JEESubject:ChemistryTopic:Chemical KineticsSubtopic:Experimental determination of order of reaction

⏱ 2mℹ️ Source: QB

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