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The inversion of cane sugar is represent...

The inversion of cane sugar is represented by
`C_(12)H_(22)O_(11) + H_(2)O rarr C_(6)H_(12)O_(6) + C_(6)H_(12)O_(6)`
It is a reaction of

A

Second order

B

Unimolecular

C

Pseudo unimolecular

D

None of the three

Text Solution

Verified by Experts

The correct Answer is:
C
Inversion of cane sugar is a Psudo unimolecular reaction .
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For the inversion of cane, sugar, C_(12)H_(22)O_(11)+H_(2)OrarrC_(6)H_(12)O_(6)+C_(6)H_(12)O_(6) , the rate constant is 2.12xx10^(-4) L "mol"^(-1) "sec"^(-1) "at" 27^(@)C . The activation energy of the reaction is 1.07xx10^(5) K "mol"^(-1) . What is the rate constant of the reaction at 37^(@)C

Given : 1 L, 2M solution of C_(12)H_(22)O_(11) under go first order reaction. C_(12)H_(22)O_(11)+H_(2)Orarr C_(6)H_(12)O_(6)+C_(6)H_(12)O_(6) {:(,"Angle of rotation ",theta,20^(@),-10^(@),-40^(@)),(,"Time",t("min"),0,10,oo):} If osmotic pressure of solution is pi atm at (300)/(0.821)K temperature after 10 min then calculate the value of (pi)/(10) ?

For the reaction C_(12)H_(22)O_(11)+H_(2)O overset(H^(+))to C_(6)H_(12)O_(6)+C_(6)H_(12)O_(6) , write (i) Rate of reaction expression (ii) rate law equation (iii) molecularity (iv) order of reaction

The conversion of sugar C_(12)H_(22)O_(11) to CO_(2) is :

ERRORLESS-CHEMICAL KINETICS-NCERT BASED QUESTIONS (Rate Law and Rate Constant)
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