The inversion of cane sugar into glucose and fructose according to the equation ${{\rm{C}}_{12}}{{\rm{H}}_{22}}{{\rm{O}}_{11}} + {{\rm{H}}_2}{\rm{O}} \to {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6} + {{\rm{C}}_6}{{\rm{H}}_{12}}{{\rm{O}}_6}$ is an example of

JEE CHEMISTRY MCQ: CHEMICAL KINETICS

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The inversion of cane sugar into glucose and fructose according to the equation

{\mathrm{C}_{12}}{\mathrm{H}_{22}}{\mathrm{O}_{11}} + {\mathrm{H}_2}\mathrm{O} \to {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6} + {\mathrm{C}_6}{\mathrm{H}_{12}}{\mathrm{O}_6}

is an example of

View answer and explanation

Correct option: C

Pseudo unimolecular

Option C (Pseudo unimolecular) is correct because the reaction appears to be first order with respect to cane sugar, as the concentration of water is in large excess and remains nearly constant.

More Questions from CHEMICAL KINETICS

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The reaction

\mathrm{A}_2+\mathrm{B}_2 \rightarrow 2 \mathrm{AB}

follows the mechanism

\begin{aligned} & {{\text{A}}_2}\mathop \rightleftharpoons \limits_{{{\text{k}}_{ - 1}}}^{{{\text{k}}_1}} {\text{A}} + {\text{A (fast) }} \\ & \mathrm{A}+\mathrm{B}_2 \xrightarrow{\mathrm{k}_2} \mathrm{AB}+\mathrm{B} \text { (slow) } \\ & \mathrm{A}+\mathrm{B} \rightarrow \mathrm{AB} \text { (fast) } \end{aligned}

The overall order of the reaction is: For the first order reaction with the rate constant

k

, which expression gives the rate half-life period?

(Initial\mathrm{ }conc.\mathrm{ } = a)

Which of the following is a zero order reaction? For a certain reaction involving a single reactant, it is found that

C_0 \sqrt{T}

is constant, where

C_0

is the initial concentration of the reactant and

T

is the half-life. What is the order of the reaction? Arrhenius equation is The energy of activation for an uncatalysed reaction is

100 \mathrm{~kJ/mol}.

Presence of a catalyst lowers the energy of activation by

75 \%

. The

\log _{10} \frac{K_2}{K_1}

of the ratio of rate constant of catalysed and uncatalysed reactions at

27^{\circ} \mathrm{C}

is?

Assume the frequency factor is same for both reactions. (Given

2.303 \times 8.314=19.147

)