Test: Rate of a Chemical Reaction - Chemistry MCQ

Since it is an elementary reaction, its rate law r₁= k [A] ³[B]
When the concentrations are changed the new rate will be r₂= k (2[A])³([B]/2) = 4k[A]³[B]
So, r₂=4r₁.

Acid hydrolysis of ester, CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
The order of the reaction may be altered sometimes by taking reactant in excess compared to the other.
The rate law R= k [CH₃COOC₂H₅] [H₂O] however water is present in excess.
So, R= k [CH₃COOC₂H₅].

In any reaction the slowest step is the rate determining step, the rate of the overall reaction depends on this step. So, 2NO₂ → NO + NO₃(k1) is the rate determining step. Therefore the rate law R= k₁[NO₂]².

Given,
k= 3.28 × 10^-4 s^-1
The general formula to find the units for rate constant, k=(mol L^-1)^1-ns^-1 where n is the order of the reaction. The value of n must be 1 for (mol L^-1)^1-ns^-1 to become s^-1. Therefore, k=3.28 × 10^-4s^-1 represents a first order reaction.

Given, R=0.6 s^-1 and k= 0.035
For a first order reaction R= k [A]

Given,
[A] = 0.75 M, k= 0.02
The reaction belongs to pseudo first order reaction so, the unit is s^-1
R= k [A]= 0.02 × 0.75= 0.015 s^-1.

Given, [A] = 0.5 M, [B] = 0.1 M and k= 0.03
From the rate law it is evident that the order of the reaction is 1+ 0.5 = 1.5 = 3/2
Therefore the unit of k= (mol L^-1)^1-1.5 s^-1 = (L/mol)^1/2 s^-1
R= k[A]¹[B]^1/2 = 0.03 × 0.5 × 0.1^1/2 = 4.74 × 10^-2(L/mol)^1/2 s^-1.

the unit of the rate of the reaction is Lmol^-1s^-1.
For a second order reaction, rate=k[A]²
molL⁻¹s⁻¹=k(molL⁻¹)²
k=Lmol⁻¹s⁻¹

The reaction for the inversion of cane sugar is C₁₂H₂₂O₁₁ + H₂O → glucose + fructose.
In this reaction water is present in excess and belongs to a pseudo first order reaction, even though the molecularity is 2 the order of the reaction is 1 so the rate law R=k[C₁₂H₂₂O₁₁].