Chemical Kinetics

A first-order reaction has a specific reaction rate of 10^–2 s^–1. How much time will it take for 20 g of the reactant to reduce to 5 g?

• 238.6 second

• 138.6 second

• 346.5 second

• 346.5 second

The equilibrium constants of the following are.



The equilibrium constant (K) of the reaction:

A mixture of 2.3 g formic acid and 4.5 g oxalic acid is treated with conc. H₂SO₄. The evolved gaseous mixture is passed through KOH pellets. Weight (in gram) of the remaining product at STP will be

• 1.4

• 3.0

• 4.4

• 2.8

The correct difference between first and second order reactions is that

• The rate of a first-order reaction does not depend on reactant concentrations; the rate of a second-order reaction does depend on reactant concentrations

• The half-life of a first-order reaction does not depend on [A]₀; the half-life of a second-order reaction does depend on [A]₀

• The rate of a first-order reaction does depend on reactant concentrations; the rate of a second-order reaction does not depend on reactant concentrations

• A first-order reaction can catalyzed; a second-order reaction cannot be catalyzed

When initial concentration of the reactant is doubled, the half-life period of a zero order reaction

• Is halved

• Is doubled

• Remains unchanged

• Is tripled

For the chemical reaction,

2O₃ ⇌ 3O₂

The reaction proceeds as follows

O₃ ⇌ O₂ + O

O + O₃ → 2O₂ (slow)

The rate law expression will be

• r = k' [O₃]²

• r = k' [O₃]²[O₂]^-1

• r = k'[O₃][O₂]

• unpredictable

In the following graph

The slope of line AB give the information of the

• value of $\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303}$

• Value of $\frac{2.303}{{\mathrm{E}}_{\mathrm{a}}}$

• value of $-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303}$

• value of $-\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303 \mathrm{RT}}$

A first-order reaction is 50% completed in 1.26 x 10¹⁴s. How much time would it take for 100% completion?

• 1.26 x 10¹⁵ s

• 2.52 x 10¹⁴ s

• 2.52 x 10²⁸ s

• Infinite

For the chemical reaction, 2O₃ ⇌ 3O₂. The reaction proceeds as follows

O₃ ⇌ O₂ + O (fast)

O + O₃ → 2O₂ (slow)

The rate law expression will be

• r = k'[O₃]²

• r = k'[O₃]²[O₂]^-1

• r = k'[O₃][O₂]

• Unpredictable

Two similar reactions have the same rate constant at 25$°$C, but at 35$°$C, one of the reaction has a higher rate constant than the other. The appropriate reason for this is

• due to effective collisions

• due to different activation energies

• due to different threshold energies

• due to higher population of molecules