The chemical reaction, 2O₃ $\to$3O₂ proceeds as follows;

O₃ $\rightleftharpoons$O₂ + O .....(Fast)

O+O₃ $\to$ 2O₂ ....(Slow)

The rate law expression should be:

The rate of reaction becomes 2 times for every 10°C rise in temperature. How the rate of reaction will increase when temperature is increased from 30°C to 80°C?

The rate constant (K) for the reaction 2A +B $\to$ Product was found to be 2.5x10^-5 litre mol^-1 sec^-1 after 15 sec, 2.60 x10^-5 litre mol^-1sec^-1 after 30 sec and 2.55 x10^-5litre mol^-1 sec^-1 after 50 sec. The order of reaction is:

A reaction A₂ + B₂ $\to$ 2AB occurs by the following mechanism;

A₂ $\to$ A + A               .....(slow)

A + B₂ $\to$ AB + B       .....(fast)

A + B $\to$ AB               .....(fast)

Its order would be:

For a given reaction, presence of catalyst reduces the energy of activation by 2 kcal at 27$°$C. The rate of reaction will be increased by:

What fraction of a reactant showing first order remains after 40 minute if t_1/2 is 20 minute?

Rate equation for a second order reaction is:

If 2M, 1L solution of acetic acid is added to 3M, 1L ethyl alcohol then the following elementary reaction takes place.

CH₃COOH + C₂H₅OH $\to$ CH₃COOC₂H₅ + H₂0

If each solution is diluted by 1 litre then the initial rate becomes how many times:-

For the reaction 2NO₂ + F₂ → 2NO₂F, following

mechanism has been provided,

 NO₂ + F₂   $\stackrel{slow}{\to }$  NO₂F+F

NO₂ + F   $\stackrel{fast}{\to }$ NO₂F

Thus, rate expression of the above

reaction can be written as:

For the reaction:

[Cu(NH₃)₄]²⁺ + H₂O_{$\rightleftharpoons$}[Cu(NH₃)₃H₂O]²⁺ + NH₃

the net rate of reaction at any time is given by, net rate =

2.0x10^-4 [Cu(NH₃)₄]²⁺[H₂O] - 3.0x10⁵ [Cu(NH₃ )₃ H₂0]²⁺[NH₃]

Then correct statement is/are :

Rate constant of reaction can be expressed by Arrhenius equation as,

                             $K=A{e}^{\raisebox{1ex}{$-E_a$}\!\left/ \!\raisebox{-1ex}{$RT$}\right.}$   

In this equation, ${\mathrm{E}}_{\mathrm{a}}$ represents:

Consider the chemical reaction,

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

The rate of this reaction can be expressed in terms of time derivative of concentration of  ${\mathrm{N}}_2 \left(\mathrm{g}\right), {\mathrm{H}}_2\left(\mathrm{g}\right)$ and ${\mathrm{NH}}_3\left(\mathrm{g}\right)$.

The correct relationship amongest the rate expressions is: 

For a first order reaction A$\to$ Product, the initial concentration of A is 0.1 M and after 40 minute it becomes 0.025 M. Calculate the rate of reaction at reactant concentration of 0.01M:

Select the intermediate in the following reaction mechanism:

O₃(g) $\rightleftharpoons$ O₂(g) +O(g)

O(g) +O₃(g) $\to$ 2O₂(g)

A reactant with initial concentration 1.386 mol litre^-1 showing first order change takes 40 minute to become half. If it shows zero order change taking 20 minute to becomes half under the similar conditions, the ratio, K₁/K₀ for first order and zero order kinetics will be:

In a first order reaction, the concentration of the reactant is decreased from 1.0 M to 0.25M in 20 minute. The rate constant of the reaction would be:

The following mechanism has been proposed for the reaction of NO with Br₂ to form NOBr:

NO(g) + Br₂(g) $\rightleftharpoons$ NOBr₂(g)

NOBr₂(g) + NO(g) $\to$2NOBr(g)

If the second step is the rate determining step, the order of the reaction with respect to NO(g) is:

For the reaction $A \stackrel{ }{\to }nB$, at the point of intersection of two curves show, the [B] is can be given by:

The rate constant of a first order reaction is 4x10^-3 sec^-1. At a reactant concentration of 0.02 M, the rate of reaction would be:

If concentration of reactants is increased by 'X', the rate constant K becomes:

In acidic medium the rate of reaction between (BrO₃)^- and Br^- ions is given by the expression

$-\frac{d\left({\mathrm{BrO}}_3^{-}\right)}{dt}=K\left[{\mathrm{BrO}}_3^{-}\right]\left[{\mathrm{Br}}^{-}\right]{\left[{\mathrm{H}}^{+}\right]}^2$

It means:

A graph plotted between log (t) 50% vs. log (a) concentration is a straight line. What conclusion can you draw from the given graph?

For a first-order reaction A$\to$ Products, the rate of reaction at [A] = 0.2 M is 1.0 x 10^-2 mol litre^-1 min^-1. The half-life period for the reaction is-

The rate of a chemical reaction doubles for every 10°C rise of temperature. If the temperature is raised by 50°C, the rate of the reaction increases by about :

Consider the reaction:

Cl₂(aq) + H₂S(aq) → S(s) +2H⁺(aq) +2Cl^-(aq)

The rate equation for this reaction is rate = k[Cl₂][H₂S] Which of these mechanisms is/are consistent with this rate equation?

A. Cl₂ + H₂S → H⁺ + Cl^- +Cl⁺ + HS^- (slow)

cl⁺ + HS^- → H⁺ +Cl^- + S (fast)

B. H₂S $\iff$ H⁺ + HS^- (fast equilibrium)

Cl₂ + HS^- → 2Cl^- + H⁺ + S (slow)

3A $\to$ C + D, For this reaction it is observed that when initial concentration of A is 10 mole/lit then t_1/2 value of this reaction is 40 min and when initial concentration of A is 20 mole / lit, t_1/2 value has been changed into 20 min. Which of the following is true?

In the following reaction : xA →  yB

$\log \left(-\frac{d\left[A\right]}{dt}\right)= \log \left(\frac{d\left[B\right]}{dt}\right) + 0.3$ 

where -ve sign indicates the rate of disappearance of the reactant. Then x : y is -

Half-life is independent of the concentration of reactant. After 10 minutes volume of N₂ gas is 10L and after complete reaction, it is 50 L. Hence rate constant is :

The activation energies of the forward and backward reactions in the case of a chemical reaction are 30.5 and 45.4 KJ/mol respectively. The reaction is

The t_0.5 for the first order reaction.

PCl₅(g) $\to$PCl₃(g) + Cl₂(g) is 20 min. The time in which the conc. of PCl₅ reduces to 25% of the initial conc. is close to