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:

1. 3.47x10^-4 M min^-1

2. 3.47x10^-5 M min^-1

3. 1.735 x 10^-6 M min^-1

4. 1.735 x10^-4 M min^-1

Select the intermediate in the following reaction mechanism:

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

O(g) +O₃(g) $\to$ 2O₂(g)

1. O₃(g)

2. O(g)

3. O₂(g)

4. none of these

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

1. 0.5 mol^-1 litre

2. 1.0 mol/litre

3. 1.5 mol/litre

4. 2.0 mol^-1 litre

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:

1. 10min^-1

2. 6.931 min^-1

3. 0.6931 min^-1

4. 0.06931 min^-1

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) will be:

1. 1

2. 0

3. 3

4. 2

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

1. $\frac{n{A}_0}{2}$

2. $\frac{A_0}{n-1}$$$

3. $\frac{n{A}_0}{n+1}$

4. $\left[\frac{n-1}{n+1}\right]A_0$

The rate constant of a first-order reaction is\(4 \times 10^{-3} \mathrm{sec}^{-1}.\) At a reactant concentration of \(0.02~\mathrm{M},\) the rate of reaction would be:

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

1. e^K/X

2. K/X

3. K

4. X/K

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?

1. n=1, t_1/2 = 1/K.a

2. n=2, t_1/2 = 1/a

3. n=1, t_1/2 = 0.693/K

4. None of the above