For the reaction 2A + B → A₂B, the rate law is rate = k[A][B]², where k = 2.0 × 10⁻⁶ mol⁻² L² s⁻¹,
[A] = 0.1 M, and [B] = 0.2 M. What is the initial rate of the reaction?

1. 0.04$\times {10}^{-9}$ mol ${\mathrm{L}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$

2. 8$\times {10}^{-8}<mspace\; width="1em"></mspace>{\mathrm{mol}}^{-1}\mathrm{L}<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$

3. 8$\times {10}^{-9}<mspace\; width="1em"></mspace>\mathrm{mol}<mspace\; width="1em"></mspace>{\mathrm{L}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$

4. 8$\times {10}^{-7}$ mol ${\mathrm{L}}^{-1}<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$

What is the ratio of the time required for 99% completion to the time required for
90% completion of a first-order reaction?

What are the dimensions of the rate constant K in the rate law \(\text { Rate }=k\left[H_2 O_2\right]\left[I^{-}\right]\)?
 

Consider the following rate expression.

$\mathrm{Rate}=\mathrm{k}{\left[C{H}_{3}CHO\right]}^{\frac{3}{2}}$

The order of reaction and dimension of the rate constant are, respectively-

1. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$; k = ${\mathrm{mol}}^{-1}{\mathrm{Ls}}^{-1}$

2. 3; k = ${\mathrm{mol}}^{-1}{\mathrm{Ls}}^{-1}$

3. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$;  k = ${\mathrm{mol}}^{\frac{1}{2}}{\mathrm{L}}^{-\frac{1}{2}}{\mathrm{s}}^{-1}$

4. $\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$; k = ${\mathrm{mol}}^{-\frac{1}{2}}{\mathrm{L}}^{\frac{1}{2}}{\mathrm{s}}^{-1}$

If the concentration of the reactant is made twice, the new rate of reaction for the second order reaction would be-

1. 2 times

2. 4 times

3. 3 times

4. No change in the rate of the reaction

A first-order reaction takes 40 min for 30% decomposition. Half life of the reaction is-

1. 55.9 min

2. 77.9 min

3. 63.9 min

4. 80.9 min

The rate constant for a first-order reaction is $60<mspace\; width="1em"></mspace>{\mathrm{s}}^{-1}$. The time required to reduce the initial concentration of the reactant to its 1/16 value is-

A first-order reaction's 10 percent completion time at 298 K is the same as its 25 percent completion time at 308 K. The value of ${\mathrm{E}}_{\mathrm{a}}$ will be:

1. $76.64$ $J$ ${\mathrm{mol}}^{-1}$

2. $66.64$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

3. $76.64$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

4. $70.34$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

Consider the following graph:
      

The instantaneous rate of reaction at t = 600 sec will be:

$1.-\hspace{0.17em}4.75$ $\times {10}^{-4}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$2.$ $5.75\times {10}^{-5}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$3.$ $$ $6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$
$4.$ $-6.75\times {10}^{-6}$ $\mathrm{mol}$ ${\mathrm{L}}^{-1}{\mathrm{s}}^{-1}$