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 $60 {\mathrm{s}}^{-1}$. 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:

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

The slope of the given below graph is -

 $1.$ $\frac{\mathrm{k}}{2.303}$
$$
$2.$ $-\frac{\mathrm{k}}{2.303}$
$$
$3.$ $-\frac{2.303}{\mathrm{k}}$
$$
$4.$ $\frac{2.303}{\mathrm{k}}$

The nature of the reaction represented in the following graph is:

The graph between lnK and 1/T is given below:

The value of activation energy would be:

The correct statement about X in the below mentioned graph:

For a general reaction A → B, the plot of concentration of A vs time is given below:

The order of the reaction would be- 

1. Zero 

2. First

3. Half

4. Second