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

1.	is halved	2.	is doubled
3.	is tripled	4.	remains unchanged

A reaction having equal energies of activation for forward and reverse reaction has:

1. ΔG = 0

2. ΔH = 0

3. ΔH = ΔG = ΔS = 0

4. ΔS = 0

In a reaction, A + B  → Product, the rate is doubled when the concentration of B is doubled, and the rate increases by a factor of 8, when the concentrations of both the reactants (A and B) are doubled. The rate law for the reaction can be written as:

1. Rate = k[A][B]²

2. Rate = k[A]²[B]²

3. Rate = k[A][B]

4. Rate = k[A]²[B]

In a zero-order reaction for every 10 ° rise of temperature, the rate is doubled.
If the temperature is increased from 10 °C to 100 °C,
the rate of the reaction will become:

1. 256 times

2. 512 times

3. 64 times

4. 128 times

The incorrect statement regarding the order of reaction is:

During the kinetic study of the reaction, 2A + B\( \rightarrow\)C + D, following results were obtained:
 

Based on the above data which one of the following is correct?

1. rate= k[A]²[B]

2. rate= k[A][B]

3. rate= k[A]²[B]²

4. rate= k[A][B]²

The half-life period of a first-order reaction is 1386 s. The specific rate constant of the reaction is:

1. $5.0\times {10}^{-3}{s}^{-1}$

2. $0.5\times {10}^{-2}{s}^{-1}$

3. $0.5\times {10}^{-3}{s}^{-1}$

4. $5.0\times {10}^{-2}{s}^{-1}$

For the reaction, \(\mathrm{N}_2+3 \mathrm{H}_2 \rightarrow 2 \mathrm{NH}_3,\) if, \(\frac{d[NH_{3}]}{dt} \ = \ 2\times 10^{-4} \ mol \ L^{-1} \ s^{-1}\), the value of  \(\frac{-d[H_{2}]}{dt}\) would be:

In the reaction, 
${\mathrm{BrO}}_3^{-}\left(\mathrm{aq}\right)+5{\mathrm{Br}}^{-}\left(\mathrm{aq}\right)+6{\mathrm{H}}^{+}\to$ 3Br₂(l)+3H₂O(l) 
The rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ions:

1. $\frac{d\left[B{r}_2\right]}{dt}=-\frac{3}{5}\frac{d\left[B{r}^{-}\right]}{dt}$

2. $\frac{d\left[B{r}_2\right]}{dt}=-\frac{5}{3}\frac{d\left[Br\right]}{dt}$

3. $\frac{d\left[B{r}_2\right]}{dt}=\frac{5}{3}\frac{d\left[B{r}^{-}\right]}{dt}$

4. $\frac{d\left[B{r}_2\right]}{dt}=\frac{3}{5}\frac{d\left[B{r}^{-}\right]}{dt}$