During the formation of ammonia by Haber's process N₂ + 3H₂ → 2NH₃, the rate of appearance of NH₃ was measured as 2.5 x 10^-4 mol L^-1 s^-1. The rate of disappearance of H₂ will be: 

1. 2.5 x 10^-4 mol L^-1 s^-1

2. 1.25 x 10^-4 mol L^-1 s^-1

3. 3.75 x 10^-4 mol L^-1 s^-1

4. 15.00 x 10^-4 mol L^-1 s^-1

The correct expression for the 3/4th life of a first-order reaction is:

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

From the plot of log k vs 1/T, following parameter can be calculated:

1. Activation energy.

2. Rate constant of reaction .

3. Order of reaction.

4. Activation energy as well as the frequency factor.

The rate constant for a chemical reaction that takes place at 500 K is expressed as K = A e^-1000. The activation energy of the reaction will be:

1. 100 cal/mol

2. 1000 kcal/mol

3. 10⁴ kcal/mol

4. 10⁶ kcal/mol

A gaseous reaction A₂(g) → B(g) + $\frac{1}{2}C$(g) shows increase in pressure from 100 mm to 120 mm in 5 minutes. The rate of disappearance of A₂ will be : 

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

1. $\frac{2.303}{10}$log 5 min^-1

2. $\frac{2.303}{10}$ log 1.25 min^-1

3. $\frac{2.303}{10}$ log 2 min^-1

4. $\frac{2.303}{10}$ log 4 min^-1

In the following reaction: xA →  yB

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

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

If the rate of reaction doubles when the temperature is raised from 20 °C to 35 °C, then the activation energy for the reaction will be :
(R = 8.314 J mol^-1 K^-1)

1. 342 kJ mol^-1

2. 269 kJ mol^-1

3. 34.7 kJ mol^-1

4. 15.1 kJ mol^-1

For the reaction,

N₂O₅(g) → 2NO₂(g) + \(\frac{1}{2}\)O₂(g)

the value of the rate of disappearance of ${\mathrm{N}}_2{\mathrm{O}}_5$ is given as $6.\mathrm{25 }\times {10}^{-3}$ $mol$ $L^{-1}{s}^{-1}$. The rate of formation of $N{O}_2$ $and$ $O_2$ is given respectively as:

1. 6.25 x 10^-3 mol L^-1s^-1 and 6.25 x 10^-3 mol L^-1s^-1

2. 1.25 x 10^-2 mol L^-1s^-1 and 3.125 x 10^-3 mol L^-1s^-1$\mathrm{}{\mathrm{}}^{}$

3. 6.25 x 10^-3 mol L^-1s^-1 and 3.125 x 10^-3 mol L^-1s^-1

4. 1.25 x 10^-2 mol L^-1s^-1 and 6.25 x 10^-3 mol L^-1s^-1

A first-order reaction is 15 % completed in 20 minutes. The amount of time required to complete 60 % of the reaction is: