The reaction A → B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. The time taken for conversion of 0.9 mole of A to produce 0.675 mole of B is : 

If the rate of the reaction is equal to the rate constant, the order of the reaction is -

The temperature dependence of rate constant (k) of a chemical reaction is written in terms of Arrhenius equation,
k = A.e^–E*/RT. Activation energy (E*) of the reaction can be calculated by plotting : 

The radioisotope, tritium $\left({}_1{}^3\mathrm{H}\right)$ has a half-life of 12.3 years. If the initial amount of tritium is 32 mg, how many milligrams of its would remain after 49.2 years :

The decomposition of NH₃ on platinum surface is a zero order reaction. The rates of production of N₂ and H₂ if k = 2.5 × 10^–4 mol^–1 L s^–1 will be respectively -

The rate equation of a reaction is expressed as, Rate=K $({\mathrm{P}}_{{\mathrm{CH}}_3{\mathrm{OCH}}_3}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

The factor(s) that affect the rate of a chemical reaction is/are -

The correct statement about the rate constant of a reaction is-

A reaction is first order with respect to A and second-order with respect to B. The concentration of B is increased three times. The new rate of the reaction would  : 

In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

The order of the reaction with respect to A and B would be -

For a reaction, 2A + B → C + D , following observations were recorded:

The rate law applicable for the above mentioned reaction would be -

Given the following observations:

The reaction between A and B is first-order with respect to A and zero-order with respect to B. The value of X and Y are, respectively : 

The rate constant of a radioactive substance is $4 {\mathrm{years}}^{-1}$. The value of half-life will be : 

During a nuclear explosion, one of the products is ⁹⁰Sr with a half-life of 28.1 years. If 1µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, the amount of ⁹⁰Sr that will remain after 10 years in the now grown up child would be -

For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data was obtained:

The rate constant of the above reaction would be -

For the  reaction $2\mathrm{A} + \mathrm{B}\to {\mathrm{A}}_2\mathrm{B}$, rate = $\mathrm{k}\left[\mathrm{A}\right]{\left[\mathrm{B}\right]}^2$ with k = $2.0\times {10}^{-6} {\mathrm{mol}}^{-2}{\mathrm{L}}^2{\mathrm{s}}^{-1}$ and $\left[\mathrm{A}\right]=0.1 \mathrm{M}; \left[\mathrm{B}\right]=0.2 \mathrm{M}$. The initial rate of the reaction will be

For a first-order reaction, the relationship between time required for 99% completion to the time required for the completion of 90% of the reaction would be

$$\begin{gathered} 1. {\mathrm{t}}_{99\%} = 3 {\mathrm{t}}_{90\%} \\ 2. {\mathrm{t}}_{99\% }= 2 {\mathrm{t}}_{90\%} \\ 3. {\mathrm{t}}_{99\% }= 4 {\mathrm{t}}_{90\%} \\ 4. {\mathrm{t}}_{90\%} = 2{\mathrm{t}}_{99\%} \end{gathered}$$

Given that $\mathrm{Rate}=\mathrm{k}\left[{\mathrm{H}}_2{\mathrm{O}}_2\right]\left[{\mathrm{I}}^{-}\right]$. The dimension of the rate constant in the given rate law is -

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-

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

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

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-

The time required for 10% completion of a first-order reaction at 298K is equal to that required for its 25% completion at 308K. The value of ${\mathrm{E}}_{\mathrm{a}}$ is : 

The average rate of reaction based on the above graph will be :

$$\begin{gathered} 1.\frac{\left[{\mathrm{R}}_2\right]-\left[{\mathrm{R}}_1\right]}{{\mathrm{t}}_2-{\mathrm{t}}_1} \\ 2. - \left(\frac{\left[{\mathrm{R}}_2\right]-\left[{\mathrm{R}}_1\right]}{{\mathrm{t}}_2-{\mathrm{t}}_1}\right) \\ 3. \frac{\left[{\mathrm{R}}_2\right]}{{\mathrm{t}}_2} \\ 4. -\left(\frac{\left[{\mathrm{R}}_1\right]-\left[{\mathrm{R}}_2\right]}{{\mathrm{t}}_2-{\mathrm{t}}_1}\right) \end{gathered}$$

Consider the following graph.

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

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

The slope for the given graph is-

Consider the following graph:

The nature of the reaction represented in the above-mentioned graph is : 

The graph between lnK, and $\frac{1}{\mathrm{T}}$is given below:

The value of activation energy would be -

 $$\begin{gathered} 1. 207. 8 \mathrm{kJ}/\mathrm{mol} \\ 2. -207. 8 \mathrm{kJ}/\mathrm{mol} \\ 3. 210.8 \mathrm{kJ}/\mathrm{mol} \\ 4. -210.8 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$