A reaction follows the rate law: Rate = k[A][B]². If the initial concentrations of both A and B are doubled, and the temperature remains constant, the initial rate of the reaction will increase by a factor of:

The rate of reaction:

$2{\rm{NO}} + {\rm{C}}{{\rm{l}}_2} o 2{\rm{NOCl}}$ is given by the rate, equation $rate\; = k{\left[ {{\rm{NO}}} \right]^2}\left[ {{\rm{C}}{{\rm{l}}_2}} \right].$ The value of the rate constant can be increased by:

Consider the reaction ${N_2}(g) + 3{H_2}(g) o 2N{H_3}(g)$. The equality relationship between $\frac{{d\left[ {N{H_3}} \right]}}{{dt}}\alpha - \frac{{d\left[ {{H_2}} \right]}}{{dt}}$ is

If the rate of a reaction becomes eight times when the temperature is increased from $30^\circ C$ to $60^\circ C$, what is the activation energy of the reaction? (Given: $R = 8.314 J K^{-1} mol^{-1}$, $\log 2 = 0.3010$)

The rate of chemical reaction (except zero order):

The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10^\circ C$ is

In the reversible reaction

$2N{O_2}\underset{{{k_2}}}{\overset{{{k_1}}}{\leftrightarrows}}{N_2}{O_4}$

The rate of disappearance of $N{O_2}$ is equal to

For a hypothetical reaction

$A + 2B o 3C + D$

$d\left[ C \right]/dt$ is equal to

Rate constant of a reaction depends upon

For the non-equilibrium process, $A + B o products$, the rate is first order with respect to $A$ and second order respect to $B$. If $1.0\;mole$ each of $A$ and $B$ are introduced into a 1 litre vessel and the initial rate was $1.0 imes {10^{ - 2}}\;{\rm{mol}}/{\rm{litre}}\;\;{\rm{sec}}{\rm{.}}$ The rate (in ${\rm{mol}}\;{\rm{litr}}{{\rm{e}}^{ - 1}}{\sec ^{ - 1}}$ ) when half of the reactants have been used:

As the reaction progresses, the rate of reaction