Chemical Kinetics Questions

Consider an endothermic reaction, $X\to Y$ with the activation energies $E_b$ and $E_f$ for the backward and forward reactions, respectively. In general:

A reaction involving two different reactants can never be:

$t_{1/4}$ is the time taken for the concentration of a reactant to drop to $3/4$ of its initial value. If the rate constant is $K$, $t_{1/4}$ is:

A reaction was found to be second order with respect to the concentration of carbon monoxide. If the concentration of carbon monoxide is doubled, with everything else kept the same, the rate of reaction will:

Rate of a reaction can be expressed by Arrhenius equation as $k=A{e}^{-E_a/RT}$. In this equation, $E_a$ represents:

The following mechanism has been proposed for the reaction of $\text{NO}$ with ${\text{Br}}_2$ to form $\text{NOBr}$:$\text{NO}(g)+{\text{Br}}_2(g)\rightleftharpoons {\text{NOBr}}_2(g)\text{ (fast)}$${\text{NOBr}}_2(g)+\text{NO}(g)\to 2\text{NOBr}(g)\text{ (slow)}$ If the second step is the rate determining step, the order of the reaction with respect to $\text{NO}(g)$ is:

The energies of activation for forward and reverse reactions for $A_2+B_2\rightleftharpoons 2AB$ are $180{\text{kJ mol}}^{-1}$ and $200{\text{kJ mol}}^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100{\text{kJ mol}}^{-1}$. The enthalpy change of the reaction ($A_2+B_2\to 2AB$) in the presence of catalyst will be (in ${\text{kJ mol}}^{-1}$):

Consider the reaction, $2A+B\to \text{Products}$. When concentration of $B$ alone was doubled, the half-life did not change. When the concentration of $A$ alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is:

In a first order reaction, the concentration of the reactant decreases from $0.8\text{M}$ to $0.4\text{M}$ in $15$ minutes. The time taken for the concentration to change from $0.1\text{M}$ to $0.025\text{M}$ is:

The rate equation for the reaction $2\text{A}+\text{B}\to \text{C}$ is found to be: $\text{rate}=k[\text{A}][\text{B}]$. The correct statement in relation to this reaction is that the:

For a reaction $\frac{1}{2}A\to 2B$, the rate of disappearance of ‘A’ is related to the rate of appearance of ‘B’ by the expression:

The half life period of a first order chemical reaction is $6.93\text{minutes}$. The time required for the completion of $99\%$ of the chemical reaction will be ($\log 2=0.301$):

The time for half life period of a certain reaction $\text{A}\to \text{products}$ is $1\text{hour}$. When the initial concentration of the reactant '$\text{A}$ ', is $2.0{\text{mol L}}^{-1}$, how much time does it take for its concentration to come from $0.50$ to $0.25{\text{mol L}}^{-1}$ if it is a zero order reaction?

Consider the reaction:${\text{Cl}}_2(aq)+{\text{H}}_2\text{S}(aq)\to \text{S}(s)+2{\text{H}}^{+}(aq)+2{\text{Cl}}^{-}(aq)$ The rate equation for this reaction is $\text{rate}=k[{\text{Cl}}_2][{\text{H}}_2\text{S}]$.Which of these mechanisms is/are consistent with this rate equation?(A) ${\text{Cl}}_2+{\text{H}}_2\text{S}\to {\text{H}}^{+}+{\text{Cl}}^{-}+{\text{Cl}}^{+}+{\text{HS}}^{-}$ (slow)${\text{Cl}}^{+}+{\text{HS}}^{-}\to {\text{H}}^{+}+{\text{Cl}}^{-}+\text{S}$ (fast)(B) ${\text{H}}_2\text{S}\rightleftharpoons {\text{H}}^{+}+{\text{HS}}^{-}$ (fast equilibrium)${\text{Cl}}_2+{\text{HS}}^{-}\to 2{\text{Cl}}^{-}+{\text{H}}^{+}+\text{S}$ (slow)

The rate of a chemical reaction doubles for every $10^{\circ }\text{C}$ rise of temperature. If the temperature is raised by $50^{\circ }\text{C}$, the rate of the reaction increases by about:

For a first order reaction, $\text{A}\to \text{products}$, the concentration of A changes from $0.1\text{M}$ to $0.025\text{M}$ in $40$ minutes. The rate of reaction when the concentration of A is $0.01\text{M}$ is: