The half-life of a first-order reaction is 20 minutes. What fraction of the reactant remains after 80 minutes?

Consider the reaction,

$C{l_2}\left( {aq} \right) + {H_2}S\left( {aq} \right) o S\left( s \right) + 2{H^ + }\left( {aq} \right) + 2C{l^ - }\left( {aq} \right)$ the rate equation for this reaction is , $rate = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]$

Which of these mechanisms is/are consistent with this rate equation?

$\begin{array}{l}

A.;;C{l_2} + {H_2}S\left( {aq} \right) o {H^ + }C{l^ - } + C{l^ + } + H{S^ - }\left( {slow} \right)\

;;;;;C{l^ + } + H{S^ - } o {H^ + }C{l^ - } + S\left( {fast} \right)\

B.;;{H_2}S \leftrightarrow {H^ + } + H{S^ - }\left( {fast;equilibrium} \right)\

;;;;C{l_2} + H{S^ - } o 2C{l^ - } + {H^ + }S\left( {slow} \right)

\end{array}$

The enzyme catalysed reaction is faster than metal catalysed reaction because its activation energy is:

The rate of a chemical reaction depends upon:

Rate of reaction

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:

A first-order reaction has a rate constant of $9.21 imes 10^{-3} \, s^{-1}$. How long will it take for the concentration of the reactant to decrease from $1.5 \, M$ to $0.3 \, M$?

For the two gaseous reactions, following data are given,

$A o B;\;{k_1} = {10^{10}}\;{e^{ - 20,000/T}}$

$C o D;\;{k_2} = {10^{12}}\;{e^{ - 24,606/T}}$

The temperature at which ${k_1}$ becomes equal to ${k_2}$ is

If a plot of $\ln k$ versus $\frac{1}{T}$ for a reaction yields a straight line with a slope of -5000 K, what is the activation energy ($E_a$) of the reaction? (R = 8.314 J/mol·K)

The rate of a certain reaction follows the expression $r = k[X]^2[Y]$. If the concentration of X is doubled and the concentration of Y is halved, how will the initial rate change?

Which statement is correct?