If the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)

The reaction $L \rightarrow M$ starts with $10 \, gL^{-1}$. After $30$ and $90$ minutes,$5 \, gL^{-1}$ and $1.25 \, gL^{-1}$ remain,respectively. The order of the reaction is:

The reaction $CH_3COCH_{3(g)} \to C_2H_{4(g)} + H_{2(g)} + CO_{(g)}$ is a first-order reaction. If the initial pressure is $0.40 \ atm$ and the total pressure after $10 \ min$ is $0.50 \ atm$,calculate the rate constant of the reaction. $(\log 3.5 = 0.5441)$

At $27^{\circ} C$ temperature,the time required for $75 \%$ completion of a first order reaction is $20 \ s$. What will be its rate constant?

Calculate the time required for a reactant to decrease in concentration from $100 \%$ to $20 \%$,if the rate constant of the first-order reaction is $0.02303 \ hour^{-1}$. (in $hour$)

In a first order reaction,the concentration of the reactant is reduced to $1/8$ of the initial concentration in $75 \ minutes$. The $t_{1/2}$ of the reaction (in minutes) is $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.