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 $(\log 2 = 0.30, \log 3 = 0.47, \log 4 = 0.60)$.

For the gaseous reaction $A(g) \longrightarrow B(g) + C(g) + D(g)$,starting with an initial pressure of $400 \ atm$. If the total pressure after $2 \ hr$ is $800 \ atm$,the value of the rate constant is: (consider the reaction to be $1^{st}$ order): (in $hr^{-1}$)

For a first-order reaction,it takes $40 \ min$ for $90\%$ completion. The half-life of the reaction is ..... (in $min$)

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 decomposition of $N_2O_5$ occurs as $2N_2O_5 \to 4NO_2 + O_2$ and follows first-order kinetics. Hence,

If the rate constant of a first order reaction is $4.606 \times 10^{-3} \ s^{-1}$,then find the time required for $400 \ g$ of the reactant to reduce to $50 \ g$. (in $min$)