$A$ first order reaction has a rate constant of $1 \times 10^{-2} \ s^{-1}$. How much time will it take for $20 \ g$ of reactant to reduce to $5 \ g$ (in $s$)?

For a first order reaction $A \rightarrow P$,$t_{1/2}$ (half-life) is $10 \text{ days}$. The time required for $1/4^{th}$ conversion of $A$ (in days) is:

$PCl_{5(g)} \rightarrow PCl_{3(g)} + Cl_{2(g)}$
In the above first order reaction,the concentration of $PCl_{5}$ reduces from an initial concentration of $50 \ mol \ L^{-1}$ to $10 \ mol \ L^{-1}$ in $120 \ minutes$ at $300 \ K$. The rate constant for the reaction at $300 \ K$ is $X \times 10^{-2} \ min^{-1}$. The value of $X$ is $......$
$[$ Given $\log 5 = 0.6989 ]$

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

Consider the following two first order reactions occurring at $298 \ K$ with same initial concentration of $A$: $(1)$ $A \rightarrow B$; rate constant,$k=0.693 \ min^{-1}$ $(2)$ $A \rightarrow C$; half-life,$t_{1/2}=0.693 \ min$. Choose the correct option.

In the reaction $2N_2O_5 \to 4NO_2 + O_2$,the initial pressure is $500 \ atm$ and the rate constant $K$ is $3.38 \times 10^{-5} \ s^{-1}$. After $10 \ minutes$,the final pressure of $N_2O_5$ is ........ $atm$.