If the concentration is expressed in $mol \ L^{-1}$,the unit of the rate constant for a first-order reaction is........

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:

$A$ first order reaction has a rate constant of $5.78 \times 10^{-5} \ s^{-1}$. What fraction of the reactant will remain after $10 \ hours$?

In a first order reaction,the concentration of reactant decreases from $800 \ mol/dm^3$ to $50 \ mol/dm^3$ in $2 \times 10^2 \ s$. The rate constant of the reaction in $s^{-1}$ is:

Decomposition of $A$ is a first order reaction at $T \ K$ and is given by $A_{(g)} \rightarrow B_{(g)} + C_{(g)}$. In a closed $1 \ L$ vessel, $1 \ bar$ of $A_{(g)}$ is allowed to decompose at $T \ K$. After $100 \ minutes$, the total pressure was $1.5 \ bar$. What is the rate constant (in $min^{-1}$) of the reaction? (Given: $\log 2 = 0.3$)

For a first order reaction,the concentration of reactant decreases from $0.2 \ M$ to $0.1 \ M$ in $100 \ minutes$. What is the rate constant of the reaction?