Initial concentration of reactant in a first order reaction is $0.08 \text{ mol dm}^{-3}$. What concentration would remain after $40 \text{ minutes}$? (Given $\frac{[A]_0}{[A]_t} = 5.00$)

The integrated rate law equation for a first-order gas-phase reaction $A(g) \rightarrow B(g) + C(g)$ is given by (where $P_i$ is the initial pressure and $P_t$ is the total pressure at time $t$):

The rate constant of a first-order reaction is $6 \ min^{-1}$. If the initial concentration of the reactant is $0.5 \ mol \ L^{-1}$,after how many minutes will the concentration of the reactant become $0.05 \ mol \ L^{-1}$?

For a first order reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$ at constant volume and $300 \ K$,the total pressure at the beginning $(t=0)$ and at time $t$ are $P_0$ and $P_t$,respectively. Initially,only $A$ is present with concentration $[A]_0$,and $t_{1/3}$ is the time required for the partial pressure of $A$ to reach $1/3^{rd}$ of its initial value. The correct option$(s)$ is (are) (Assume that all these gases behave as ideal gases)

In a first-order reaction,the concentration of the reactant decreases from $0.80 \, mol \, L^{-1}$ to $0.06 \, mol \, L^{-1}$ in $45 \, min$. Calculate the half-life $(t_{1/2})$. (in $, min$)

The rate constant of the reaction $2 NO_2Cl_{(g)} \longrightarrow 2 NO_{2(g)} + Cl_{2(g)}$ is $4.7672 \text{ minute}^{-1}$. Calculate the half-life of the reaction.