The rate constant for a first order reaction is $60 \text{ s}^{-1}$. How much time (in seconds) will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

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: $\log 2 = 0.3$)

Consider the following first-order gas-phase reaction: $A_{(g)} \to B_{(g)} + C_{(g)}$. At time $t$,the total pressure is $p_t \ atm$. Derive the integrated rate equation for this reaction.

For a first order reaction,the time required for completion of $90 \%$ reaction is '$x$' times the half life of the reaction. The value of '$x$' is $........$. (Given: $\ln 10 = 2.303$ and $\log 2 = 0.3010$)

Which one of the following graphs is not applicable for a $1^{st}$ order reaction $(R \rightarrow P)$?

$t_{1/4}$ can be taken as the time taken for the concentration of a reactant to drop to $3/4$ of its initial value. If the rate constant for a first order reaction is $K$,the $t_{1/4}$ can be written as (in $/K$)