$t_{1/4}$ 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 $/K$)

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

For a first-order gas-phase reaction $A(g) \to B(g) + C(g)$,let $p_i$ be the initial pressure of gas $A$ and $p_t$ be the total pressure of the reaction mixture at time $t$. The expression for the rate constant $(k)$ is:

$A \rightarrow P$ is a first order reaction. At $300 \ K$,this reaction was started with $[A] = 0.5 \ mol \ L^{-1}$. The rate constant of the reaction was $0.125 \ min^{-1}$. The same reaction was started separately with $[A] = 1 \ mol \ L^{-1}$ at $300 \ K$. The rate constant (in $min^{-1}$) now is:

$A_{(g)} \rightarrow B_{(g)} + C_{(g)}$ is a first order reaction.

Time	$t$	$\infty$
$P_{\text{system}}$	$P_t$	$P_{\infty}$

The reaction was started with reactant $A$ only. Which of the following expressions is correct for the rate constant $k$?

For a first-order reaction,which of the following statements is correct?