An organic compound undergoes first order decomposition. The time taken for its decomposition to $\frac{1}{8}$ and $\frac{1}{10}$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}}$? $[\log 2 = 0.30]$

The half-life of a first order reaction is $30 \, min$. The time required for $75 \, \%$ completion of the same reaction will be $..... \, min$

The half-life period for a first order reaction is $693 \ sec$. The rate constant for this reaction would be: (in $sec^{-1}$)

The decomposition of benzene diazonium chloride is a first order reaction. The time taken for the decomposition of $\frac{1}{4}$ and $\frac{1}{10}$ of its initial concentration are $t_{1/4}$ and $t_{1/10}$ respectively. The value of $\frac{t_{1/4}}{t_{1/10}} \times 100$ is (Given: $\log 2 = 0.3, \log 3 = 0.477$)

$A$ reaction has a half-life of $1 \, \text{min}$. The time required for $99.9 \, \%$ completion of the reaction is ......... $\text{min}$. (Round off to the nearest integer)
[Use: $\ln 2 = 0.69, \ln 10 = 2.3$]

Azo isopropane decomposes according to the equation:
$((CH_3)_2CHN)_2N_{2(g)} \xrightarrow{250 - 290 \ ^oC} N_{2(g)} + C_6H_{14(g)}$
It is found to be a first order reaction. If the initial pressure is $P_o$ and the total pressure of the mixture at time $t$ is $P_t$,then the rate constant $K$ is given by: