Calculate the half-life of the first-order reaction $C_2H_4O_{(g)} \to CH_{4(g)} + CO_{(g)}$. If the initial pressure of $C_2H_4O_{(g)}$ is $80 \ mm$ and the total pressure at the end of $20 \ minutes$ is $120 \ mm$,find the half-life in $\text{min}$.

For any reaction,if we plot a graph between time $t$ and $\log (a - x)$,a straight line is obtained. The order of reaction is

For a first order reaction,the rate constant is given as $\log_{10} K = 12 - \frac{6 \times 10^3}{T}$. What will be the value of temperature if its half-life period is $6.93 \times 10^{-3} \, \text{min}$ (in $, K$)?

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:

$A ( g ) \rightarrow 2 B ( g ) + C ( g )$ is a first order reaction. The initial pressure of the system was found to be $800 \ mm \ Hg$ which increased to $1600 \ mm \ Hg$ after $10 \ min$. The total pressure of the system after $30 \ min$ will be . . . . . . $mm \ Hg$. (Nearest integer)

What is the half-life of a first-order reaction if the time required to decrease the concentration of the reactant from $0.8 \text{ M}$ to $0.2 \text{ M}$ is $12 \text{ hours}$ (in $\text{ hours}$)?