For a reaction $A \longrightarrow P$,a plot between $[A]_0$ $Vs$ $\frac{1}{t_{1/2}}$ is a straight line having a positive slope. When the initial concentration is $1 \times 10^{-2} \ M$,its half-life period is found to be $20 \ min$. When the concentration of $A$ is $2 \times 10^{-2} \ M$,then the half-life will be (in $min$)?

For a second-order reaction where the initial concentrations of both reactants are equal,it takes $3000 \ s$ for the reaction to be $60\%$ complete. How much time (in $s$) will it take for the reaction to be $20\%$ complete?

The mechanism for the reaction is given below:
$2P + Q \to S + T$
$P + Q \to R + S$ (slow)
$P + R \to T$ (fast)
The rate law expression for the reaction is:

The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:

For a particular reaction,the variation of the rate constant with temperature is given by $\ln k_t = \ln k_0 + \left( \frac{\ln (2.5)}{10} \right) \times t$ $(t \geqslant 0 \, ^{\circ}C)$. Here,$k_0$ is the rate constant at $0 \, ^{\circ}C$ and $k_t$ is the rate constant at $t \, ^{\circ}C$. The temperature coefficient of the reaction,assuming that the rate constant increases by the same factor for every $10 \, ^{\circ}C$ rise in temperature,is:

For the reaction $NO_2 + CO \rightarrow CO_2 + NO$,the rate law is given as $\text{Rate} = K [NO_2]^2$. What is the number of $CO$ molecules participating in the slow step?