Half life of a first order reaction is $900 \ \text{min}$ at $400 \ K$. Find its half life at $300 \ K$. Given: $\frac{E_a}{2.303 \ R} = 1.3056 \times 10^3 \ K$. (in $\text{min}$)

What does $P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}$ indicate in the rate equation?

What is the activation energy for the reverse of this reaction?
$N_2O_{4(g)} \to 2NO_{2(g)}$
Data for the given reaction is:
$\Delta H = +54 \ kJ$ and $E_a = +57.2 \ kJ$
Answer in $kJ$.

The energy profile diagram for a multi-step reaction,$A$ $\xrightarrow{1} B$ $\xrightarrow{2} C$ $\xrightarrow{3} D,$ is given below. The rate-determining step of the reaction is:

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

The rate constant of a reaction at temperature $200 \ K$ is $10$ times less than the rate constant at $400 \ K.$ What is the activation energy $({E_a})$ of the reaction ($R = $ gas constant) (in $R$)?