At temperature $T \ K$,the rate constant of a reaction is $1/10$th of the rate constant at temperature $2T \ K$. What will be the activation energy of this reaction (in $RT$)?

The rate constant of a reaction is $2 \times 10^{-3} \ min^{-1}$ at $300 \ K$. When the temperature is increased by $20 \ K$,the rate constant becomes three times its initial value. Calculate the activation energy $(E_a)$ of the reaction. Also,determine the rate constant at $310 \ K$.

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} e^{-(25000/RT)} \, s^{-1}$. It means

The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.