The activation energy of a certain reaction is $87 \ kJ \ mol^{-1}$. When the temperature is decreased from $37^{\circ} C$ to $15^{\circ} C$,what is the ratio of the rate constant at $37^{\circ} C$ to that at $15^{\circ} C$ for this reaction (in $/ 1$)?

The rate constant for the first order decomposition of a certain reaction is described by the equation $\ln k (s^{-1}) = 14.34 - \frac{1.25 \times 10^{4} \ K}{T}$. The energy of activation for this reaction is

Define the following terms:
$1.$ Collision
$2.$ Frequency of collision

For a reaction $E_a = 0$ and $k = 3.2 \times 10^8 \ s^{-1}$ at $300 \ K$. The value of frequency factor at $400 \ K$ would be

The temperature at which the rate constants of the two gaseous reactions given below become equal is . . . . . . $K$. (Nearest integer).
$X \longrightarrow Y \quad k_1 = 10^6 e^{\frac{-30000}{T}}$
$P \longrightarrow Q \quad k_2 = 10^4 e^{\frac{-24000}{T}}$
Given: $\ln 10 = 2.303$

For the following reactions:
$A \xrightarrow{700 \ K}$ Product
$A \xrightarrow[\text{catalyst}]{500 \ K}$ Product
it was found that $E_{a}$ is decreased by $30 \ kJ/mol$ in the presence of a catalyst. If the rate remains unchanged,the activation energy for the catalysed reaction is (Assume pre-exponential factor is same):