Catalysts are used to increase the rate of a chemical reaction. Because it

What is the value of the slope of the plot $\ln K$ versus $\frac{1}{T}$ for a reaction having $E_{a} = 33.256 \ J \ mol^{-1}$?

The decomposition of formic acid on a gold surface follows first-order kinetics. If the rate constant at $300 \ K$ is $1.0 \times 10^{-3} \ s^{-1}$ and the activation energy $E_a = 11.488 \ kJ \ mol^{-1}$,the rate constant at $200 \ K$ is ............ $\times 10^{-5} \ s^{-1}$.
(Round off to the Nearest Integer).
(Given: $R = 8.314 \ J \ mol^{-1} K^{-1}$)

Thermodynamic feasibility of the reaction alone cannot decide the rate of the reaction. Explain with the help of one example.

For a first-order reaction at $300 \, ^\circ C$,the activation energy is $35 \, kcal \, mol^{-1}$ and the frequency factor is $1.45 \times 10^{11} \, s^{-1}$. Calculate the rate constant.

For reaction $A \rightarrow P$, rate constant $k = 1.5 \times 10^3 \text{ s}^{-1}$ at $27^\circ\text{C}$. If activation energy for the above reaction is $60 \text{ kJ mol}^{-1}$, then the temperature (in $^\circ\text{C}$) at which rate constant $k = 4.5 \times 10^3 \text{ s}^{-1}$ is . . . . . . .