Which statement best describes the relationship between activation energy and reaction rate?

Which increases on increase of temperature?

Which increases on increase of temperature?

For a reaction with an activation energy ($E_a$) of 50 kJ/mol, how much would a 10°C increase in temperature increase the rate constant, according to the Arrhenius equation (assuming $ext{R} = 8.314 ext{ J mol}^{-1} ext{K}^{-1}$) at room temperature (approx. 300K)?

The slope of Arrhenius plot $\left( {\ln \,k\,{\rm{v/s}}\frac{1}{T}} \right)$ of first order reaction is $- 5 imes {10^3}K.$ The value of ${E_a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314{\rm{ J}}{{\rm{K}}^{ - 1}}{\rm{ mo}}{{\rm{l}}^{ - 1}}$]

The slope of Arrhenius plot $\left( {\ln \,k\,{\rm{v/s}}\frac{1}{T}} \right)$ of first order reaction is $- 5 imes {10^3}K.$ The value of ${E_a}$ of the reaction is. Choose the correct option for your answer. [Given $R = 8.314{\rm{ J}}{{\rm{K}}^{ - 1}}{\rm{ mo}}{{\rm{l}}^{ - 1}}$]

If the slope of an Arrhenius plot for a particular reaction is $-6.5 imes 10^3 \ K$, determine the activation energy ($E_a$) for this reaction, given that the gas constant $R$ is $8.314 \ \rm{J} \ {\rm{K}}^{-1} \rm{mol}^{-1}$.

The pre-exponential factor (A) in the Arrhenius equation is related to the:

In an Arrhenius plot for a reaction, the slope is found to be $-12 imes 10^3 K$. If the gas constant $R$ is $8.314 \ \rm{J} \ {\rm{K}}^{-1} \rm{mol}^{-1}$, what is the activation energy ($E_a$) of the reaction in $\rm{kJ/mol}$?

The slope of the Arrhenius plot $\left( \ln k \ \rm{v/s} \frac{1}{T} \right)$ of a first-order reaction is $-7 imes 10^3 \ K$. The value of $E_a$ of the reaction is (Given $R = 8.314 \ \rm{J} \ {\rm{K}}^{-1} \rm{mol}^{-1}$). Choose the correct option.

For a second-order reaction, the slope of the Arrhenius plot $\left( \ln k \ \rm{v/s} \frac{1}{T} \right)$ is determined to be $-10 imes 10^3 \ K$. Calculate the activation energy ($E_a$) for this reaction. (Use $R = 8.314 \ \rm{J} \ {\rm{K}}^{-1} \rm{mol}^{-1}$).