The following equation is obtained for a first order reaction at $300 \ K$.
$\log_{10} \frac{k}{A} = 0.00174$
What is the activation energy (in $J \ mol^{-1}$) of the reaction?
$(R = 8.314 \ J \ mol^{-1} \ K^{-1})$

The time required for $10 \%$ completion of a first order reaction at $298 \,K$ is equal to that required for its $25 \%$ completion at $308 \,K$. If the value of $A$ is $4 \times 10^{10} \,s^{-1}$,calculate $k$ at $318 \,K$ and $E_a$.

For a given reaction,the overall enthalpy change is $+100 \ kJ/mol$ and the activation energy of the reverse reaction is $+200 \ kJ/mol$. The activation energy for the forward reaction would be......$kJ/mol$.

The first order rate constant for a certain reaction increases from $1.667 \times 10^{-6} \ s^{-1}$ at $727 \ ^oC$ to $1.667 \times 10^{-4} \ s^{-1}$ at $1571 \ ^oC$. The rate constant at $1150 \ ^oC$,assuming constancy of activation energy over the given temperature range is [Given : $log \ 19.9 = 1.299$ ]

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 $\text{min}$)

In the Arrhenius equation,the fraction of molecules having energy equal to or greater than the activation energy at a given temperature is represented by .....