The activation energy of a reaction is zero. Its rate constant at $280 \ K$ is $1.6 \times 10^{-6} \ s^{-1}$,the rate constant at $300 \ K$ is

The rate of a reaction $A$ doubles on increasing the temperature from $300 \, K$ to $310 \, K$. By how much should the temperature of reaction $B$ be increased from $300 \, K$ so that its rate doubles,if the activation energy of reaction $B$ is twice that of reaction $A$ (in $, K$)?

The rate coefficient $(k)$ for a particular reaction is $1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$ at $100 \ ^oC$ and $1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$ at $150 \ ^oC$. What is the energy of activation $(E_a)$ (in $kJ \ mol^{-1}$) for this reaction? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

The temperature dependence of the rate constant $(k)$ of a chemical reaction is expressed by the Arrhenius equation,$k = A \cdot e^{-E^*/RT}$. The activation energy $(E^*)$ of the reaction can be calculated by plotting:

What does $P \cdot Z_{AB} \cdot e^{-\frac{E_a}{RT}}$ indicate in the rate equation?

With an increase in temperature,the rate constant $k$: