For an exothermic reaction $A \rightarrow B$,the activation energy of $A$ is $17 \, kJ/mol$. The heat of reaction is $40 \, kJ$. Calculate the activation energy for the reverse reaction $B \rightarrow A$.

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

For a chemical reaction,if the temperature is increased from $25\,^{\circ}C$ to $55\,^{\circ}C$,the rate of reaction will change by a factor of (Assume $\mu = 3$).

For a first-order reaction $A \rightarrow P$,the rate constant $k$ depends on temperature $(T)$ according to the equation $\log \, k = - \frac{2000}{T} + 0.6$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

The rate constant of a reaction depends upon

Explain why proper orientation is required for any chemical reaction to occur.