Question

The activation energy of a reaction is zero. The rate constant of the reaction:
A. increases with the increase in temperature
B. decreases with the decrease in temperature
C. decreases with the increase in temperature
D. is nearly independent of temperature

Answer 1

Hint: Activation energy is defined as the amount of energy required to bring out the reaction. Both the terms (activation energy and temperature) are related by the expression, $\text{k=A}{{\text{e}}^{-\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}$. Put the values given in the question to obtain the required relation. Rate of reaction depends on activation energy and temperature.

Complete step by step answer:
Let us first discuss the equation, there is an equation named as Arrhenius equation which shows the dependence of rate of reaction on temperatures. Mathematically, it is represented as $\text{k=A}{{\text{e}}^{-\frac{{{\text{E}}_{\text{a}}}}{\text{RT}}}}$ , k is the rate of reaction, A is Arrhenius constant or collision frequency or collision factor, ${{\text{E}}_{\text{a}}}$ is the activation energy, T is the temperature and R is universal gas constant with a constant value of $8.314\text{ J}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$.
Now, coming back to the question, we are given that the activation energy of a reaction is zero, it means that ${{\text{E}}_{\text{a}}}$ is zero. Now, put ${{\text{E}}_{\text{a}}}=0$, in the formula.
Then, the expression will be $\text{k=A}{{\text{e}}^{-\frac{0}{\text{RT}}}}$. It is clear that ${{\text{e}}^{-\frac{0}{\text{RT}}}}={{\text{e}}^{-0}}$. Any value having zero as its power gives 1 as its answer. Finally, the expression is $\text{k=A}\times \left( 1 \right)$ or $\text{k=A}$. It means that the rate of reaction, when activation energy is zero will have the value equal to the value of the collision frequency not temperature.

The activation energy of a reaction is zero. The rate constant of the reaction is nearly independent of temperature. The correct option is (D).

Additional Information: The important relation between two different rate of reactions $\left( {{\text{k}}_{1}}\text{ and }{{\text{k}}_{2}} \right)$ at two different temperatures $\left( {{\text{T}}_{1}}\text{ and }{{\text{T}}_{2}} \right)$. In this equation, there is no need for a collision constant to find the rates, temperatures and activation energy. The relation is $\text{ln}\left( \frac{{{\text{k}}_{2}}}{{{\text{k}}_{1}}} \right)=-\frac{{{\text{E}}_{\text{a}}}}{\text{R}}\left( \frac{1}{{{\text{T}}_{2}}}-\frac{1}{{{\text{T}}_{1}}} \right)$.

Note: In the real world, there is no reaction which has zero activation energy. This is because when activation energy is zero, then, the rate of reaction equals collision frequency. That means there will be effective collisions between the particles or molecules which is not possible, as the occurrence of a reaction also depends on successful collisions not all collisions. The activation energy is always positive. As it is defined as the amount of energy required to bring out the reaction.