To solve this problem, let's analyze each statement based on the Arrhenius equation:
[tex]\[
k = A e^{-E_a / RT}
\][/tex]
Where:
- [tex]\( k \)[/tex] is the rate constant of the reaction.
- [tex]\( A \)[/tex] is the frequency factor.
- [tex]\( E_a \)[/tex] is the activation energy.
- [tex]\( R \)[/tex] is the ideal gas constant.
- [tex]\( T \)[/tex] is the absolute temperature.
### Analysis of Each Statement:
1. At constant temperature, reactions with lower activation energies proceed more rapidly.
Let's consider the effect of activation energy [tex]\( E_a \)[/tex] at a constant temperature [tex]\( T \)[/tex].
The term [tex]\( e^{-E_a / RT} \)[/tex] decreases as [tex]\( E_a \)[/tex] increases because the exponent becomes more negative. Consequently, if [tex]\( E_a \)[/tex] is lower, the value of [tex]\( e^{-E_a / RT} \)[/tex] is higher, making [tex]\( k \)[/tex] larger. A larger [tex]\( k \)[/tex] means a faster reaction. So, reactions with lower activation energies indeed proceed more rapidly at constant temperature.
2. At constant temperature, reactions with lower activation energies proceed less rapidly.
This statement is the opposite of the first one. As described, lower activation energies result in higher rate constants, meaning the reaction will proceed more rapidly, not less rapidly.
3. At constant energy of activation, reactions at lower temperatures proceed more rapidly.
If the activation energy [tex]\( E_a \)[/tex] is constant, consider the effect of temperature [tex]\( T \)[/tex]. As [tex]\( T \)[/tex] decreases, the value of [tex]\( RT \)[/tex] decreases, which makes the exponent [tex]\( -E_a / RT \)[/tex] more negative, hence reducing the value of [tex]\( e^{-E_a / RT} \)[/tex]. Thus, the rate constant [tex]\( k \)[/tex] becomes smaller, meaning the reaction proceeds less rapidly at lower temperatures.
4. At constant energy of activation, reactions with smaller values of [tex]\( A \)[/tex] proceed more rapidly.
Here, if [tex]\( E_a \)[/tex] is constant and we vary [tex]\( A \)[/tex], a smaller [tex]\( A \)[/tex] would directly make [tex]\( k \)[/tex] smaller since [tex]\( k = A e^{-E_a / RT} \)[/tex]. A smaller [tex]\( k \)[/tex] means a slower reaction, not a more rapid one.
Based on the above analysis, the correct statement is:
1. At constant temperature, reactions with lower activation energies proceed more rapidly.
Therefore, the true answer is:
```
1
```
