TeXes Science Practice Test

In the context of reaction rate, what does the slope of a graph showing Ln k versus 1/T represent?

1. R/Es

2. -Ea/R

3. k/R

4. A/R

The correct answer is: -Ea/R

The slope of a graph plotting the natural logarithm of the rate constant \( k \) against the inverse of temperature \( 1/T \) represents the negative value of the activation energy \( E_a \) divided by the universal gas constant \( R \). This relationship is derived from the Arrhenius equation, which describes how the rate constant \( k \) of a chemical reaction depends on temperature. The Arrhenius equation is commonly expressed in the form: \[ k = A e^{-E_a/(RT)} \] Taking the natural logarithm of both sides gives: \[ \ln k = \ln A - \frac{E_a}{R} \cdot \frac{1}{T} \] From this, it becomes evident that when \( \ln k \) is plotted against \( 1/T \), the slope of the resulting linear graph is \(-\frac{E_a}{R}\). This means that the slope indicates how changes in temperature influence the rate constant, with a steeper slope signifying a higher activation energy for the reaction. Understanding this relationship is crucial in physical chemistry, as it allows researchers to infer the activation energy needed for a reaction to proceed based on experimental data.