The accompanying graph shows plots of ln k versus 1>T for two different reactions. The plots have been extrapolated to the y-intercepts. Which reaction (red or blue) has (a) the larger value for Ea,

Verified step by step guidance

1. Identify the Arrhenius equation in its linear form: \( \ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A \).
2. Recognize that the slope of the line in the plot of \( \ln k \) versus \( \frac{1}{T} \) is equal to \( -\frac{E_a}{R} \).
3. Compare the slopes of the two lines (Reaction 1 and Reaction 2) on the graph. The steeper the slope, the larger the magnitude of \( -\frac{E_a}{R} \), and thus the larger the activation energy \( E_a \).
4. Observe that Reaction 2 has a steeper slope compared to Reaction 1.
5. Conclude that Reaction 2 has the larger activation energy \( E_a \) compared to Reaction 1.

Key Concepts

Arrhenius Equation

The Arrhenius equation describes the temperature dependence of reaction rates, expressed as k = A * e^(-Ea/RT), where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin. This equation shows that as temperature increases, the rate constant k increases, indicating a faster reaction.

Activation Energy (Ea)

Activation energy (Ea) is the minimum energy required for a chemical reaction to occur. It represents the energy barrier that reactants must overcome to form products. In the context of the Arrhenius equation, a higher Ea results in a steeper slope in the ln k vs. 1/T plot, indicating that the reaction is less sensitive to temperature changes.

Slope of the ln k vs. 1/T Plot

In a plot of ln k versus 1/T, the slope is equal to -Ea/R, where R is the gas constant. A steeper slope indicates a larger activation energy, meaning that the reaction requires more energy to proceed. By comparing the slopes of the two reactions in the graph, one can determine which reaction has a higher activation energy.