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Chapter 57: Problem 16

What is the mathematical relationship between \(E_{\mathrm{a}}(\) forward \(), E_{\mathrm{a}}(\) reverse \()\), and \(\Delta H ?\)

Short Answer

Expert verified
The mathematical relationship between \(E_{a}(\mathrm{forward})\), \(E_{a}(\mathrm{reverse})\), and \Delta H is \(E_{a}(\mathrm{forward}) - E_{a}(\mathrm{reverse}) = \Delta H\).

Step by step solution

01

Identify the activation energy of the forward reaction

The activation energy of the forward reaction, denoted as \(E_{a}(\mathrm{forward})\), is the energy barrier that needs to be overcome for the reactants to convert into products. It's usually given a positive value.
02

Identify the activation energy of the reverse reaction

The activation energy of the reverse reaction, denoted as \(E_{a}(\mathrm{reverse})\), is the energy barrier that needs to be overcome for the products to revert back into the reactants. It is also typically given a positive value.
03

Identify the enthalpy change of the reaction

The enthalpy change (\Delta H) is the heat change at constant pressure. \Delta H is positive for endothermic reactions, and negative for exothermic reactions.
04

State the relationship

The relationship between \(E_{a}(\mathrm{forward})\), \(E_{a}(\mathrm{reverse})\), and \Delta H can be represented mathematically as: \(E_{a}(\mathrm{forward}) - E_{a}(\mathrm{reverse}) = \Delta H\). This means that the difference between the activation energy of the forward reaction and the reverse reaction is equal to the enthalpy change of the reaction.

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Most popular questions from this chapter

The bond between the \(\mathrm{N}\) atom and \(\mathrm{Br}\) atom must be weakened or broken for reaction 2 to proceed upon collision of two ONBr molecules. Note that a violent collision between two ONBr molecules is more likely to weaken or break the bonds than a gentle collision. Based on this idea, explain why there will be more collisions that will be effective in weakening or breaking the \(\mathrm{N}-\mathrm{Br}\) bond at high temperature compared to low temperature.

Provide two major reasons that all collisions between molecules might not be effective in causing a reaction.
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