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Chapter 12: Problem 59

The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is \(125 \mathrm{~kJ} / \mathrm{mol}\), and \(\Delta E\) for the reaction is \(-216 \mathrm{~kJ} / \mathrm{mol}\). What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow\right.\) \(\left.\mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)

Short Answer

Expert verified
The activation energy for the reverse reaction is \(-91 \mathrm{~kJ/mol}\).

Step by step solution

01

1. Write down the formula relationship between activation energies and change in reaction energy

The formula relating activation energies for the forward and reverse reactions (\(E_{a(f)}\) and \(E_{a(r)}\), respectively) and the change in energy for the reaction (\(\Delta E\)), is: \[ E_{a(r)} = E_{a(f)} + \Delta E \]
02

2. Plug in the given values

We are given the activation energy for the forward reaction, \(E_{a(f)} = 125 \mathrm{~kJ/mol}\), and the change in energy for the reaction, \(\Delta E = -216 \mathrm{~kJ/mol}\). Plug these values into the formula: \[ E_{a(r)} = 125 \mathrm{~kJ/mol} - 216 \mathrm{~kJ/mol} \]
03

3. Calculate the activation energy for the reverse reaction

Perform the subtraction to find the activation energy for the reverse reaction: \[ E_{a(r)} = -91 \mathrm{~kJ/mol} \] The activation energy for the reverse reaction is \(-91 \mathrm{~kJ/mol}\).

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Most popular questions from this chapter

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