The time for which the oxygen atom remains adsorbed on a tungsten surface is 0.36 at 2550K and 3.49 at 2360K. The activation energy of desorption of oxygen atom is ______kJ/mol. (answer should be an integer)Correct answer is between '595,600'. Can you explain this answer?

Question

Answer

Calculating Activation Energy of Desorption of Oxygen Atom

Given:

Time for which oxygen atom remains adsorbed on tungsten surface at 2550K = 0.36

Time for which oxygen atom remains adsorbed on tungsten surface at 2360K = 3.49

To find: Activation energy of desorption of oxygen atom (in kJ/mol)

Formula:
$$\ln\frac{t_1}{t_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1} ight)$$

where,

$t_1$ and $t_2$ are the times for which the oxygen atom remains adsorbed on tungsten surface at temperatures $T_1$ and $T_2$ respectively

$E_a$ is the activation energy of desorption of oxygen atom

$R$ is the gas constant ($8.314$ J/K/mol)

Solution:

Substituting the given values in the formula:

$$\ln\frac{0.36}{3.49} = \frac{E_a}{8.314}\left(\frac{1}{2360} - \frac{1}{2550} ight)$$

Simplifying the equation:

$$E_a = -8.314 imes \frac{\ln\frac{0.36}{3.49}}{\frac{1}{2360} - \frac{1}{2550}}$$

Calculating the value of $E_a$:

$$E_a \approx 596 ext{ kJ/mol}$$

Answer: The activation energy of desorption of oxygen atom is 596 kJ/mol.

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