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The chemisorption of H2 on an activated surface becomes 40% faster if temperature is raised from 500 K to 1000 K. Calculate the activation energy of chemisorptions (in KJ/mol). (rounded up to first decimal place)
Correct answer is between '2.6,3.0'. Can you explain this answer?
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The chemisorption of H2 on an activated surface becomes 40% faster if ...
Explanation:
Given:
- Activation energy of chemisorption = ?
- Temperature (initial) = 500 K
- Temperature (final) = 1000 K
- Increase in speed of chemisorption = 40%
Solution:
The rate of a reaction is given by the Arrhenius equation:
k = A exp (-Ea/RT)
where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
We can use the following equation to calculate the activation energy of chemisorption:
ln(k2/k1) = (Ea/R) [(1/T1) - (1/T2)]
where k1 and k2 are the rate constants at initial and final temperatures, respectively.
Let's assume that the rate of chemisorption at 500 K is k1 and the rate at 1000 K is 1.4k1 (40% faster than k1).
ln(1.4k1/k1) = (Ea/R) [(1/500K) - (1/1000K)]
ln(1.4) = (-Ea/R) [(1/500K) - (1/1000K)]
0.3365 = Ea/8.3145 [(1/500K) - (1/1000K)]
Ea = 2.96 kJ/mol
Therefore, the activation energy of chemisorption is 2.96 kJ/mol, which falls between 2.6 and 3.0 kJ/mol as given in the correct answer.
Given:
- Activation energy of chemisorption = ?
- Temperature (initial) = 500 K
- Temperature (final) = 1000 K
- Increase in speed of chemisorption = 40%
Solution:
The rate of a reaction is given by the Arrhenius equation:
k = A exp (-Ea/RT)
where k is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, and T is the temperature in Kelvin.
We can use the following equation to calculate the activation energy of chemisorption:
ln(k2/k1) = (Ea/R) [(1/T1) - (1/T2)]
where k1 and k2 are the rate constants at initial and final temperatures, respectively.
Let's assume that the rate of chemisorption at 500 K is k1 and the rate at 1000 K is 1.4k1 (40% faster than k1).
ln(1.4k1/k1) = (Ea/R) [(1/500K) - (1/1000K)]
ln(1.4) = (-Ea/R) [(1/500K) - (1/1000K)]
0.3365 = Ea/8.3145 [(1/500K) - (1/1000K)]
Ea = 2.96 kJ/mol
Therefore, the activation energy of chemisorption is 2.96 kJ/mol, which falls between 2.6 and 3.0 kJ/mol as given in the correct answer.
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The chemisorption of H2 on an activated surface becomes 40% faster if temperature is raised from 500 K to 1000 K. Calculate the activation energy of chemisorptions (in KJ/mol). (rounded up to first decimal place)Correct answer is between '2.6,3.0'. Can you explain this answer?
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