Answer:
27.3 kJ/mol
Explanation:
You would use the Arrhenius Equation to solve this problem.
[tex]ln(\frac{k_{2} }{k_{1} } )=\frac{E_{a} }{R} (\frac{1}{T_{1} } -\frac{1}{T_{2} } )[/tex]
Plug the rate constants into k1 and k2. Plug the temperatures into T1 and T2. R is a constant and is equal to 8.314. Solve for Ea.
Your answer should be 27,303.03 J/mol or 27.3 kJ/mol.
Hope this helps! <3
The activation energy will be equal to "12.40 kJ/mol". To understand the calculation, check below.
According to the question,
For 1st order, rate constant = 0.060 s⁻¹
Temperature, T₁ = 298 K
T₂ = 331 K
Increased rate constant = 0.18s⁻¹
We know the formula,
→ ln k = ln A - [tex]\frac{E_a}{RT}[/tex]
By substituting the rate constant value,
ln 0.06 = ln A - [tex]\frac{E_a}{298 \ R}[/tex] ...(equation 1)
ln 0.18 = ln A - [tex]\frac{E_a}{318 \ R}[/tex] ...(equation 2)
From "equation 1 and 2", we get
Activation energy, 1.09 = [tex]\frac{83 \ E_a}{113538 \ R}[/tex]
[tex]E_a[/tex] = 12.40 kJ/mol
Thus the above answer is correct.
Find out more information about activation energy here:
https://brainly.com/question/24749252
