(b) If the diffusivity is observed to double when the temperature is increased by 10 degrees C, what is the activation energy for diffusion? Express your answer in units of eV.
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To find the activation energy for diffusion, we can use the Arrhenius equation.
The Arrhenius equation relates the temperature dependence of a reaction rate constant to the activation energy:
k = A * exp(-Ea/RT)
Where:
- k is the reaction rate constant
- A is the pre-exponential factor
- Ea is the activation energy
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
In this problem, we know that the diffusivity doubles when the temperature is increased by 10 degrees Celsius. We can express this change as a ratio of diffusivity at two different temperatures:
D2/D1 = 2
Taking the natural logarithm of both sides, we have:
ln(D2/D1) = ln(2)
Now, let's use the relationship between diffusivity and the reaction rate constant:
D = k * L^2 / t
Where:
- D is the diffusivity
- k is the reaction rate constant
- L is the characteristic length
- t is time
For simplicity, let's assume that L and t remain constant throughout the temperature change.
Now, let's plug in these values into the Arrhenius equation for two different temperatures, T2 and T1:
ln(D2/D1) = ln(k2/k1)
We know that the difference between T2 and T1 is 10 degrees Celsius, which is equivalent to 10 Kelvin:
T2 = T1 + 10
Substituting this into the Arrhenius equation, we have:
ln(D2/D1) = ln(k2/k1) = (-Ea/R) * (1/T2 - 1/T1)
Since D2/D1 = 2, the equation becomes:
ln(2) = (-Ea/R) * (1/T2 - 1/T1)
To find the activation energy Ea, we can rearrange the equation:
Ea = -R * ln(2) / (1/T2 - 1/T1)
Substituting the values for R, T2, and T1:
Ea = - (8.314 J/(mol·K)) * ln(2) / (1/(T1+10) - 1/T1)
To express the activation energy in electron volts (eV), we can use the conversion factor:
1 eV = 1.60218 x 10^-19 J
So, the final step is to convert the activation energy from Joules to eV by dividing by the conversion factor.
I hope this explanation helps you understand how to determine the activation energy for diffusion based on the given information!