Question

The diffusivity of ni in mgo is 1.23 x 10-16 m2/s at 1200˚c and 1.45 x 10-14 m2/s at 1800˚c. calculate the activation energy and the pre-exponential constant

Answer 1

Answer : The Activation energy, Ea = 201.820 KJ/mol

The Pre-exponential constant,
`D_(0)` = 1.7658 ×
`10^(-9)`
`m^(2)/s`

Solution : Given,

Diffusivity of Ni at
`1200^(0)C` ,
`D_(1)` = 1.23 ×
`10^(-16)`
`m^(2)/s`

Diffusivity of Ni at
`1800^(0)C` ,
`D_(1)` = 1.45 ×
`10^(-14)`
`m^(2)/s`

Temperature,
`T_(1)` =
`1200^(0)C` = 1200 + 273 = 1473 K

Temperature,
`T_(2)` =
`1800^(0)C` = 1800 + 273 = 2073 K

Value of R = 8.314 J/mol/K

Formula used :

`D=D_(0)*\text{exp}\left ((-Ea)/(RT) \right )` ..........(1)

This formula convert into logarithm term for the calculation of activation energy. So the formula is,

`ln(D_(1))/(D_(2))=(Ea)/(R)\left [(1)/(T_(2))-(1)/(T_(1)) \right ]` ........(2)

Now put all the values in above formula (2), we get

`ln(1.23* 10^(-16))/(1.45* 10^(-14))=(Ea)/(8.314)\left [(1)/(2073)-(1)/(1473) \right ]`

Rearranging the terms, we get the value of Activation energy, (Ea) as

Ea = 201.820 KJ/mol

Now, we have to calculate the value of Pre-exponential constant,
`D_(0)` bye using formula (1)

`D=D_(0)*\text{exp}\left ((-Ea)/(RT) \right )`

now put all the values in formula,we get

`1.23*10^(-16)m^(2)/s =D_(0)*\text{exp}\left (\frac{-201820J/mol} {8.314J/mol/K*1473K} \right )`

Rearranging the terms, we get the value of Pre-exponential constant,
`D_(0)` as

`D_(0)` = 1.7658 ×
`10^(-9) m^(2)/s`