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Alexxandr [17]
3 years ago
10

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
Chemistry
1 answer:
zubka84 [21]3 years ago
7 0

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}\times\text{exp}\left (\frac{-Ea}{RT} \right )      ..........(1)

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

ln\frac{D_{1}}{D_{2}}=\frac{Ea}{R}\left [\frac{1}{T_{2}}-\frac{1}{T_{1}} \right ]  ........(2)

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

ln\frac{1.23\times 10^{-16}}{1.45\times 10^{-14}}=\frac{Ea}{8.314}\left [\frac{1}{2073}-\frac{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}\times\text{exp}\left (\frac{-Ea}{RT} \right )

now put all the values in formula,we get

1.23\times10^{-16}m^{2}/s =D_{0}\times\text{exp}\left (\frac{-201820J/mol} {8.314J/mol/K\times1473K} \right )

Rearranging the terms, we get the value of Pre-exponential constant, D_{0} as

D_{0} = 1.7658 × 10^{-9} m^{2}/s







           

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