Chemistry, 09.07.2019 04:00 NerdyE

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

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Answer from: Gunn2368

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 × $10^{-9} m^{2}/s$

Answer from: Quest

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Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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