Question

The diffusion coefficients for species A in metal B are given at two temperatures: T (°C) D (m2/s) 1030 6.11 × 10-17 1250 6.02× 10-16 (a) Determine the value of the activation energy Qd (in J/mol). J/mol (b) Determine the value of D0. m2/s (Use scientific notation.) (c) What is the magnitude of D at 1160°C? m2/s (Use scientific notation.)

Answer 1

Answer:

(a). The value of the activation energy is 171565.80 J/mol.

(b). The value of D₀ is $1.652\times10^{10^{-8}}\ m^2/s$

(c). The magnitude of D at 1160°C is $5.067\times10^{-8}\ m^2/s$

Explanation:

Given that,

Temperatures,

$T_{1}=1030^{\circ}C$

$T_{2}=1250^{\circ}C$

Diffusion coefficients are

$D_{1}=6.11\times10^{-17}$

$D_{2}=6.02\times10^{-16}$

(a). We need to calculate the value of the activation energy

Using relation between diffusion coefficient and temper

$ln(D_{1})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{1}})$....(I)

$ln(D_{2})=ln(D_{0})-(\dfrac{Q_{d}}{RT_{2}})$.....(II)

Divided equation (II) by (I)

$ln(\dfrac{D_{2}}{D_{1}})=\dfrac{Q_{d}}{R}\times(\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}})$

Put the value into the formula

$ln(\dfrac{6.02\times10^{-16}}{6.11\times10^{-17}})=\dfrac{Q_{d}}{8.314}\times(\dfrac{1}{1030+273}-\dfrac{1}{1250+273})$

$Q_{d}=\dfrac{2.2877\times8.314\times1984469}{220}$

$Q_{d}=171565.80\ J/mol$

(b). We need to calculate the value of D₀

Using relation between diffusion coefficient and temperature

$ln(D_{0})=ln(D)+(\dfrac{Q_{d}}{RT_{1}})$

Put the value of  in to the formula

$ln(D_{0})=ln(6.11\times10^{-17})+(\dfrac{171565.80}{8.314\times1030+273})$

$ln(D_{0})=-17.9182$

$D_{0}=e^{-17.9182}$

$D_{0}=1.652\times10^{10^{-8}}\ m^2/s$

Now again put the value in the equation (II)

$ln(D)=ln(1.652\times10^{10^{-8}})-(\dfrac{171565.80}{8.314\times1160+273})$

$ln(D_{0})=-16.79776$

$D_{0}=e^{-16.79776}$

$D_{0}=5.067\times10^{-8}\ m^2/s$

Hence, (a). The value of the activation energy is 171565.80 J/mol.

(b). The value of D₀ is $1.652\times10^{10^{-8}}\ m^2/s$

(c). The magnitude of D at 1160°C is $5.067\times10^{-8}\ m^2/s$