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The diffusion coefficients for species A in metal B are given at two temperatures:
T (°C) D (m2/s) 1030
6.69 × 10-17 1290
6.56× 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 1180°C_________m2/s .

Respuesta :

ElennaF

Answer:

a) 149 kJ/mol, b) 6.11*10^-11 m^2/s ,c) 2.76*10^-16 m^2/s

Explanation:

Diffusion is governed by Arrhenius equation

[tex]D = D_0e^{\frac{-Q_d}{RT} }[/tex]

I will be using R in the equation instead of k_b as the problem asks for molar activation energy

I will be using

[tex]R = 8.314\ J/mol*K[/tex]

and

°C + 273 = K

here, adjust your precision as neccessary

Since we got 2 difusion coefficients at 2 temperatures alredy, we can simply turn these into 2 linear equations to solve for a) and b) simply by taking logarithm

So:

[tex]ln(6.69*10^{-17})=ln(D_0) -\frac{Q_d}{R*(1030+273)}[/tex]

and

[tex]ln(6.56*10^{-16}) = ln(D_0) -\frac{Q_d}{R*(1290+273)}[/tex]

You might notice that these equations have the form of  

[tex]d=y-ax[/tex]

You can solve this equation system easily using calculator, and you will eventually get

[tex]D_0 =6.11*10^{-11}\ m^2/s\\ Q_d=1.49 *10^3\ J/mol[/tex]

After you got those 2 parameters, the rest is easy, you can just plug them all   including the given temperature of 1180°C into the Arrhenius equation

[tex]6.11*10^{-11}e^{\frac{149\ 000}{8.143*(1180+273)}[/tex]

And you should get D = 2.76*10^-16 m^/s as an answer for c)

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