Question

The diffusion coefficients for copper diffusing in gold are given at two temperatures (hint: logarithm):

T(K)	D(m2/s)
1273	9.4 x 10-16
1473	2.4 x 10-14

(a) Determine the values of D₀ and the activation energy Q_d?

(b) What is the magnitude of D at 1100°C?

Answer 1

(a) The formulae for the temperature dependence of diffusion coefficients for copper diffusing in gold is

RT = Qd / (lnD₀ - lnD)

R = universal gas constant,

Qd = activation energy

When T = 1273 K and D = 9.4 x 10^-16 m2/s

1273KxR = Qd / [lnD₀ - ln(9.4 x 10^-16 m2/s)] ----- (1)

When T = 1473 K and D = 2.4 x 10^-14 m2/s

1473KxR = Qd / [lnD₀ - ln(2.4 x 10^-14 m2/s)] ----- (2)

By dividing equation(1) and (2) we get

1273 / 1473 = [lnD₀ - ln(2.4 x 10^-14 m2/s)] / [lnD₀ - ln(9.4 x 10^-16 m2/s)]

By solving the above equation we get

D₀ = 2.17x10^-5 m2/s (answer)

Putting the value of D₀ in equation (1) we get

1273Kx8.314 JK-1mol-1 = Qd / [ln(2.17x10^-5 m2/s) - ln(9.4 x 10^-16 m2/s)]

=> Qd = (1273Kx8.314 JK-1mol-1) x [ln(2.17x10^-5 m2/s) - ln(9.4 x 10^-16 m2/s)] = 252.55 KJ (answer)

(b) T = 1100 C = 1100+273 = 1373 K

Hence 1373Kx8.314 JK-1mol-1 = 252550J / [ln(2.17x10^-5 m2/s) - lnD]

=> D = 5.35x10^-15 m2/s (answer)