Question

The diffusion coefficients for species A in metal B are given at two temperatures: T (°C) D (m2/s) 1040 6.22 × 10-17 1280 6.11× 10-16 (a) Determine the value of the activation energy Qd (in J/mol). Entry field with incorrect answer 1.04e5 J/mol (b) Determine the value of D0. Entry field with incorrect answer 3.71e-22 m2/s (Use scientific notation.) (c) What is the magnitude of D at 1160°C? Entry field with incorrect answer 3.69e-18 m2/s (Use scientific notation.)

Answer 1

diffusion coefficient follow as

D= DoEXP(-Qd/RT)

T D VALUE(m2/s)

1040°C= 1040+273K= 1313K    6.22 × 10-17

1280°C= 1280+273K= 1553K 6.11× 10-16

R = 8.314 Jmol-1K-1

From above information ,

we get two equation,

6.22 × 10-17 = Do exp(-Qd/8.314* 1313) ------1

6.11× 10-16 = Do exp(-Qd/ 8.314* 1553)------ 2

dividing Eqn 2/1

6.11× 10-16/6.22 × 10-17 =Do exp(-Qd/8.314* 1313)/Do exp(-Qd/ 8.314* 1553)

0.982315 = exp(-Qd/10916.28)/ exp(-Qd/12911.64)

0.982315=exp( -Qd/10916.28+Qd/12911.64)

ln 0.982315= Qd(1/12911.64-1/10916.28)

-0.017843248 = Qd (0.00007744949-0.00009160629)

-0.017843248 = Qd(-0.0000141568)

so, Qd= -0.017843248/(-0.0000141568)

so, Qd = 1260.40122061 joule /mol

b) value of Do

6.22 × 10-17 = Do exp(-Qd/8.314* 1313)

6.22 × 10-17 = Do exp(-1260.40/10916.28)

6.22*10^-17= Do exp(-0.11546068996)

6.22*10^-17 = Do* 0.89095559504

so, Do = 6.22*10^-17 / 0.89095559504

so, Do= 6.981268* 10^-17

c) D at 1160°C= 1160+273K= 1433K

D= DoEXP(-Qd/RT)

=  6.981268* 10^-17 EXP(1260.40122061/8.314* 1433)

=6.981268* 10^-17 EXP (0.10579194566)

=  6.981268* 10^-17 * 1.1115905812

=7.76* 10^-17