Question

Hydrogen is diffusing through solid nickel. At a temperature of 358 K, the diffusivity is 1.16 x 10-8 cm2/s. At a temperature of 438 K, the diffusivity is 1.05 x 10-7 cm2/s. What is the activation energy in J/mol?

Answer 1

Answer:

Q_d=35881 J/mol

So the activation energy is 35881 J/mol.

Explanation:

Consider the following equations:

$lnD_{1} =lnD_{o} -\frac{Q_{d} }{R*T_{1} }$

$lnD_{2} =lnD_{o} -\frac{Q_{d} }{R*T_{2} }$

Solving the above two equation to find the Q_d in term of diffusivity and temperature we will get:

$Q_{d}=-R*\frac{lnD_{1} -lnD_{2} }{\frac{1}{T_{1} }-\frac{1}{T_{2} } }$

where:

Q_d is the activation energy

D_1 is the diffusivity at T_1

D_2 is the diffusivity at T_2

$Q_{d}=-8.31*\frac{ln1.16*10^-12 -ln1.05*10^-11 }{\frac{1}{358 }-\frac{1}{438 } }$

Q_d=35881 J/mol

So the activation energy is 35881 J/mol.