Question

The diffusion coefficients for species A in metal B are given at two temperatures: T (°C)D (m2/s) 1020 7.23 x 10 1290 7.09% 10-16 (a) Determine the value of the activation energy Qd (in נ/mol). J/mol (b) Determine the value of Do m2/s (Use scientific notation.) (c) What is the magnitude of D at 1180°C? m2/s (Use scientific notation.)

Answer 1

First remember the formula that involves diffusion and activation energy, as:

$D = D_o e^{\frac{Q_d}{RT}}$

Now in order to know the activation energy, Qd, we would first have to equal two equations, given that the Do for both should be the same. It will be as follows:

First solve for Do

$D_o = \frac{D}{e^{\frac{Q_d}{RT}}}$

Now make two equations equal as:

$\frac{D_1}{e^{\frac{Q_d}{RT_1}}} = \frac{D_2}{e^{\frac{Q_d}{RT_2}}} \rightarrow \frac{D_1}{D_2} = \frac{e^{\frac{Q_d}{RT_1}}}{e^{\frac{Q_d}{RT_2}}} \rightarrow ln\frac{D_1}{D_2} = ln\frac{e^{\frac{Q_d}{RT_1}}}{e^{\frac{Q_d}{RT_2}}}$

$ln\frac{D_1}{D_2} =\frac{Q_d}{RT_1} - \frac{Q_d}{RT_2} \rightarrow ln\frac{D_1}{D_2} \times (RT_2 \cdot RT_1) = Q_d(RT_2) - Q_d(RT_1)$

$\rightarrow ln\frac{D_1}{D_2} \times (RT_2 \cdot RT_1) = Q_d(RT_2 - RT_1) \rightarrow \frac{ln\frac{D_1}{D_2} \times (RT_2 \cdot RT_1)}{RT_2 - RT_1} = Q_d$

Now substitute in the equation:

$Q_d= \frac{ln\frac{D_1}{D_2} \times (RT_2 \cdot RT_1)}{RT_2 - RT_1} = \frac{ln\frac{7.23 \times 10^{-17}}{7.09 \times 10^{-16}} \times ((8.314 \frac{J}{molK})(1563K) \cdot (8.314 \frac{J}{molK})(1293K))}{(8.314 \frac{J}{molK})(1563K) - (8.314 \frac{J}{molK})(1293K)}$

b) The value of Do can be determined using the new value for activation energy

$D_o = \frac{D}{e^{\frac{Q_d}{RT}}} = \frac{7.23 \times 10^{-17}}{e^{\frac{-142074.34 J/mol}{(8.314\frac{J}{molK})(1293K)}}} = 3.97 \times 10^{-11} m^2 /s$

c) to solve D at 1180°C, do as follows:

$D = D_o e^{\frac{Q_d}{RT}} = (3.97 \times 10^{-11})e^{\frac{-142074.34J/mol}{8.314J/molK(1453K)}} = 3.098 \times 10^{-16} m^2 /s$

**Remember that temperatures must be in Kelvin, so to each celsius measure you have to add 273 to convert them to K.