Question 27.3: Assuming that the rate of heterogeneous nucleation of potass......
Assuming that the rate of heterogeneous nucleation of potassium chloride is consistent with an apparent interfacial tension of 2.5 ergs/cm², determine the nucleation rate as a function of s at a temperature of 80°F (300 K).
Use Eq. (27.11). The molecular weight of KCI is 74.56. The density of the crystal is 1.988 g/cm³ • Since KCl dissociates into two ions, \rm{K^+ \ and \ Cl^-, ν = 2}. Then
B^{\circ}=10^{25} \exp \left[-\frac{16 \pi V_M^2 N _a \sigma_a^3}{3(R T)^3 ν^2 s^2}\right] (27.11)
V_M=\frac{74.56}{1.988}=37.51 \ cm ^3 / g \ mol \quad \sigma_a=2.5 \ ergs/ cm ^2
The exponent in Eq. (27.11) is
-\frac{16 \pi(37.51)^2 \times 6.0222 \times 10^{23} \times 2.5^3}{3\left(300 \times 8.3134 \times 10^7\right)^3\left(2^2 s^2\right)}=-\frac{0.03575}{s^2}
For B° = 1, the value of sis given by
1=10^{25} e^{-0.03575 / s^2}=e^{57.565} e^{-0.03575 / s^2}
\frac{0.03575}{s^2}=57.565 \quad s=\sqrt{\frac{0.03575}{57.565}}=0.02492
From the equation
B^{\circ}=e^{57.565} e^{-0.03575 / s^2}
the value of B° can be calculated for magnitudes of s around 0.025. The results are shown in Table 27.1. The explosive increase in B° as s increases is apparent.
| TABLE 27.1 | |||
| s | B° | s | B° |
| 0.023 | 4.47 \times 10^{-5} | 0.0255 | 13.3 |
| 0.024 | 1.11 \times 10^{-2} | 0.027 | 5.04 \times 10^{3} |
| 0.02492 | 1 | 0.029 | 3.46 \times 10^{6} |