In Example 6.5, we found Ni nanopartibles of 2 nm radius would melt at 587 K, assuming the system contained pure Ni liquid. Consider now 2 nm Ni nanoparticles in equilibrium with a liquid mixture. How would the actual melting temperatature compare to the value using the assumptions in Example 6.5? Explain.

Example 6.5

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The field of nanotechnology is an emerging area for chemical and biological engineers. Consider nickel nanoparticles for use as a catalyst to grow carbon nanotubes. The normal melting temperature, Tm, of nickel is 1728 K. In a nanoparticle, the curved surface leads to a force acting tangentially to the surface of the particles, which changes the Gibbs energy of the solid. The magnitude of this effect is determined by the surface tension, . The differential change in Gibbs energy can be written to account for surface tension in terms of curvature, or inverse radius (1/R). In this case, the fundamental property relation given by Equation (5.9) can be modified as follows: dg, = sidT+ vdP + 2v0,d The following properties are available for Ni: surface tension of solid Ni: density of solid Ni: Enthalpy of fusion: (1) R Ahfus. Ni Ni : 1.75 = Phi = 8,900 = -17.48 [] [$] kJ mol