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Since 1891 lighting lamps have been manufactured in the Netherlands. The improvement today in comparison to the first lamp is enormous , especially with the intorduction of the gas discharge lamps. The life- time has increased by orders of magnitude . the colour is also an important aspect. Rare earth metal compunds like `CeBr_(3)` are now included to reach a colour temperature of 6000K in the lamp the compounds are ionic solids at room temperature , and upon heating they sublime partially to give a vapour of neutral metal halide molecules . To achieve a high vapour pressure, the sublimation enthalpy should be as low as possible. Calculate the enthalpy of sublimation fo `CsCeBr_(4)` (in integers) Use the Born-Lande formula for all steps in the process and report the separate energies also (be aware of the signs). The `CeBr_(4)^(-)` anion is a teteahedron and in which the ratio between the edge and the destance between a corner of the tetrahedron and the centre of gravity (body - radius) amounts to `(2sqrt(6))//3 = 1.33` . The Born exponent of the CsBr is 11. The radius of Cs is 0.181 nm. |
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Answer» Correct Answer - Steps 1 : The lattice energy of `CsCeBr_(4)` with opposite sign is: `H_(1) =(139xx1xx1xx1.75)/(0.617) xx(10)/(11) KJ "mol"^(-1)` step 2: `H_(2) = 4 xx(139.3.1)/(0.297)(10)/(11) -6 xx(139 xx1xx1)/(0.297 xx (2)/(3)xx sqrt(6) xx(10)/(11) KJ "mol"^(-1) = 3543 KJ "mol"^(-1)` Step 3 : The electronic energy in the gas phase of `CeBr_(3)` is (see answer 8.2): `H_(3) =-3xx(139xx 1xx1)/(0.297 xx sqrt(3))xx(10)/(11) KJ "mol"^(-1) =- 3092 KJ "mol"^(-1)` Step 4 : The electrostatic energy in the gas phase of CsBr is `H_(4) =-(139xx1xx1)/(0.363) xx(10)/(11) KJ mol^(-1) =-348 KJ "mol"^(-1)` Total sum . `H_("total") = H_(1) + H_(2) + H_(3) +H_(4) =461 KJ "mol"^(-1)` |
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