1- an element crystalizes in a body centered cubic lattice. theedge of the unit

Question

1- an element crystalizes in a body centered cubic lattice. theedge of the unit cell is 3.36 angstroms, and the density of thecrystal is 7.97g/ cm3. what is the atomicweight of the element in grams per mole?

2- the following is a quote from an article in the august 18,1998, issue of the new york times about the breakdown of celluloseand starch: " a drop of 18 degrees Farhenheit from 77 degreesF° to 59degrees F° lowers the reaction rate six times; a 36degree drop from 77 F° to 41F°bproduces a forty folddecrease in the rate." calculate the activation energy for thebreakdown process based on the first and the second estimate of theeffect of temperature on rate. are these values consitent? assumehere that k gives a good approximation of the rate of thereaction

Explanation / Answer

(1)No . of atoms for a bcc unit cell = 2 Edje length , a = 3.36 * 10^-10 cm Volume of the unit cell V = a^3 = 37.933 * 10^-30cm^3 Density of the crystal d = 7.97 g / cm^3 Mass of the unit cell , M = No .of atoms in the unit cell *Mass of each atom                                      = 2* m mass of an atom m = atomic mass / avagadro Number                               = m / N Avagadro number N = 6.023*10^23 Density of the unit cell d = mass of the unit cell /Volume of the unit cell                                   d = 2* ( m/ N ) / V           So m =dVN / 2                   = 910.456 * 10^-7 g (2) F = (9/5)C + 32      C = (5/9) [ F-32] Convertion of 77 oF : ------------------------ C = (5/9) [ 77-32] = 25 oC Convertion of 59 oF : ------------------------ C = (5/9) [ 59-32] =15 oC Convertion of 59 oF : ------------------------ C = (5/9) [ 59-32] =15 oC We know that ln ( K/ K' ) = ( Ea/R ) [ (1/T ) - ( 1/T )] Given K = initial rate constant          K' =final rate constant = K /6         T =25 oC =25 + 273 = 298 K           T' = 15oC = 15+ 273 = 288 K          R = gasconstant = 8.314 J / mol - K Plug the values we get ln(K'/K) = ( Ea / 8.314 ) [ (1/298 ) -(1/288) ]                                    ln(1/6) =  ( Ea / 8.314 ) [ (0.003355)-(0.00347) ]                                    -1.7917 = - (Ea / 8.314 )*0.000115                                                            Ea = 129532.12 J                                                = 129.53 KJ Similarlly the second one

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