Tutor profile: Douglas R.
Subject: Physical Chemistry
The data given below are for the adsorption of nitrogen on alumina at 77.3 K. Show that the data can be fitted by the BET equation in the range of adsorption and calculate the volume of monolayer formed on the surface of the alumina. At 77.3 K, the saturation pressure is P* = 733.59 torr. The volumes are corrected to STP and referred to 1g of alumina. P(torr) 37.67 74.20 114.54 142.0 185.34 V (cm3/g, STP) 23.14 28.1 33.1. 36.35 41.49
The BET (after Brunauer, Emmett and Teller) equation is usually employed to determined the specific surface area of porous solids from the adsorption data of inert gases. The BET equation indicates the volume of gas needed to form a monolayer on the surface of the sample and can be represented in linear form as follows: (z/(1-z))*(1/V) = ((c-1)*z/(c*Vmono)) + 1/(c*Vmono ) Where, V is the volume of gas adsorbed at pressure p, Vmono is the amount of gas corresponding to one monolayer, z is the ratio of gas pressure and the pressure at saturation (p/p*), and c is a constant. From the above data: z = p/p* = 37.67 torr/733.59 torr = 0.051 (z/(1-z))*(1/V) = (0.051/(1-0.051))*(1/23.14 cm^3 ⁄ g) = 0.0023 g ⁄ cm^3 For the rest of the data: z (p/p*) 0.051 0.10 0.16 0.19 0.25 z/V(1-z) 0.0023 0.0039 0.0056 0.0066 0.0082 A plot of z/V(1-z) against z will give a straight line, showing the good fit between the BET equation and the data. From the fitting the slope and intercept values are 0.0297 and 0.0009, respectively. From the intercept: 1/(c*Vmono) = 0.0009 → c*Vmono = 1111.11 g ⁄ cm^3 From the slope: (c-1)/(c*Vmono) = 0.0297 → c = 0.0297 * 1111.11 + 1 = 34 And, Vmono = 1111.11/34 = 32.68 g ⁄ cm^3
Subject: Inorganic Chemistry
Determine the free space percentage of a bcc unit cell structure.
Unit cells are the smallest repeating atoms in a crystal. The body centered unit cell (bcc) has atoms in each corner of a cube comprising 8 in total and one atom in the center of a cube. The percentage of free space is determined by: % of free space = 100% - packing efficiency Body centered unit cells contain 2 atoms per unit cell (Z = 2). The packing efficiency is given by the following expression: packing efficiency =(⋕of atoms (Z) * volume of one atom)/(volume of the unit cell)*100% volume of one atom = 4/3*μ*r^3 volume of the unit cell = a^3 Where 'r' is the radius of the atom and 'a' the length of one of the cube sides. Considering that the atoms touch each other along the cube diagonal (hypothetical distance AD from opposite corners A and D), half of each corner atom (r + r) and the central atom (2r): AD = r + 2r + r = 4r Considering C and B as the adjacent corners from A and D, respectively, from the triangle ABC: AB^2 = AC^2 + CB^2 → AB = √(a^2+a^2 ) = √2 a Considering now the triangle ADC: AD^2 = BD^2 + AB^2 → AD = √(a^2+(√2 a)^2 ) = √3 a Then we have: a = 4/√3 r The packing efficiency is: packing efficiency = (2*4 ⁄ 3 *π*r^3)/(4*r / √3)^3 *100% = aprox 68% And, % of free space = 100% - 68% = 32%
Subject: Chemical Engineering
Hydrogen and oxygen gases are introduced into a pug flow reactor operated in steady state conditions, and water is produced after ignition: H2 + 1 ⁄ 2 O2→ H2O Conversion of the reaction is 80%. Calculate the mass of components leaving the reactor if incoming mass flows of oxygen and hydrogen are 48 kg/s and 4 kg/s, respectively.
The first step is to determine which compound is the limiting reactant. Since 1 mol of hydrogen is necessary to react with half mol of oxygen to form 1 mol of water, we need to calculate the number of available moles for each compound. For H2: (4 kgH2 ⁄ s/(2 kgH2 ⁄ kmolH2))*(1kmolH2O/1kmolH2) = 2 kmol/s H2O For O2: (48 KgO2 ⁄ s/(32 kgO2 ⁄ kmolO2))*(1kmolH2O/1 ⁄ 2 kmolO2) = 3 kmol/s H2O Hydrogen is the limiting reactant since it is only possible to produce 2 kmol/s of H2O from its incoming flow. The mass of components leaving the reactor are calculated from the general mass balance equation. For example for H2: mH2in - mH2out + mH2gen - mH2cons = mH2acc (In) - (Out) + (Generation) - (Consumption) = (Accumulation) Where mH2in is the mass flow rate entering the reactor, mH2out is the mass flow rate leaving the reactor, mH2gen is the mass production rate, mH2cons is the mass consumption rate, and mH2acc is the accumulation rate of H2 in the reactor. Since H2O is a product and H2 and O2 reactants, mH20in = 0, mH2gen = 0 and mO2gen = 0. In addition, there isn’t accumulation in a pug flow reaction in steady state conditions. Therefore the mass of components leaving the reactor is: mH2out = mH2in - mH2cons =4 kgH2 ⁄ s - 0.8*4 kgH2 ⁄ s = 0.8 kgH2 ⁄ s mO2out = mO2in -mO2cons = 48 kgO2 ⁄ s - 16kgO2/2kgH2*0.8*4 kgH2 ⁄ s =22.4 kgO2 ⁄ s mH2Oout = mH2Ogen = 18kgH2O/2kgH2 *0.8*4 kgH2 ⁄ s = 28.8 kgO2 ⁄ s The mass of oxygen consumed and water generated a derived from the moles of the limiting reactant (H2) as depicted in the equation 1. (1kmolH2O/1kmolH2) * (18kgH2O/1kmolH2O)/(2kgH2/1kmolH2 ) = 18kgH2O/2kgH2 (1/2kmolO2/1kmolH2) * (32kgO2/1kmolO2)/(2kgH2/1kmolH2 ) = 16kgO2/2kgH2