Awaste gas contains 0.1 mol% toluene in air and occupies a volume of 2,120 m3 at 298 k and 116 kpa. in an effort to reduce the toluene content of this gas, it is exposed to 113 kg of activated carbon, initially free of toluene. the system is allowed to reach equilibrium at constant temperature and pressure. assuming that the air does not adsorb on the carbon, calculate (a) the equilibrium mole fraction of toluene in the gaseous phase and (b)the amount of toluene adsorbed by the carbon. the adsorption equilibrium for this system is given by the freundlich isotherm: w = 0.208(p*)^0.11. in this equation w is the carbon equilibrium adsorptivity, in kg of toluene/kg of carbon, and p* is the equilibrium toluene partial pressure, in pa. (mw of toluene = 92)

Be sure to answer all parts. the catalytic destruction of ozone occurs via a two-step mechanism, where x can be any of several species: (1) x + o3 → xo + o2 [slow] (2) xo + o → x + o2 [fast] (a) write the overall reaction. o3 + o → 2 o2 o3 + xo → x + 2 o2 xo + x + o3 + o → xo + x + 2 o2 x + o3 → xo + o2 (b) write the rate law for each step (using k for the rate constant). reaction 1: reaction 2: k[x][o2] k[xo][o2] k[x][o3] k[xo][o3] k[xo][o2] k[x][o] k[x][o2] k[xo][o] (c) x acts as a and xo acts as a catalyst intermediate intermediate catalyst (d) high-flying aircraft release no into the stratosphere, which catalyzes this process. when o3 and no concentrations are 3 ă— 1012 molecule/cm3 and 9.9 ă— 109 molecule/cm3, respectively, what is the rate of o3 depletion? the rate constant k for the rate-determining step is 6 ă— 10â’15 (cm3)2/moleculeâ·s. give your answer in scientific notation. use one significant figure in your answer. ă— 10 molecule/s