Question

The atmosphere slowly oxidizes hydrocarbons in a number of steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reactions of a number of such steps for methane gas is as follows:
CH4(g)+5O2(g)+5NO(g)→CO2(g)+H2O(g)+5NO2(g)+2OH(g)
Suppose that an atmospheric chemist combines 165 mL of methane at STP, 890 mL of oxygen at STP, and 56.0 mL of NO at STP in a 2.2 −L flask. The reaction is allowed to stand for several weeks at 275 K.

Part A

If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the reactants in the flask at 275 K?

Part B

If the reaction reaches 92.0% of completion (92.0% of the limiting reactant is consumed), what are the partial pressures of each of the products in the flask at 275 K?

Enter your answers numerically separated by commas.

Part C

What is the total pressure in the flask?

Answer 1

Answer: page:01 According to the given information, to determine the following problems. Given, The reaction is, CHuegi + 50ucg) + SN00g) - Colig? + H2.069) + 5N021g)+241 .. The number of moles of each species in the 2.2.1 flask by using the ideal gas law. for methane;- [che] (0.165x) xllabon) 0.08206 xatoni molk / 243K = 0.165 moles 22. 40938 20.00.436 moles CHu for oxygen :- (03) 0.890 14xlatm) 0108966 xaton/mol;K (273*

page toa i 0.890 mol 22.40 238 0:0394 moles og - for oxide - (NO) :(0.0560 11 xl att 0.08206 Xatril molex 273K 0.0560 mol 23.40938 -0.00259 mol (NO) Since, one mole of chu reads with smoles of Og and smoles of No. No is limiting regent. Because, the sweaction goes only 92.04% to completion, Sone of the oractands are present as well Pinal moles of paroducts. bet,

page:03 10.00259 molesto) x Cimol Coals moletno) x 0.920 = 0.00 asq x 0.2 mol com x 01920 14.86 xco" mal con ande -10.00 asq moles Ho)x(Imol Hgo Is molesto)x0.920 = 0.00259 x oig mol H20 x01920 2 u. 76x104 mol Hao and, (0.002sq moles Holx (s moler Nelsmoles no) *0.920 = 0.00959 x inol NoL x 0.920 - 10.00a38 mol Noel and, -10.000sq naol to) x.Camol ons molesnto) X0-920 0.00259 x0.mol OH X 0.920 19.63 xlo“ mol. OH

final pressure of products : Page:04 (G. 176X164 not com} x 10.08206 Xatminot.K) xa15k/arax. vog ? 4.46 xlou coq. X 0.08206 atm x 195 0.00uks atm Coa? for Hao- (4.46 xlo" noot H90 1x (0.08206 Xatmimetik) x 245k/212K 5.4096 XCOW H2OX 0108206 alm x 195 - 0.00u88 atm Hio for Noai -10.00238 met nog)x (0.08206 Kalm Inoet k) y ayskla.2x a 0.00238 Noq x 0108206 atm x 125 = 10.0244 atm Noel for oH; +(9:53 x 104 mol obl) x (008 206 Katm (monotox) x 275k/3.21 39.53 XLÓW OH X 0.08 206 atm X125 - 10.00 97.1 atm oh]

pageios final moles og reactands: for methane ; CHU 30.00-136 molche - (0.00 asq molho x (Imol Cheel Smolnte)x0192) = 0.00136 chu mol - (0.00159 xo.a mol CH4 * 0,92) 0.00736 mol chee- 494656 Xcó mol Chee - 16.00 688 mol CH4 for oxygen : 02 010394 mol od -10.00 259 molio x (smoloa fs motato) x0x72] = 0.039+ molog - 0.00259 XImol oq x0,92 40.039+moloq – 2.8828X60% molo -0.0343 mol og for oxide :- NO À 0.00259 mol No - (0.002sqmol NO X0192) 0.00259 mol No - 2.3828 x103 mol no = (4.04 x 10" mol No

page:06 final pressure of reactants : for chur 2.0.00688 hol Chux (0.08206 Xatmimetik 1x 2-56% .28 • Đ• 00688 Cuu x bogos q & 12 - 0:00 68% Clu X 4D: asis a -loioti atm Chel -for 023- = 0.0 373 mol oq x (0.08206 Katm Imol.k) x 245% 12:26 = 0.0313 Od X 0.08206 atm X 125 c. 0.0373 Os x 10. a575 atm =/0.382 atmoa for No - (2.07 x 164 mot No) x 10.08206 katm. mol.k) x 275k12.2 = 9:07 X104 NO x 0.08206 gtm x 125 = 2:01 XLÓWNO X 10.8595 atm =10.00 212 atm No

page: 07 Therefore, total pressure in flask: final pressure of products + final pressure of reactan's b488 +0.00488 + 0.0944 +0100947 +0.071+ 0.382 +0:00 212 ] atm co.uaq atm. Ricrotal pressure in flask = 0,499 atm