Question

Exercise 3.2. Consider the following gas-phase reaction: CO(g) + 3 H2(g) + CH4(g) + H2O(g) with the equilibrium constant Kc = 3.88. A reaction mixture consisting of 2.45 mol CO(g), 3.16 mol H2(g) and 1.27 mol H2O(g) is placed in a 10.0-L reaction vessel and heated to 1200 K. Using the final notebook from Example 3.1 as a template, create a Mathematica notebook to calculate the equilibrium concentrations (in mol) of all four gases: (a) CO, (b) Hz, (c) CH4, (d) H20, and (e) perform the check for Kc. Save the notebook as "Exercise_3.2.nb". Warning! All calculation steps must be performed in your Mathematica notebook, and all items (a-e) must be clearly labelled in the Print[functions, or the score for this exercise will be zero! Answers: (a) 2.06 mol (b) 2.00 mol (c) 0.387 mol (d) 1.66 mol (e)... 00 nhase reaction:

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Answer 1

des) Rxni, co (g) + 3H₂ (9) - Mulg) & 40(0) -Step I find Concentration of all species (gas) at initial Concentration = moles volume a) [co] = 2045mol = 0.245M ol 6) [p] = 3016 mal = 0.316M C) [420] 1024me - 0.127M 5) UUUUU LOL 00127M 1027mol LOL d) [chul- omol z OM lol Steh I find direction of Reaction Q = [che] [120] 0x00127) [co] [Hz] (0.245) (0.316)3 Q = 0 . Q Lke so Ron is forward

steh 11) find Concentration of all species at equiliburun CO(g) + 3H₂(g) - Cha(8) + H2O(9) Enedal 0.245 0.316 0 0 0127 change - x - x x x 0.245.x 0. 316-30 00127 + x TALALALALLLLLLL k = [hu] [120] [co] [HQ] 3 3.88 = (x ) C00127 tic) Co. 245-x) (0.316-3x)3 3. 88 (0.316-3x)² (0.245 -se) = 0012 7x + x² as you solve this equation. you will find. x = 0.0387 ole as result. [col= 0.245-000387 = 0.206 [H₂] = 0.316-3X0.0387= 0.2 [Cha] = 0.0387 [120] = 0.127+ 0.0387 = 0.166M 8o. Convent it to moes . (males = concentration) no= [60] Vo = 0.206th x 1022 2.06m. Voleme 2 0.2x10= 2mal - nenez 0.0387 X10= 0.387, ny0016 6X10 =1066 UUUUU 10387810= 0.38t, "no (O.16hxio