Question

Consider the reaction CO(g) + 3H2(9)—CH (9) +H20(9) for which AH° = -206.1 kJ and A Sº = -214.7 J/K at 298.15 K. (1) Calculate the entropy change of the UNIVERSE when 1.982 moles of CO(g) react under standard conditions at 298.15 K. A Suniverse JK (2) Is this reaction reactant or product favored under standard conditions? (3) If the reaction is product favored, is it enthalpy favored, entropy favored, or favored by both enthalpy and entropy? If the reaction is reactant favored choose 'reactant favored'.

Answer 1

CO(g) + 3H₂CJ) CH4 (8) + H₂O(g) 4 Suniverse = 45 45system + 4S surrounding. Ssurrounding = not n-number of moles 10982*-206.1)x102 I JK 298.15 = 1:3700 x103 g/k. .. As universe af-214-7 +(1370)} Ik 2 - 1584.7 g/k 2-|-584.7 kJ/K tys) ② If an (cribbs free energy of a reaction is -ve then the reaction is favored under standard condition : 4hz 4H-TAS. - - 206100 - 299,157- = - 206100 + 64012.80 I =- 142087.19 3 . The products are farroarred at standard condition as the the values are -ve for the forward reaction they As is re souT4s is tre. in this are due to high -ve 4H value. so the reaction is product forrows, is enthalpy farrow-

2CH. (4) + 702 (9) 400,(3) + 6H003). A S surroundings - 2.082 *(-2,855480') wik 298.15 =-19.9366 kJ/K ③. AS AS niverse a 4 System + 4 Ssurrounding + 92.70 +1-199366) OK = - 19843-9 Ilk (Hrs) Ah=4H-TAS. - (-2.858x80°) x ro - 298-15(** 92-70) 2- 2855600 - 27638-505 ] = - 2882638.505 J As ah is-ve so the reaction is product forrows unde stood standard conditions. As is positive so Tas is also the and a H is -ve both the AH & as terms. snake Ah-ver so the product forour and both the enthalpy & entropy favour as