Question

Consider the reaction 2H2S(g) 302(g) >2H20(I) + 2SO2(g) -1.124x103 kJ and AS° = -390.7 J/K at 298.15 K for which AHo (1) Calculate the entropy change of the UNIVERSE when 2.258 moles of H2s(g) react under standard conditions at 298.15 K ASuniverse J/K (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

Given The reaction 212-3699759.95 > 214960jy *25.173 440 =-11243103 kg and AS - 390.7 Tempesuture (7) = 298-15% Given Hos 10) = 2.258 moles, Firstly, we have to calculate the Entropy, change of the system. under standard conditions the story for the Beaction is - 3 90.7 Jl. -This for 2 mol of Heo We have 2.258 mol of Heo reaching Thus the entropy change of the systemis Assystem = 290-4) 2258) 2

-882-2006 2 --441.1003 As system 4411 Ilk.] Now, calculate the change in Entropy of the sunrondings. to use the Erthalpy change of the reaction here, Aagain remember that the Enthalpa Change is for nel of seeding AS - Hsystem Surrounding - 611248103) 12.258) 1376298-15) - - 2537.992 596.3 +/- 4256123) = 4.256x103J1k the two values to calculate the Then ble use Entropy change of the univege.

+ As System its surrounding As. As universe system .: Asuniverse - 4441.15k) +(4.256x1833k) = 3814.9 = 3.81x103J/ks. As convesse = 3-81x70*3lk. 7 Consider the Gibb's free Energy AG = 4.#- I AS AG =(-1-12 43410 73429815) 1-390.7) = -1.124x1034 92337.055 (2) = 91.2x10 kr T AG= 91.2x 70 3 kJ) The chibbs free Energy change is the Therefore, the reaction is non-Spontaneous at this temperature

As Explained above "Reactant favoured." part :-(2) Given The reaction is 600 (9) + 6H20 15 C6H12O6t 60219) 4H º = 2.801X 103 65 44 4.50 - 259.0 Jik Temperature (7) = 2.98-15 | I Given con = 1.603 So, foo 1.603 males of com At = 1.603 molx (2.8014 103 kJ/ 6mol = (1.603 mol)(466-833] TAH° = 748.333 kJ Asº = 1:603 mol (-259.87 47 = (4.603 mol) (-43.1666) =-69.1961 ok . (ASO = -69.1961 512

Aro (system) 5 - 69.1961 55 = 748.333 746 sten = 778333.33 J As surroundings = 4H surrounding Surrounding = AH System Tsurrounding 778 333.33 298.15 = 2610.542 Jk 1Ssurroundings = 2610.542 36"}] AS univege = Afurroundings + Assystem. = 2610.542 Jkt 69.1965 = 2541.346 55+ : Asuniverse = 2541.346 56-7 AGO =AH-TASO = 748. 333 kJ - (298.15k) (-69.1961#)

748.33 + 206 30.8172 = 21379.14 J. .: AGO = 21379.145 . It is +ve, the reaction is non- Spontancars That is the reaction is Reactant favoured." (3) The reaction is "heactant favoured"
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