Question

rrect Question 39 0/2 pts 57.) Ethylene gas (C2H4) is a ripening agent for fruit. Calculate the Delta H for the formation of ethylene gas according to the following reaction: C2H6(g) yields C2H4 (g) + H2 (8) Based on the enthalpies of the reactions: 2 C2H6(E) + 702() yields 4 CO2 (8) + 6 H20 (1) DH = -3119.4 kJ C2H4 (8) + 302(g) yields 2 CO2(g) + 2 H20 (1) DH = -1410.9 kJ 2H2(g) + O2(e) yields 2 H20 (1) = -571.66 kJ DH Since Canvas will only accept a numerical answer, do not include units. Include a negative sign if the value is negative.

Answer 1

57) Given equations 2 CzH6 (9) + 70,(9) -> 4CO2(9) + 6H204) a= -3119.4KJ C2H49) + 302 (9) ► 2002 (9) + 2H204) (H = -141049 KJ 2H2 (9) + O2 (9) + 2H200) 4H= -571066I<J Required equation : Cette CzH4 (9) + H2(9) AH=? To get the required equation 0x -③ – x2 Oxv> Cz H69) + 7020) — 269/9 +34206) 243 - 1559.765 @ C2H49) + 3% (9) 2007 (9) +242004) alta - 1410.9kJ 3 H₂9) + 2/2 (9) → H₂64) AH = - 285:83 kJ 0x - - Ca H6(9) C2H4 (9) +H2 (9) 4H = - 1559.7 + 14 10.9 + 285.83 . [C2 H6 (9) - (altu 9) + H₂ (9) alt= +137 kJ/- T + 137.03KJ 56d) combustion reaction of C6H6 (1) Given value (-6535 KI) is heat of combustion for 2 moles of benzene. .: 2 CGH El) +15 029) 712 CO2(9) + 6H2O 4 AH = -6535 KT From the ca 2x 78=156 g a benzene combustion gives 6535kJ Heat log - 10 x 6535 = 418.91kJ 156 (418.91 KJ Heat is evolved if log of benzene is burned Entually of the reverse reaction is 4H + 6535KJ