For the reaction C2H4(g) + H2O(g) CH3CH2OH(g) G° = -9.6 kJ and S° = -125.7 J/K...

For the reaction C2H4(g) + H2O(g) CH3CH2OH(g) G° = -9.6 kJ and S° = -125.7 J/K at 286 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 286 K. The standard enthalpy change for the reaction of 2.19 moles of C2H4(g) at this temperature would be kJ.

Expert Solution

C2H4(g) + H2O(g)-----------> CH3CH2OH(g) G° = -9.6 kJ = -9600J and S° = -125.7 J/K at 286 K

G⁰ = H⁰ - TS

-9600 = H⁰ -286*-125.7

-9600 = H⁰ +35950

H⁰ = -9600-35950

H0 = -45550J

= -45.55KJ

C2H4(g) + H2O(g) ------>CH3CH2OH(g) H⁰ -45.55KJ

1 mole of C2H4 react with H2O to gives CH3CH2OH produce -45.55KJ

2.19 moles of C2H4 react with H2O to gives CH3CH2OH produce = -45.55*2.19/1

= -99.7545KJ >>answer

queen_honey_blossom answered 1 year ago

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