What is the Gibbs Free Energy change for the reaction C_3 H_g (g) + 5O_2 rightar

Question

What is the Gibbs Free Energy change for the reaction C_3 H_g (g) + 5O_2 rightarrow 3 CO_2 (g) + 4H_2 O(t) (A) -2111 kJ (B) (C) -108, 932 kJ (D) For which should Delta H degree and Delta G degree be about (A) 2NO_2 (g) rightarrow N_2 O_4 (g) (B) 2H_2 (g) + O_2 (g) rightarrow 2H_2 O (g) (C) 2Al (s) + Fe_2 O_3 (s) rightarrow 2 Fe(s) + Al_2 O_3 (s) (D) 2Na (s) + 2H_2 O (l) rightarrow 2NaOH (aq) + H What is the entropy change for this reaction? C_3 H_a (g) + 5O_2 (g) rightarrow 3 CO_2 (g) + 4H_2 O(l) (A) -191 J middot K^-1 (B) -373 J middot K^-1 (C) +373 J middot K^-1 (D) -191 kJ middot K^-1

Explanation / Answer

12. DG0rxn = DH0rxn - TDS0rxn

DH0rxn = 4*DH0fH2O+ 3*DH0fCO2 - dH0fC3H8+dh0fO2

         = (4*-286 + 3*-394) -(-104+5*0)

         = -2222 kj

DSrxn = 4*S0H2O+3*S0fCO2 - S0fC3H8 + 5* S0fO2

        = (4*70 + 3*214)-(270 + 5*205)

        = -373 j.k-1

Temperature(T) = 25 C = 298 k

DG0rxn = -2222*10^3 - (298*-373)

        = -2111 kj

13 is not visible completely

14.

DSrxn = 4*S0H2O+3*S0fCO2 - S0fC3H8 + 5* S0fO2

        = (4*70 + 3*214)-(270 + 5*205)

        = -373 j.k-1

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