Calculate Delta G^0 for this reaction: S_8(s) + 8H_2O(l) + 12O_2(g) rightarrow 8

Question

Calculate Delta G^0 for this reaction: S_8(s) + 8H_2O(l) + 12O_2(g) rightarrow 8H_2SO_4(l) from the following. S_8(s) + 8O_2(g) rightarrow 8SO_2(g) Delta G^0 = -300.1 kJ 2SO_2(g) + O_2(g) rightarrow 2SO_3(g) Delta G^0 = -71.1 kJ SO_3(g) + H_2O(1) rightarrow H_2SO_4(l) Delta G^0 =-81.8 kJ -1, 238.9 kJ -3.339.6 kJ -1, 523.3 kJ -911.7 kJ -453.0 kJ The reaction N_2O_4(g) irreversible 2NO_2(g) has Delta H^0_trn of 85.9 kJ and Delta S^0_rxn of 271 J/K. Assuming that Delta H^0_rxn and Delta S^0_rxn are temperature independent, at what temperature does this reaction have an equilibrium constant of 1? 317 K 19 K 0.321 K 0.317 K -317 K What is the equilibrium constant at 298 K for a reaction with a standard free energy change of 0.50 kJ? 1.0 1.6 0.63 1.2 0.82

Explanation / Answer

7. The three equations are

S8 (s) + 8 O2 (g) ------> 8 SO2 (g); G0 = -300.1 kJ ….(R1)

2 SO2 (g) + O2 (g) ------> 2 SO3 (g); G0 = -71.1 kJ …..(R2)

SO3 (g) + H2O (l) ------> H2SO4 (l); G0 = -81.8 kJ …..(R3)

Multiply (R2) by 4 and (R3) by 8 and add to (R1); we get

S8 (s) + 8 O2 (g) + 8 SO2 (g) + 4 O2 (g) + 8 SO3 (g) + 8 H2O (l) -----> 8 SO2 (g) + 8 SO3 (g) + 8 H2SO4 (l)

Cancel out common terms to write

S8 (g) + 12 O2 (g) + 8 H2O (l) ------> 8 H2SO4 (l)

This is the given equation. Add the G0 terms to get the G0 (reaction) as

G0 (reaction) = G0 (R1) + 4*G0 (R2) + 8*G0 (R3) = [(-300.1) + 4*(-71.1) + 8*(-81.8)] kJ = (-300.1 -284.4 – 654.4) kJ = -1238.9 kJ (ans)

Ans: (A) – 1238.9 kJ

8. The given reaction is

N2O4 (g) <====> 2 NO2 (g); H0rxn = 85.9 kJ; S0rxn = 271 J/K.

We know that

G0rxn = -R*T*ln Keq where G0rxn = standard Gibb’s free energy change for the reaction and Keq = equilibrium constant for the said reaction.

When Keq = 1, G0rxn = 0.

Again G0rxn = H0rxn – T*S0rxn

Given G0rxn = 0,

0 = (85.9 kJ) – T*(271 J/K)

===> T*(271 J/K) = 85.9 kJ

===> T = 85.9 kJ/(271 J/K) = (85.9*103 J)/(271 J/K) = 316.974 K 317 K (ans)

Ans: (A) 317 K

9. We know that

G0 = -R*T*ln K where G0 = standard Gibb’s free energy change and K = equilibrium constant.

Given, G0 = 0.50 kJ and T = 298 K, we have,

(0.50 kJ) = -(8.314 J K-1)*(298 K)*ln K

===> (0.50*1000 J) =-(2477.572 J)*ln K

===> ln K = -(0.50*1000)/(2477.572) = -0.2018

===> K = e-0.2018 = 0.817 0.82 (ans)

Ans: (E) 0.82

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