Please show all work so that i can get a better understanding A.) Estimate G° rx

Question

Please show all work so that i can get a better understanding

A.) Estimate G°rxn for the following reaction at 449.0 K.

CH2O(g) + 2 H2(g) CH4(g) + H2O(g) H°= -94.9 kJ; S°= -224.2 J/K

B.) Which of the following statements is TRUE?

C.) Calculate Grxn at 298 K under the conditions shown below for the following reaction.

2 Hg(g) + O2(g) 2 HgO(s) G° = -180.8 kJ

P(Hg) = 0.025 atm, P(O2) = 0.037 atm

D.)Above what temperature does the following reaction become nonspontaneous?

FeO(s) + CO(g) CO2(g) + Fe(s) H= -11.0 kJ; S = -17.4 J/K

Explanation / Answer

A)

CH2O(g) + 2 H2(g) CH4(g) + H2O(g) H°= -94.9 kJ; S°= -224.2 J/K

dG = dG - T*dS = (-94.9) - (298)(-224.2/1000) = 5.7658 kJ/mol

B.) Which of the following statements is TRUE?

a. Endothermic processes decrease the entropy of the surroundings, at constant T and P. True, since they decrease Heat in surrounding

b. Exothermic processes are always spontaneous. --> False, impossible to determine without dG

c. Endothermic processes are never spontaneous. --> False, impossible to determine without dG

d. Entropy is not a state function. --> False, it is

e. None of the above are true. --> False, at least one was

C.) Calculate Grxn at 298 K under the conditions shown below for the following reaction.

2 Hg(g) + O2(g) 2 HgO(s) G° = -180.8 kJ

P(Hg) = 0.025 atm, P(O2) = 0.037 atm

dG = dG° + RT*ln(Q)

dG = (-180.8*10^3) - (8.314*298)*ln((0.025^2)(0.037))

dG = -154352.91915

dG = -154.35 kJ/mol

D.)Above what temperature does the following reaction become nonspontaneous?

FeO(s) + CO(g) CO2(g) + Fe(s) H= -11.0 kJ; S = -17.4 J/K

dG = dH - T*dS

dG < 0 so

dH - T*dS < 0

dH/dS < T

-11000 / -17.4 < T

T = 632 K

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