According to the following equation and delta G degree_f values (right), Delta G

Question

According to the following equation and delta G degree_f values (right), Delta G degree for the combustion of one mole of ethyl alcohol (C_2H_5OH) at 25 degree C is___. C_2H_5OH(g) + 3O_2(g) rightarrow 2CO_2(g) + 3H_2O(g) -1.65 times 10^3 kJ -798 kJ -843 kJ -448 kJ -1.30 times 10^3 kJ For a particular reaction, Delta H = -31.0 kJ/mol and Delta S = -80.5 J/K-mol. At 385 K, the value of Delta G for this reaction is___and the reaction is___. 62 kJ/mol, spontaneous in the forward direction -62 kJ/mol, spontaneous in the reverse direction 0.0 kJ/mol, at equilibrium -30.9 kJ/mol, spontaneous in the forward direction 30.9 kJ/mol, spontaneous in reverse direction The molar heats of fusion and vaporization of benzene are 10.9 and 31.0 kJ/mol, respectively. Entropy changes for solid to liquid and liquid to vapor transitions are 39.1 J/K-mol and 87.8 J/K-mol, respectively. At 1 atm, the melting and boiling points of benzene are___and___. respectively. Not enough information 319 K, 419 K 273 K, 373 K 279 K, 353 K 224 K, 398 K Urea (NH_2CONH_2), used commonly as a fertilizer, is involved in an equilibrium reaction with carbon dioxide and water in the soil as follows: CO_2(g) + 2NH_3(g) doubleheadarrow NH_2CONH_2(aq) + H_2O(g). The equilibrium constant at 25.0 degree C for this reaction is___, given that the standard free energy change at that temperature is -13.6 kJ. 242 5.49 260 4.25 times 10^-3 520 The following reaction is___at 1000 degree C, and at that temperature the Delta G degree is___. CaCO_3(s) doubleheadarrow CaO(s) + CO_2(g) spontaneous, -1300 kJ spontaneous, -673.3 kJ spontaneous, -24.1 kJ not spontaneous, 24.1 kJ not spontaneous, 673.3 kJ

Explanation / Answer

1)

we know that

dGo rxn = dGof products - dGof reactants

consider the given reaction

C2H5OH + 3 02 ---> 2 C02 + 3 H20

in this case

dGo rxn = ( 2 x dGof C02) + ( 3 x dGof H20) - (dGof C2H5OH) - ( 3 x dGof 02)

using the given values

dGo rx = ( 2 x -394.4) + ( 3 x -228.6) - (-174.9) - ( 3 x 0)

dGo rxn = -1299.7

dGo rxn = -1300 kJ

so

the answer is e) -1.3 x 10^3 kJ

2)

we know that

dG = dH - TdS

substituting the given values

we get

dG = ( -31 x 1000) - ( 385 x -80.5)

dG = 0

we know that

if dG < 0 , the reaction is spontaneous in the forward direction

if dG = 0 , the reaction is at equilibrium

if dG > 0 , the reaction is spontaneous in the reverse direction

so

the answer is C) 0 kJ/mol , at equilibrium

3)

at melting point

solid ---> liquid

we know that at melting point , solid and liquid are at equilibrium

so

at equilibrium , we know that dG = 0

also

dG = dH - TdS

so

dG = dH - TdS = 0

dH = TdS

T = dH / dS

so

at melting point

Tf = 10.9 x 1000 / 39.1

Tf = 278.77 K

so

the melting point of benzene is 279 K

similarly for boiling point

T = dH / dS

Tb = 31 x 1000 / 87.8

Tb = 353 K

so

the boiling point of benzene is 353 K

so

the answer is d) 279 K , 353 K

4)

we know that

dGo = -RT lnKeq

given

dGo = -13.6 x 1000 J

temperature (T) = 25 C = 25 + 273 = 298 K

so

-13.6 x 1000 = -8.314 x 298 x lnKeq

Keq = 242

so

the value of equilibrium constant is 242

so

the answer is a) 242

5)

we know that

dHo rxn = dHof products - dHof reactants

in this case

dHo rxn = ( dHof CaO) + ( dHof C02) - (dHof CaC03)

dHo rxn = ( -635.1) + (-393.5) - (-1206.6)

dHo rxn = 178 kJ

now

dSo rxn = ( So products ) - So reactants

in this case

dSo rxn = ( So Ca0) + (So C02) - (So CaC03)

dSo rxn = ( 38.2) + ( 213.7) - 92.9

dSo rxn = 159 J / K

now

dGo = dHo - TdSo

given

temperature (T) = 1000 C = 1000 + 273 = 1273 K

so

dGo = ( 178 x 1000) - ( 1273 x 159)

dGo = -24.407 kJ

now

we know that

if dGo > 0 , the reaction is not spontaneous

if dGo < 0 the reaction is spontaneous

so

the answer is C) spontaneous , -24.1 kJ

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