A) A student determines the value of the equilibrium constant to be 2.57×10 14 f

Question

A) A student determines the value of the equilibrium constant to be 2.57×1014 for the following reaction.
2HBr(g) + Cl2(g) --> 2HCl(g) + Br2(g)
Based on this value of Keq:
delta G° for this reaction is expected to be (greater, less) ________ than zero.
Calculate the free energy change for the reaction of 1.81 moles of HBr(g) at standard conditions at 298K.
delta G°rxn = ______ kJ

B) A student determines the value of the equilibrium constant to be 3.81×10-18 for the following reaction.
Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g)
Based on this value of Keq:
delta G° for this reaction is expected to be (greater, less) ______ than zero.
Calculate the free energy change for the reaction of 2.08 moles of Fe3O4(s) at standard conditions at 298K.
deltaG°rxn = ______ kJ

C) A student determines the value of the equilibrium constant to be 1.87×10-18 for the following reaction.

N2(g) + 2O2(g) ---> 2NO2(g)
Based on this value of Keq:
delta G° for this reaction is expected to be (greater,less) _____ than zero.
Calculate the free energy change for the reaction of 2.27 moles of N2(g) at standard conditions at 298K.
delta G°rxn = ___ kJ

Explanation / Answer

1)

2HBr(g) + Cl2(g) --> 2HCl(g) + Br2(g) K =  2.57×1014

since K > 1, this must favours products

dG < 1

dG = -RT*ln(K)

dG = -8.314*298*ln(2.57*10^14) = -82206.079 J = -82.20 kJ

this is based in 2 mol of HBr so..

for

1.81 moles of HBr(g)

-82.20 * 1.81/2 = -74.391 kJ

2)

Fe3O4(s) + 4H2(g) --> 3Fe(s) + 4H2O(g) K = 3.81×10-18

since K < 1 , this must favours reactants

dG > 0

dG = -RT*ln(K)

dG = -8.314*298*ln(3.81*10^-18) = 99372.6937 J/mol per 1 mol of Fe3O4

we need 2.08 so--> 2.08*99372.6937 = 206695.20 J = 206.69 kJ

3)

K = 1.87*10^-18

dG!° = -RT*ln(K)

dG° = -8.314*298*ln(1.87*10^-18) = 101135.95 J

this is per 1 mol of N2

so for 2.27:

2.27*101135.95 = 229578.6065 J = 229.578 kJ

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