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
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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