A student determines the value of the equilibrium constant to be 6.99×104 for the following reaction.
S(s,rhombic) + 2CO(g)SO2(g) + 2C(s,graphite)
Based on this value of Keq:
G° for this reaction is expected to be (greater, less) than zero. ________
Calculate the free energy change for the reaction of 1.74 moles of S(s,rhombic) at standard conditions at 298K. G°rxn = kJ
S(s,rhombic) + 2CO(g) -----------> SO2 (g) + 2 C (s,graphite)
We know that,
ΔGo = -RT lnKeq
ΔGo = -8.314 x 10-3 kJ/mole.K x 298 K ln(6.99×104)
ΔGo = -27.64 kJ/mole
ΔGo < 0
For 1 mole, ΔGo = -27.64 kJ
For 1.74 mole, ΔGo = 1.74 x (-27.64 kJ)
= -48.09 kJ
A student determines the value of the equilibrium constant to be 6.99×104 for the following reaction....
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