Question

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

Answer 1

S(s,rhombic) + 2CO(g) -----------> SO₂ (g) + 2 C (s,graphite)

We know that,

ΔG^o = -RT lnK_eq

ΔG^o = -8.314 x 10^-3 kJ/mole.K x 298 K ln(6.99×10⁴)

ΔG^o = -27.64 kJ/mole

ΔG^o < 0

For 1 mole, ΔG^o = -27.64 kJ

For 1.74 mole, ΔG^o = 1.74 x (-27.64 kJ)

= -48.09 kJ