Question

1) Consider the following reaction: 3C(s) + 4H2(g) → C3H8(g); AH° = -104.7 kJ; AS = -287.4J/K at 298 K What is the equilibriu
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Answer #1

ΔHo = -104.7 KJ

ΔSo = -287.4 J/K

= -0.2874 KJ/K

T = 298 K

use:

ΔGo = ΔHo - T*ΔSo

ΔGo = -104.7 - 298.0 * -0.2874

ΔGo = -19.0548 KJ

Now we have:

T = 298 K

ΔGo = -19.0548 KJ/mol

ΔGo = -19054.8 J/mol

use:

ΔGo = -R*T*ln Kc

-19054.8 = - 8.314*298.0* ln(Kc)

ln Kc = 7.6909

Kc = 2.188*10^3

Answer: 2.19*10^3

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