Homework Answers
T= 25.0 oC
= (25.0+273) K
= 298 K
ΔGo = 8.25 KJ/mol
ΔGo = 8250 J/mol
use:
ΔGo = -R*T*ln Kc
8250 = - 8.314*298.0* ln(Kc)
ln Kc = -3.3299
Kc = 3.58*10^-2
T = 298 K
Δ n = number of gaseous molecule in product - number of gaseous molecule in reactant
Δ n = 1
use:
Kp= Kc (RT)^Δ n
Kp = 3.58*10^-2*(0.08206*298.0)^(1)
Kp = 0.8754
Answer: 0.875
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