Using standard free energy of formation values given in the introduction, calculate the equilibrium constant kp of the reaction cl2(g)+2no(g)⇌2nocl(g)
1 answer:
d G = 2 (dG NOCl) - 2 (dG NO)
= 2 * 66.08 - 2 * 87.6 = -43.04 kJ /mol
dG = - 2.303 RT log K
-43.04 = - 2.303 RT log K
7.54 = log K
K = 3.42 * 10^7
Kp = k (RT)^dn
d n = 2 - 1 -2 = -1
Kp = (3.42 * 10^7) (8.314 * 298)^-1
Kp = 1.31 * 10^4
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